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# Python. The cmath module. Working with complex numbers. Trigonometric functions
## The cmath module. Working with complex numbers. Trigonometric functions. Hyperbolic functions. Classification functions
### Contents
Search other websites:
##### 1. Trigonometric functions
The cmath module contains 6 trigonometric functions that process complex numbers:
• cmath.acos(x) – returns the inverse cosine of the x argument;
• cmath.asin(x) – returns the arcsine of the x argument;
• cmath.atan(x) – defines the arctangent of the argument x;
• cmath.cos(x) – returns the cosine of the x argument;
• cmath.sin(x) – returns the sine of the x argument;
• cmath.tan(x) – returns the tangent of the x argument.
### ⇑
##### 1.1. Functions cmath.acos(), cmath.asin(), cmath.atan(). Get inverse cosine, arcsine, arctangent of an argument
Functions
```cmath.acos(x)
cmath.asin(x)```
return the inverse cosine and inverse sine of the argument x, respectively.
Each of the functions has two branch cuts: one extends right from 1 along the real axis to ∞, continuous from below. The other extends left from -1 along the real axis to -∞, continuous from above.
Function
`cmath.atan(x)`
allows you to get the arctangent of the argument x.
The function has two branch cuts. One extends from 1j along the imaginary axis to ∞j, continuous from the right. The other extends from -1j along the imaginary axis to -∞j, continuous from the left.
Example.
```# Trigonometric functions acos(), asin(), atan()
# 1. Include module cmath
import cmath
# 2. Get a complex number from the keyboard
re = int(input('re = '))
im = int(input('im = '))
z = complex(re, im)
# 3. Calculate the inverse cosine of argument z and display it
arc_cos = cmath.acos(z)
print('arc_cos = ', arc_cos)
# 4. Calculate the inverse sine of z
arc_sin = cmath.asin(z)
print('arc_sin = ', arc_sin)
# 5. Calculate the inverse tangent of z
arc_tan = cmath.atan(z)
print('arc_tan = ', arc_tan)```
Test example
```re = 3
im = -4
arc_cos = (0.9368124611557198+2.305509031243477j)
arc_sin = (0.6339838656391766-2.305509031243477j)
arc_tan = (1.4483069952314644-0.15899719167999918j)```
### ⇑
##### 1.2. Functions cmath.cos(), cmath.sin(), cmath.tan(). Get cosine, sine and tangent of an argument
Functions
```cmath.cos(x)
cmath.sin(x)
cmath.tan(x)```
return the cosine, sine, and tangent of the argument x.
Example.
```# Functions cos(), sin(), tan()
# Include module cmath
import cmath
# Create a complex number z = 7-8j
z = complex(7, -8)
# Calculate cosine, sine, tangent
sin_res = cmath.sin(z)
cos_res = cmath.cos(z)
tan_res = cmath.tan(z)
print('sin_res = ', sin_res)
print('cos_res = ', cos_res)
print('tan_res = ', tan_res)```
Test example
```sin_res = (979.2248346123021-1123.6753468137035j)
cos_res = (1123.675599719735+979.2246142178511j)
tan_res = (2.2295633684101687e-07-0.9999999692244822j)```
### ⇑
##### 2. Hyperbolic functions
The cmath module contains the implementation of the following hyperbolic functions that operate on complex numbers:
• cmath.acosh(x) – returns the inverse hyperbolic cosine of the x argument;
• cmath.asinh(x) – returns the hyperbolic arcsine of the argument x;
• cmath.atanh(x) – returns the hyperbolic arctangent of the argument x;
• cmath.cosh(x) – returns the hyperbolic cosine of the argument x;
• cmath.sinh(x) – returns the hyperbolic sine of the argument x;
• cmath.tanh(x) – Returns the hyperbolic tangent of the x argument.
### ⇑
##### 2.1. Functions cmath.acosh(), cmath.asinh(), cmath.atanh(). Hyperbolic inverse cosine, arcsine, arctangent
To calculate the hyperbolic arccosine of the argument x, use the function
`cmath.acosh(x)`
For this function, there is one branch cut, which extends left from 1 along the real axis to -∞, continuous from above.
To calculate the hyperbolic inverse sine of the argument x, use the function
`cmath.asinh(x)`
There are two branch cuts: One extends from 1j along the imaginary axis to ∞j, continuous from the right. The other extends from -1j along the imaginary axis to -∞j, continuous from the left.
To calculate the hyperbolic arctangent of the argument x, use the function
`cmath.atanh(x)`
There are two branch cuts. One extends from 1 along the real axis to ∞, continuous from below. The other extends from -1 along the real axis to -∞, continuous from above.
Example.
The example implements the calculation of the hyperbolic arc cosine, hyperbolic arc sine and hyperbolic arc tangent. It also demonstrates the output of the result for the hyperbolic arccosine in a convenient form with an accuracy of 2 decimal places.
```# Trigonometric functions acosh(), asinh(), atanh()
# 1. Include module cmath
import cmath
# 2. Get a complex number
re = int(input('re = ')) # real part
im = int(input('im = ')) # imaginary part
z = complex(re, im)
# 3. Calculate the inverse hyperbolic cosine of the argument z
# and display it to 2 decimal places
# 3.1. Get a value
arc_cosh = cmath.acos(z)
# 3.2. Form components of a complex number
re_str = '%.2f' % arc_cosh.real # real part of complex number arc_cosh
im_str = '%.2f' % arc_cosh.imag # imaginary part
# 3.3. Display in a convenient form
if arc_cosh.imag>0:
print('arc_cosh = ' + re_str + '+' + im_str + 'j')
elif arc_cosh.imag==0:
print('arc_cosh = ' + re_str)
else:
print('arc_cosh = ' + re_str + im_str + 'j')
# 4. Calculate and output the hyperbolic arctangent
arc_tanh = cmath.atanh(z)
print('arc_tanh = ', arc_tanh)
# 5. Calculate and output the hyperbolic arcsine
arc_sinh = cmath.asinh(z)
print('arc_sinh = ', arc_sinh)```
Test example
```re = 3
im = -4
arc_cosh = 0.94+2.31j
arc_tanh = (0.1175009073114339-1.4099210495965755j)
arc_sinh = (2.2999140408792695-0.9176168533514787j)```
### ⇑
##### 2.2. Functions cmath.cosh(), cmath.sinh(), cmath.tanh(). Hyperbolic cosine, sine, tangent
To calculate the hyperbolic cosine, sine and tangent, respectively, use the functions
```cmath.cosh(x)
cmath.sinh(x)
cmath.tanh(x)```
Example.
```# Trigonometric functions cosh(), sinh(), tanh()
# 1. Include the cmath module
import cmath
# 2. Get a complex number
re = int(input('re = ')) # real part
im = int(input('im = ')) # imaginary part
z = complex(re, im)
# 3. Calculate the hyperbolic cosine of the argument z
# and print it with precision 2 decimal places
# 3.1. Get a value
cosh = cmath.cosh(z)
# 3.2. Generate the components of a complex number
re_str = '%.2f' % cosh.real # the real part of the complex number cosh
im_str = '%.2f' % cosh.imag # imaginary part
# 3.3. Display in a convenient form
res_str = re_str
if cosh.imag>0:
res_str += '+'
res_str += im_str + 'j'
print('cosh = ' + res_str)
# 4. Calculate and output the hyperbolic tangent
tanh = cmath.tanh(z)
print('tanh = ', tanh)
# 5. Calculate and output hyperbolic sine
sinh = cmath.sinh(z)
print('sinh = ', sinh)```
Test example
```re = 4
im = -3
cosh = -27.03-3.85j
tanh = (0.999355987381473+0.0001873462046294784j)
sinh = (-27.016813258003932-3.8537380379193773j)```
### ⇑
##### 3. Classification functions
The classifications functions of the cmath module include the following:
• cmath.isinfinite(x) – determines whether there is a finite real and imaginary part of a complex number x;
• cmath.isinf(x) – determines whether there is an infinite real or imaginary part of a complex number x;
• cmath.isnan(x) – determines whether the real or imaginary parts of the complex number x are of value Nan;
• cmath.isclose (x) – determines the closeness of two values to each other.
### ⇑
##### 3.1. cmath.isfinite(). Finite complex number
Function
`cmath.isfinite(x)`
returns True if the real and imaginary parts of the complex number are finite. Otherwise, the function returns False.
Example.
```# Function isfinite()
# 1. Include the cmath module
import cmath
# 2. Get a complex number
re = int(input('re = ')) # real part
im = int(input('im = ')) # imaginary part
z = complex(re, im)
# 3. Using the isfinite() function
res = cmath.isfinite(z)
if res:
print('Both parts are finite.')
else:
print('Both parts are not finite.')```
Test example
```re = 5
im = -2
Both parts are finite.```
### ⇑
##### 3.2. cmath.isinf(). Finity of the components of a complex number
Function
`cmath.isinf(x)`
returns True if the real (x.real) or imaginary (x.imag) parts of the complex number are infinite. Otherwise, False is returned.
Example.
```# Function isinf()
# 1. Include the cmath module
import cmath
# 2. Get a complex number
re = int(input('re = ')) # real part
im = int(input('im = ')) # imaginary part
z = complex(re, im)
# 3. Using isinf() function
res = cmath.isinf(z)
if res:
print('Both parts are not finite.')
else:
print('Both parts are finite.')```
Test example
```re = 5
im = -3
Both parts are finite.```
### ⇑
##### 3.3. cmath.isnan(). Checking the components of a complex number for the value of Nan
Function
`cmath.isnan(x)`
returns True if one of the parts (real or imaginary) of the complex number is Nan. Otherwise, the function returns False.
Example.
```# Function isnan()
# 1. Include module cmath
import cmath
# 2. Get a complex number
re = int(input('re = ')) # real part
im = int(input('im = ')) # imaginary part
z = complex(re, im)
# 3. Using the isnan() function
res = cmath.isnan(z)
if res:
print('One of the parts is equal to the Nan value.')
else:
print('Both parts are not equal to Nan.')```
Test example
```re = 2
im = -7
Both parts are not equal to Nan.```
### ⇑
##### 3.4. cmath.isclose(). Determining the closeness of two values to each other
The cmath.isclose() function allows you to determine the closeness of two values to each other based on a given precision. According to the Python documentation, the syntax for declaring a function is as follows:
`cmath.isclose(a, b, rel_tol=1e-09, abs_tol = 0.0)`
here
• a, b – complex numbers that are compared with each other. If the values are close (equal with the specified precision), then the function returns True, otherwise the function returns False;
• rel_tol – optional parameter, sets the relative error during comparison;
• abs_tol – an optional parameter that defines the absolute error between the values of the elements.
Example.
```# Function isclose()
# Include the cmath module
import cmath
# Case 1. Two complex numbers are not equal
z1 = complex(2.5, 3.2) # z = 2.5+3.2j
z2 = complex(2.5000001, 3.2) # z = 2.5000001+3.2j
# Using the isclose() function
res = cmath.isclose(z1, z2) # False
if res:
print('z1 == z2')
else:
print('z1 != z2')
# Case 2. Two complex numbers are equal
z1 = complex(7.2, -1.7) # z1 = 7.2-1.7j
z2 = complex(7.2, -1.7) # z2 = 7.2-1.7j
# Use the isclose() function
res = cmath.isclose(z1, z2) # True
if res:
print('z1 == z2')
else:
print('z1 != z2')
# Case 3. Two complex numbers are equal with the specified relative precision
z1 = complex(3.001, 2.8)
z2 = complex(3.0, 2.8)
# Using the isclose() function, specifies a relative precision of 0.01
res = cmath.isclose(z1, z2, rel_tol=0.01) # z1 == z2
if res:
print('z1 == z2')
else:
print('z1 != z2')
# Case 4. Two complex numbers are equal with specified absolute precision
z1 = complex(1.0001, 5.5)
z2 = complex(1.0, 5.5)
# Function call isclose (), absolute precision 0.001 is specified
res = cmath.isclose(z1, z2, abs_tol=0.001) # z1 == z2
if res:
print('z1 == z2')
else:
print('z1 != z2')```
Test example
```z1 != z2
z1 == z2
z1 == z2
z1 == z2```
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# Solve Discrete Math Problems Online Free latest 2023
You are searching about Solve Discrete Math Problems Online Free, today we will share with you article about Solve Discrete Math Problems Online Free was compiled and edited by our team from many sources on the internet. Hope this article on the topic Solve Discrete Math Problems Online Free is useful to you.
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## Introduction to Sudoku
Have you ever played with a Rubik’s Cube? You may have mastered the Rubik’s cube and are looking for a bigger challenge. Try the puzzle game called Sudoku. Sudoku is a number placement puzzle that requires logical skills and patience. It is a fantastic puzzle game that can be found in newspapers, books, and on puzzle and game websites.
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How do you solve a Sudoku puzzle?
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https://alevelmathtuition.sg/2015/10/27/a-level-h2-math-preparation-tip-1-calculus/
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# A level H2 Math preparation tips #1: Differentiation
## Calculus (Differentiation, Integration and applications)
• Calculus is the heaviest weightage among all the pure math “modules” in the exam. Over the years the total marks for this module is between 50 to 60. Hence I always emphasise to my students that this module is a MUST to master.
• A level exams will not test directly on techniques of differentiation. They will ask application questions only. So far over the years they have been testing on tangents / normals and maximum / minimum problems. No rate of change questions (yet).
• Tangents / Normals
For the past 8 years, 7 years have questions on tangents / normals asked in the form of parametric equation (since implict differentiation can be asked through Maclaurin’s series). Furthermore, the recent years’ questions has been algebraic intensive. E.g. they ask you to find equation of a tangent at a point with parameter “p” instead of say “2” to make the workings nicer. Also it is quite common for them to merge graphing techniques (sketch parametric curves) and integration (area under curve through parametric form) as 1 big question.
• Maximum / minimum problem
It is quite common that they will give you an object such as a box of various shapes and find the value of one of the lengths for maximum volume or surface area. Students may find the first part on formulating the equation to be harder than the differentiation itself. My suggestion is: if the equation is mentioned in the question as a “show” question, just use the answer to continue with the maximum / minimum computation which will usually have more marks. DO NOT give up the entire say 9 marks question just because you could not form the equation which is just approx 3 marks.
• Maclaurin’s series
For Macluarin’s series there are two main approaches: successive differentiation and using standard series from MF 15. Over the years, binomial expansion has been tested quite frequently. Also if you see the phrase “sufficiently small” and the question involves a triangle and angle –> small angle approx (using sine or cosine rule). Usually small angle approx questions is followed by binomial expansion too.
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https://studylib.net/doc/11606397/stat-496--spring-2004-homework-assignment-%235
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STAT 496, Spring 2004 Homework Assignment #5
```STAT 496, Spring 2004
Homework Assignment #5
1. An experiment is performed to determine the importance of four factors on the efficiency of a distillation column that is used to separate two chemical compounds. The
four factors are: A - Relative Volatility that expresses a chemical relationship between
two items, B - Bottom Composition that indicates the ratio of one compound to the
total volume of the two components at the bottom of the distillation column, C - Top
Composition that indicates the ratio of one compound to the total volume of the two
components at the top of the distillation column, and D - Feed Composition which is
the ratio of the two components where they enter the system. The sixteen combinations of these four factors, each at two levels, is run in a completely randomized design.
Below are the factor levels and the data.
Factor
A: Relative Volatility
B: Bottom Composition
C: Top Composition
D: Feed Composition
Trmt.
Comb.
1
a
b
ab
c
ac
bc
abc
d
bd
abd
cd
acd
bcd
abcd
XA
−1
+1
−1
+1
−1
+1
−1
+1
−1
+1
−1
+1
−1
+1
−1
+1
XB
−1
−1
+1
+1
−1
−1
+1
+1
−1
−1
+1
+1
−1
−1
+1
+1
XC
−1
−1
−1
−1
+1
+1
+1
+1
−1
−1
−1
−1
+1
+1
+1
+1
Low (−1)
1.1
0.005
0.70
0.30
XD
−1
−1
−1
−1
−1
−1
−1
−1
+1
+1
+1
+1
+1
+1
+1
+1
High (+1)
2.0
0.200
0.99
0.60
Efficiency
88
11
31
4
138
18
80
11
84
12
31
5
140
19
85
11
Run
Order
(11)
( 2)
( 6)
( 4)
(10)
(14)
(13)
( 8)
(16)
( 5)
(15)
(12)
( 9)
( 7)
( 1)
( 3)
(a) For each factor, calculate the mean efficiency when that factor is at the high level
and at the low level. Display these means in four main effects plots, one for each
factor. Comment on the apparent size of each main effect.
(b) Construct interaction plots for the four factors. Pay particular attention to the
combinations of A - Relative Volatility with B - Bottom Composition, A - Relative
Volatility with C - Top Composition, B - Bottom Composition with C - Top
Composition. Comment on the apparent presence or absence of interaction for
these three combinations.
(c) Compute the overall grand sample mean and the 15 estimated effects. Be sure to
indicate whether you are computing estimated full effects of half effects.
(d) Construct a normal probability plot for the 15 estimated effects. Label on your
plot effects that appear to be significant.
(e) You should be able to pick up some pseudo replication by eliminating a factor
and all its interactions with the other three factors. This should give you a 23
experiment with 2 replicates. With this reduced model, compute M SrepError and
the standard error of an estimate.
(f) Which of the remaining estimated effects are significant? non-significant?
(g) Give the most parsimonious (include only significant terms) prediction equation.
Be sure to explicitly define each variable in the equation.
(h) In order to maximize the efficiency of this distillation column, what settings of
the four factors would you recommend?
(i) Give a predicted value, and prediction interval, for the filtration rate at the choice
of factor levels you give in (h). Be sure to indicate what you are using for the
estimate of error variation and why.
(j) Using the prediction equation in (g) construct plots of residuals versus predicted
values and residuals versus run order. Are there any discernible patterns in these
plots?
Note: You may use JMP, or another computer program, to assist you with any of the
graphing or calculations needed to answer the questions above.
```
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https://questioncove.com/updates/5435cc54e4b088157637b198
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Mathematics
OpenStudy (calculusxy):
Mrs. Escalante was riding a bicycle on a bike path. After riding 2/3 of a mile, she discovered that she still needed to travel 3/4 of a mile to reach the end of the path. How long is the bike path?
OpenStudy (texaschic101):
2/3 + 3/4 -- common denominator is 12 8/12 + 9/12 = 17/12 = 1 5/12 The bike path is 1 5/12 miles. ( 1 and 5/12 miles)
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http://www.mathworks.com/matlabcentral/fileexchange/19-delta-sigma-toolbox/content/delsig/dsexample2.m
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Code covered by the BSD License
Delta Sigma Toolbox
Richard Schreier (view profile)
• 1 file
• 4.59459
14 Jan 2000 (Updated )
High-level design and simulation of delta-sigma modulators
Editor's Notes:
This file was selected as MATLAB Central Pick of the Week
dsexample2
```function adc = dsexample2
% Design example for a bandpass modulator.
% Design parameters
order = 8;
osr = 32;
nlev = 2;
f0 = 0.125;
Hinf = 1.5;
form = 'CRFB';
% Derived parameters
M = nlev-1;
clc
fprintf(1,'\t\t\t%dth-Order Bandpass Example\n\n',order);
fig_pos = { [ 10 595 600 215],
[ 640 595 600 215],
[ 10 345 250 205],
};
% NTF synthesis and realization
fprintf(1,'Doing NTF synthesis and realization... ');
design_step = 1;
ntf = synthesizeNTF(order,osr,2,Hinf,f0); % Optimized zero placement
[a,g,b,c] = realizeNTF(ntf,form);
b = [b(1) zeros(1,length(b)-1)]; % Use a single feed-in for the input
ABCD = stuffABCD(a,g,b,c,form);
fprintf(1,'Done.\n');
figure(design_step); clf
set(design_step,'position',fig_pos{design_step});
DocumentNTF(ABCD,osr,f0);
drawnow;
% Time-domain simulations
fprintf(1,'Doing time-domain simulations... ');
design_step = design_step+1;
figure(design_step); clf;
set(design_step,'position',fig_pos{design_step});
% Example spectrum
subplot('position', [0.05,0.1,0.6 0.8]);
PlotExampleSpectrum(ntf,M,osr,f0);
title('Example Spectrum');
% SQNR plot
subplot('position',[.74 .18 .25 .65]);
if nlev==2
[snr_pred,amp_pred] = predictSNR(ntf,osr);
plot(amp_pred,snr_pred,'-');
hold on;
end
[snr,amp] = simulateSNR(ntf,osr,[],f0,nlev);
fprintf(1,'Done.\n');
plot(amp,snr,'og');
figureMagic([-100 0], 10, 2, [0 100], 10, 2, [7 3], 'Time-Domain Simulations');
xlabel('Input Level (dBFS)');
ylabel('SQNR (dB)');
[peak_snr,peak_amp] = peakSNR(snr,amp);
msg = sprintf('peak SQNR = %4.1fdB \n@ amp=%4.1fdB ',peak_snr,peak_amp);
text(peak_amp-10,peak_snr,msg,'hor','right', 'vertical','middle');
msg = sprintf('OSR=%d ',osr);
text(0,5,msg,'hor','right');
title('SQNR Plot');
drawnow;
% Dynamic range scaling
fprintf(1,'Doing dynamic range scaling... ');
design_step = design_step+1;
figure(design_step); clf;
set(design_step,'position',fig_pos{design_step});
ABCD0 = ABCD;
[ABCD umax] = scaleABCD(ABCD0,nlev,f0);
[a,g,b,c] = mapABCD(ABCD,form);
fprintf(1,'Done.\n');
fprintf(1,'Verifying dynamic range scaling... ');
u = linspace(0,0.95*umax,30);
N = 1e4; N0 = 50;
test_tone = cos(2*pi*f0*[0:N-1]);
test_tone(1:N0) = test_tone(1:N0) .* (0.5-0.5*cos(2*pi/N0*[0:N0-1]));
maxima = zeros(order,length(u));
for i = 1:length(u)
ui = u(i);
[v,xn,xmax] = simulateDSM( ui*test_tone, ABCD, nlev );
maxima(:,i) = xmax(:);
if any(xmax>1e2)
fprintf(1,'Warning, umax from scaleABCD was too high.\n');
umax = ui;
u = u(1:i);
maxima = maxima(:,1:i);
break;
end
end
fprintf(1,'Done.\n');
colors = hsv(order);
cmd = 'legend( handles';
handles = [];
for i = 1:order
handles(i) = plot(u,maxima(i,:),'--','color',colors(i,:));
if i==1
hold on;
end
cmd = [ cmd ',''State ' num2str(i) '''' ];
plot(u,maxima(i,:),'o','color',colors(i,:));
end
hold off; grid on;
text(umax/2,0.05,'Input Amplitude','Hor','cen','Vert','mid');
figureMagic([ 0 umax],[],[], [0 1],0.1,2, [3 3],'State Maxima');
set(gca,'position', [0.1 0.07, 0.85, 0.85]);
cmd = [cmd ',4);'];
eval(cmd);
% The next step would be to size capacitors for each integrator state based
% on noise (kT/C) limits and area.
% Simulations modelling the effects of finite op-amp gain and capacitor
% errors should also be performed.
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https://newsupland.in/ticket-fine-codechef-solution/
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crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00387.warc.gz
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Ticket Fine CodeChef Solution
We Are Discuss About CODECHEF SOLUTION
Ticket Fine CodeChef Solution
Problem
On a certain train, Chef-the ticket collector, collects a fine of Rs. X if a passenger is travelling without a ticket. It is known that a passenger carries either a single ticket or no ticket.
P passengers are travelling and they have a total of Q tickets. Help Chef calculate the total fine collected.
Input Format
The first line contains a single integer T, the number of test cases. T lines follow. Each following line contains three integers separated by spaces, whose description follows.
• The first integer, X, is the fee in rupees.
• The second integer, P, is the number of passengers on the train.
• The third integer, Q, is the number of tickets Chef collected.
Output Format
The output must consist of T lines.
• The i^{th} line must contain a single integer, the total money(in rupees) collected by Chef corresponding to the i^{th} test case.
Constraints
• 1 \le T \le 10
• 1 \le X \le 10
• 0 \le Q \le P \le 10
Sample 1:
Input
Output
4
4 1 1
2 10 7
8 5 4
9 7 0
0
6
8
63
Explanation:
Test case 1: The total number of passengers travelling without ticket are 1 – 1 = 0. So the total fine collected is 0 \cdot 4 = 0 rupees.
Test case 2: The total number of passengers travelling without ticket are 10 – 7 = 3. So the total fine collected is 3 \cdot 2 = 6 rupees.
Test case 3: The total number of passengers travelling without ticket are 5 – 4 = 1. So the total fine collected is 1 \cdot 8 = 8 rupees.
Test case 4: The total number of passengers travelling without ticket are 7 – 0 = 7. So the total fine collected is 7 \cdot 9 = 63 rupees.
Ticket Fine CodeChef Solution
Yhaa You have done it but next? if YOU Want to Get Others Please Visit Here JOIN NOW
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https://cadabra.science/qa/84/derivatives-with-multiple-indices?show=85
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# Derivatives with multiple indices
+1 vote
I have a sort of specific example in mind, and I am wondering if you have any advice on implementing it. I am trying to do some calculations in superspace, and I want to automate spinor derivatives. Specifically, I have the $\theta$-expansion of some superfield, and I want to perform some spinor derivatives and restrict to $\theta = 0$.
I tried defining an abstract derivative $D^{\alpha}_{A}$, with $\alpha$ a spinor index and $A$ an R-symmetry index. Then, I was going to define some substitution rules for how $D$ acts on the $\theta$ variables, but I found that Cadabra would interpret any derivative with multiple indices as a product of derivatives. This appears to be intentional from the description of the product_rule algorithm. So, do you have any suggestions for doing something like this? Thanks.
edit: I found a way to almost do it by following your Poincare algebra example, i.e. defining $D$ as an object that doesn't commute with $\theta$ and defining some substitution rules for their commutator. However, this suggests the problem: I actually want an anti-commutator. More precisely, I want a $\mathbb{Z}_2$ graded commutator, and I want to be able to assign a degree to each of my objects. Is this possible?
edited
If you go that route (which is probably what I would do as well), what prevents you from using substitution rules for anti-commutators? Maybe you can show a bit of detail about where you get stuck?
It would be nice to be able to treat sets of indices as a 'composite' index, and admittedly there isn't anything very satisfactory at the moment in Cadabra. I have thought about how to do this in the past, but haven't collected a sufficiently large set of different 'sample use cases' to be convinced that I could build a useful general purpose functionality for this. It would certainly be possible, without a lot of effort, to build in some functionality for composite indices (indices with more than one label, for things like your superspace derivative, but also to e.g. use SU(2)xSU(2) spinor indices to write representations of the Poincare algebra). Am not entirely convince though that that would suffice. Any thoughts/comments/ideas welcome.
Thanks for the reply! Is there a built-in anti-commutator function? Even if so, I'm not sure it would solve my issue. I think I really need a graded commutator. For example, to produce things like $[D, \theta_1 \theta_2]_- = [D, \theta_1]_+ \theta_2 - \theta_1 [D, \theta_2]_+$, where all three objects are anti-commuting and the subscript $(+)-$ refers to an (anti-)commutator. Is there a way to have Cadabra do this?
There is \anticommutator which behaves similarly to \commutator. For the purpose of representing a graded algebra, I guess it would suffice to make product_rule produce commutators if both or one of the elements are commuting, and anti-commutators if they are both anti-commuting. Right now product_rule always generates commutators if you start from a commutator, and anti-commutators if you start from an anti-commutator.
Could possibly add that relatively easily as an option to product_rule if it helps you.
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https://ubiq.co/database-blog/how-to-calculate-rolling-average-in-redshift/
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# How to Calculate Moving Average in Redshift
Rolling Average or Moving Average is a useful metric that helps you keep track of average value over a moving period (e.g average sales for past 7 days). Calculating moving average over time gives more reasonable trend, compared to plotting daily numbers. Since there is no built-in function to calculate moving average in Redshift, here’s the SQL query to do it.
## How to Calculate Moving Average in Redshift
Here are the steps to calculate moving average in Redshift. Let’s say you have the following table that contains daily sales information in Redshift.
```# create table sales(order_date date,sale int);
# insert into sales values('2020-01-01',20),
('2020-01-02',25),('2020-01-03',15),('2020-01-04',30),
('2020-01-05',20),('2020-01-10',20),('2020-01-06',25),
('2020-01-07',15),('2020-01-08',30),('2020-01-09',20);
# select * from sales;
+------------+------+
| order_date | sale |
+------------+------+
| 2020-01-01 | 20 |
| 2020-01-02 | 25 |
| 2020-01-03 | 15 |
| 2020-01-04 | 30 |
| 2020-01-05 | 20 |
| 2020-01-10 | 20 |
| 2020-01-06 | 25 |
| 2020-01-07 | 15 |
| 2020-01-08 | 30 |
| 2020-01-09 | 20 |
+------------+------+
```
Let’s say you want to calculate moving average in Redshift for past 5 days. Redshift (which is basically Postgresql) makes this really easy with the help of Redshift Window Functions. Here’s the SQL query to calculate moving average for past 5 days. We will look at it in detail below.
```SELECT a.order_date,a.sale,
AVG(a.sale)
OVER(ORDER BY a.order_date ROWS BETWEEN 4 PRECEDING AND CURRENT ROW) AS avg_sales
FROM sales a ;
```
If you want to round the results, you can use ROUND function as shown to calculate running average in Redshift
```SELECT a.order_date,a.sale,
round(AVG(a.sale)
OVER(ORDER BY a.order_date ROWS BETWEEN 4 PRECEDING AND CURRENT ROW),2) AS avg_sales
FROM sales a;
order_date | sale | avg_sales
------------+------+---------------
2020-01-01 | 20 | 20.00
2020-01-02 | 25 | 22.50
2020-01-03 | 15 | 20.00
2020-01-04 | 30 | 22.50
2020-01-05 | 20 | 22.00
2020-01-06 | 25 | 23.00
2020-01-07 | 15 | 21.00
2020-01-08 | 30 | 24.00
2020-01-09 | 20 | 22.00
2020-01-10 | 20 | 22.00
```
Let’s look at the above query in detail. AVG function calculates average value of sale column. However, when we use it along with WINDOW function OVER it calculates average value only for the window that we define.
First we use ORDER BY on our data to ensure that rows are sorted chronologically. Then we define our window for average using OVER function, and mention ROWS BETWEEN 4 PRECEDING AND CURRENT ROW. That is, for each row, calculate average for preceding 4 rows and current row. As the window frame changes for each row, only preceding 4 days and current date will be used.
You can also add filters by adding WHERE clause in the above SQL query.
``` SELECT a.order_date,a.sale,
round(AVG(a.sale)
OVER(ORDER BY a.order_date ROWS BETWEEN 4 PRECEDING AND CURRENT ROW),2) AS avg_sales
FROM sales a
WHERE condition;```
If you want to calculate moving average in Redshift for past 30 days/1 month, modify the above query to use PRECEDING 29 ROWS AND CURRENT ROW
```SELECT a.order_date,a.sale,
round(AVG(a.sale)
OVER(ORDER BY a.order_date ROWS BETWEEN 29 PRECEDING AND CURRENT ROW),2) AS avg_sales
FROM sales a;
```
Bonus Read : How to Create Pivot Table in PostgreSQL
## How to Calculate Moving Average in Redshift for Past 3 Months
Let’s say you have monthly sales data instead of daily data, and want to calculate rolling average for past 3 months.
```# create table monthly_sales(order_month date,sale int);
postgres=# insert into monthly_sales values('2019-12-01',20),
('2020-01-30',25),('2020-02-28',15),('2020-03-31',30),
('2020-04-30',20),('2020-05-31',20),('2020-06-30',25),
('2020-07-31',15),('2020-08-31',30),('2020-09-30',20);
postgres=# select * from monthly_sales;
order_month | sale
-------------+------
2019-12-01 | 20
2020-01-30 | 25
2020-02-28 | 15
2020-03-31 | 30
2020-04-30 | 20
2020-05-31 | 20
2020-06-30 | 25
2020-07-31 | 15
2020-08-31 | 30
2020-09-30 | 20
```
We use the same logic to calculate moving average in Redshift, in this case. First ORDER BY order_month column to ensure the rows are chronologically sorted. Then calculate average for PRECEDING 2 ROWS AND CURRENT ROW
``` SELECT a.order_month,a.sale,
round(AVG(a.sale)
OVER(ORDER BY a.order_month ROWS BETWEEN 2 PRECEDING AND CURRENT ROW),2) AS avg_sales
FROM monthly_sales a ;
order_month | sale | avg_sales
-------------+------+-----------
2019-12-01 | 20 | 20.00
2020-01-30 | 25 | 22.50
2020-02-28 | 15 | 20.00
2020-03-31 | 30 | 23.33
2020-04-30 | 20 | 21.67
2020-05-31 | 20 | 23.33
2020-06-30 | 25 | 21.67
2020-07-31 | 15 | 20.00
2020-08-31 | 30 | 23.33
2020-09-30 | 20 | 21.67
```
Bonus Read : How to Calculate Running Total in Redshift
You can also add filters by including WHERE clause in the above SQL query.
``` SELECT a.order_month,a.sale,
round(AVG(a.sale)
OVER(ORDER BY a.order_month ROWS BETWEEN 2 PRECEDING AND CURRENT ROW),2) AS avg_sales
FROM monthly_sales a
WHERE condition;```
You can customize the above query to calculate moving average in Redshift, as per your requirements.
After you calculate moving average in Redshift, you can use a charting tool to plot it on a line chart and share it with your team. Here’s an example of a line chart that visualizes moving average, created using Ubiq.
If you want to create charts, dashboards & reports from Redshift database, you can try Ubiq. We offer a 14-day free trial.
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View Course Path
# Ideal Filter Types, Requirements, and Characteristics
Contents
## What is an ideal filter?
You’ve heard of filter coffee. It’s a very popular type of coffee where you filter out the coarse coffee grains to get the silky-smooth concoction that is enjoyed all over the world. Similarly, in digital signal processing, we send a signal with a mix of several different frequencies through a filter to get rid of the unwanted frequencies and give us the desired frequency response. The picture below shows us the response when a low pass filter is applied, and we get a practical response from it. No filter can entirely get rid of all the frequencies that we set as a limit. There will be certain discrepancies, and some unwanted frequencies will give rise to a transition band.
However, for calculation purposes, we consider an ideal filter which does not take into account these practical discrepancies. Hence, it has a steep transition from the pass-band to the stop-band.Like the picture above. The practical filter will look more like the graph below.
## What are the characteristics of an ideal filter?
There are many different types of filters based on what you want your frequency response to be. Therefore, the characteristics of each of these types of filters differ. But each of these filters will definitely have a pass-band and a stop-band. Pass-band, as we have discussed, is the frequencies that the filter allows to be kept in the response. Stop-band includes the frequencies that are not in the said limits of the filter response.
A filter is essentially a system like the one in the diagram below
Where A is the amplitude of the signal, ω is the angular frequency of the input signal, and Φ is the phase of the system.
The frequency response of a filter is the ratio of the steady-state output of the system to the sinusoidal input. It is required to realize the dynamic characteristic of a system. It represents the magnitude and phase of a system in terms of the frequency.
The phase of the frequency response or the phase response is the ratio of the phase response of the output to that of the input of any electrical device which takes in an input modifies it and produces an output. Hence, the operating frequency of the filter can be used to further categorize these filters into low pass filter, high pass filter, band-pass filter, band-rejection filer, multipass filter, and multi-stop filter.
## Low-pass filter
A certain cut-off frequency, ωC radians per second is chosen as the limit, and as the name suggests, the portion with low frequency is allowed to pass. Hence, the frequencies before ωC are what consists of the pass-band and the frequencies after ωC are attenuated as part of the stop-band. This is pictorially depicted below:
The transfer function of an ideal low pass filter is given by
H(jω) = 1 for |ω| < ωc ; 0 for ω >= ωc
Circuit for a Low Pass Filter
As these filters are ideal, there will be no presence of the transition band, only a vertical line at the cutoff frequency. Low Pass Filters are often used to identify the continuous original signal from their discrete samples. They tend to be unstable and are not realizable as well.
## High-pass filter
The High Pass Filter allows the frequencies above the cutoff frequency to pass, which will be the passband, and attenuates the frequencies below the cutoff frequency, consisting of the stop-band. An ideal high pass filter’s response ought to look like the figure below.
The transfer function of an ideal low pass filter is given by
H(jω) = 1 for |ω| >= ωc ; 0 for ω < ωc
Circuit for a High Pass Filter
Once again, there will be no transition band due to the precise cutting off of the signal in an ideal filter. An ideal high pass filter passes high frequencies. However, the strength of the frequency is lower. A highpass filter is used for passing high frequencies, but the strength of the frequency is lower as compared to cut off frequency. High pass filters are used to sharpen images.
## Band-pass filter
The band-pass filter is actually a combination of the low pass and high pass filter, both the filters are used strategically to allow only a portion of frequencies to pass through hence forming the pass-band and the all frequencies that do not fall in this range are attenuated
The transfer function of the Band Pass Filter is given by
H(jω) = 1 for |ω1| < ω < |ω2| ; 0 for ω < |ω1| and ω > |ω2|
Circuit for a Band Pass Filter
The circuit as shown below is the high pass filter followed by the low pass filter, which forms the Band Pass Filter.
This band has two cut-off frequencies, the lower cut off frequency, and the upper cut off frequency. So, it is known as a second-order filter. Band-pass filters can also be formed using inverting operational amplifiers. These types of filters are used primarily in wireless transmitters and receivers.
Band-pass filters are widely used in wireless transmitters and receivers. It limits the frequency at which the signal is transmitted, ensuring the signal is picked up by the receiver. Limiting the bandwidth to the particular value discourages other signals from interfering.
## Band Rejection Filter/Band Stop Filter
Band rejection filter is used when you want to retain the majority of the frequency but reject a specific band of frequencies. It is basically the opposite of a band-pass filter. If the stop-band is very narrow, it is sometimes referred to as a notch filter. And similar to the band-pass filter, this filter has two cut-off frequencies making it a second-order filter. This filter allows all frequencies leading up to the first cut-off frequency(lower cut-off frequency) to pass and all frequencies after the second cut-off frequency(upper cut-off frequency) to pass making that the pass-band. And all frequencies between the lower and upper cut-off frequency is the stop-band.
The transfer function of a Band Rejection filter is as shown
H(jω) = 0 for |ω1| < ω < |ω2| ; 1 for ω < |ω1| and ω > |ω2|
Circuit for a Band Stop Filter
The circuit of a Band Rejection filter consists of a resistor, inductor, and capacitor all in parallel. This is also the same as keeping a low pass filter in parallel with a high pass filter.
The band Rejection filter is heavily used in Public Address Systems, Speaker System and in devices that require filtering to obtain good audio. It is also used to reduce static radio related devices. The frequencies that bring in the static are attenuated by giving their frequencies as the limits.
## Multi-pass filter
Multi-pass filters are basically several band-pass filters put together. It is used when you want only certain bands of frequencies to pass through and want to attenuate the majority. The Band Pass Filter itself is achieved by having the circuit of a high pass filter connected to that of a low pass filter in series, connecting this to another pair of filters will give us two bands of frequencies that are allowed to pass through with two second-order cut off frequencies each making its response look something like the following graph.
## Multi-stop filter
Multi-stop filters are several band-stop filters put together. It is used when you want certain bands of frequencies to not be passed and when you want the majority of the frequencies to be attenuated. The Band Stop Filter is achieved by having a circuit of high pass filter and low pass filter connected in parallel. Hence, this would give us two pairs of second-order cut off frequencies as well, as depicted by the graph below.
## Why can’t we design ideal filters?
To get an ideal response would mean there should be no unwanted components at all. The calculations to ensure the frequency response of an ideal filter should be precise, to know how to do this you can take a look at the post on filter approximation under the same section. So taking the Fourier transform of an ideal filter, we get an infinite time response. And the other way around, a signal with infinite frequency response is because of a finite time response. Hence, if we allow for infinite time, that would mean there will be an infinite delay in the filter. This would mean you will be getting the output as zero for a long, long time before getting an actual result.
This means an ideal filter is non-causal, which means it will not be zero before time zero is reached. And also it is not rational; it is not able to be written as a rational transfer function with a finite degree(n) for such a sudden transition.
For example, let’s say the frequency response required is:
H(jω) = 1 for |ω| < ωc ; 0 for ω >= ωc
Calculating the impulse response in time domain, we get
$h(t)=\frac { 1 }{ 2\pi } \int _{ -\infty }^{ \infty }{ H(j\omega ){ e }^{ j\omega t }d\omega =\int _{ { -\omega }_{ C } }^{ { \omega }_{ C } }{ { e }^{ j\omega t }d\omega =\frac { { \omega }_{ C } }{ \pi } sinc( } { \omega }_{ C }t) }$
The sinc function suggests that the frequency response exists for all values from -∞ to ∞. Therefore, it is non-realizable as it requires infinite memory.
So, to avoid waiting for an eternity to get a frequency response, we do not design ideal filters.
## Filter Approximation Techniques
Designing a filter involves approximation and realization. Let us take a Low Pass Filter for discussion. As mentioned, the pass-band allows the lower frequencies within the cut-off frequency and attenuates the rest of the frequencies with an immediate transition from pass-band to stop-band.
For a practical Low Pass Filter, however, the transition from pass-band to stop-band will be gradual or to be precise at a rate of -20dB/decade for every order of the filter.
Hence, realistic filters may have ripples in the pass-band or the stop, in some cases, even both. But, one thing is for sure, it is impossible to design an entirely flat response or such a steep transition
However, to try and overcome this problem of ripples or change the output to the desired response, we can use one of the five filter approximation techniques available.
Filter approximation techniques are special processes using which we can design our target filter with the closest response to an ideal filter. Each of these techniques has a close-to-ideal response in certain regions of the frequency band. Some approximation methods give you a flat pass-band. Whereas some give you a sharp transition band. Let’s get acquainted with some of the filter approximation techniques. We’ll dive deeper into filter approximation techniques in this post.
First, we have the Butterworth filter which gives a flat pass-band (also called maximally flat approximation for this reason), a stop-band with no ripples and a roll off of -20n dB/decade where n is the order of the filter. These filters are usually used for audio processing.
Secondly, we have the Chebyshev filter which gives a pass-band with ripples, a stop-band with no ripples and a roll-off faster than the Butterworth filter. This is the type of filter one should go for if a flat response is not their priority but a faster transition instead. The number of ripples is equal to the order of the filter divided by 2. The amplitude of each of these ripples is also the same. Hence, it is also known as an Equal Ripple Filter.
Next, we have the inverse Chebyshev which has the same roll off as Chebyshev and also a flat pass-band, fast transition but ripples in the stop-band. This filter can be opted for when there is no concern with the attenuated band. However, one should ensure the ripples in the stop-band do not surpass the rejection threshold.
Then, we have the Elliptic filter which has a ripple in both pass-band and stop-band but the fastest transition out of all the filters.
The Bessel filter has the opposite response, which is a flat response in the pass-band, no ripple in the stop-band but the slowest roll-off among all approximation filters.
Here is a summary,
Sources: 1, 2, 3
## One thought on “Ideal Filter Types, Requirements, and Characteristics”
1. George says:
This post was very useful, thank you Keerthana!
This site uses Akismet to reduce spam. Learn how your comment data is processed.
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# Mathematical Symbola
Place a mathematical symbol between the numerals 5 and 9 in such a way that the resulting number is greater than 5 but smaller than 9.
# Similar Riddles
##### Word of 12 alphabets
I am a word of 12 alphabets.
* Alphabets 12, 4, 7, 2, 5 : Eastern beast of burden
* Alphabets 1, 8, 10, 9 are street made famous by Sinclair Lewis.
* Alphabets 11, 3, 6 are past.
On whole it means a person suffering from delusions of greatness.
What Is it ?
Asked by Neha on 05 Jul 2021
##### Fill The Tank
Which tank will fill first?
Asked by Neha on 13 Jul 2021
##### Fox, Chicken and corn Riddle
A man has to get a fox, a chicken, and a sack of corn across a river.
He has a rowboat, and it can only carry him and one other thing.
If the fox and the chicken are left together, the fox will eat the chicken.
If the chicken and the corn are left together, the chicken will eat the corn.
How does the man do it?
Asked by Neha on 04 May 2021
##### Magic Algebra
You are given with the following sum. Each of the letters can be decoded as a digit. If we tell you that D = 5, then can you solve it entirely?
DONALD
+GERALD
=ROBERT
Asked by Neha on 29 Jun 2021
##### Building Code
A man desired to get into his work building, however he had forgotten his code.
However, he did recollect five pieces of information
* Fifth number + Third number = 14
* The fourth number is one more than the second number.
* The first number is one less than twice the second number.
* The second number and the third number equals 10.
* The sum of all five numbers is 30.
What is the code ?
Asked by Neha on 14 May 2021
##### Never eat for breakfast
What two things can you never eat for breakfast?
Asked by Neha on 17 May 2021
##### Rich River Riddle
Why is a river so rich ?
Asked by Neha on 24 Dec 2020
##### I Look at you
I look at you, you look at me, I raise my right, you raise your left. What am I?
Asked by Neha on 17 May 2021
##### Tricky Teller
A boy was at a carnival and went to a booth where a man said to the boy, "If I write your exact weight on this piece of paper then you have to give me \$50, but if I cannot, I will pay you \$50." The boy looked around and saw no scale so he agrees, thinking no matter what the carny writes he'll just say he weighs more or less. In the end the boy ended up paying the man \$50. How did the man win the bet?
Asked by Neha on 27 Jan 2023
##### My two tell you who I am
I am 5 letters long.
My first two tell you who I am
My first 3 could be a medicine
My last three reversed could be a young boy.
My 4th, 3rd and 2nd in that order could be a fruit drink.
If you have me you may hang me round your neck
WHAT AM I ?
Asked by Neha on 14 May 2021
### Amazing Facts
###### Challenging
There is a cryptic organization called Cicada 3301 that posts challenging puzzles online, possibly to recruit codebreakers and linguists.
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# Pick Up and Go Resources-
Resources for busy teachers of Maths and PE. In most cases you can just download, print and teach! If you would like to purchase more than one product at a time and there is not a bundle for it email me for a custom bundle to get 15% off the total price at pickupandgoresources@gmail.com
177k+Views
Resources for busy teachers of Maths and PE. In most cases you can just download, print and teach! If you would like to purchase more than one product at a time and there is not a bundle for it email me for a custom bundle to get 15% off the total price at pickupandgoresources@gmail.com
12 Resources
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14 Resources
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#### Multiplying and dividing negative numbers BINGO
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This negative numbers BINGO activity is a really fun way of practising multiplying and dividing with negative numbers. Skills include multiplying with numbers between -12 and 12 and dividing with numbers that are multiples of -12 to 12. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! This product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This negative numbers maths bingo is useful for revision, prior learning checks, assessment for learning activities, a starter, a plenary or just a fun activity at the end of the lesson. Click here for more negative numbers activities, including mazes and worksheets.
#### Rounding to the nearest whole number BINGO (numbers to 100)
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This Rounding To The Nearest Whole Number BINGO activity is a really fun way of practising rounding. Skills include rounding numbers up to 100 with one decimal place to the nearest whole number. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! Try out a Rounding BINGO for free here. This product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This rounding maths bingo is useful for revision, starters, plenaries, assesssment for learning activities or a fun activity at the end of the lesson. Click here for more rounding activities, including mazes, dominos and treasure hunts.
#### Rounding to the nearest 10 with 4 digit numbers BINGO
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This rounding to the nearest 10 BINGO activity is a really fun way of practising rounding. Skills include rounding numbers from 1000 to 9999 to the nearest ten. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! Try out a Rounding BINGO for free here. This product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This rounding maths bingo is useful for revision, prior learning checks, assessment for learning activities, a starter, a plenary or just a fun activity at the end of the lesson. Click here for more rounding activities, including mazes, dominos and treasure hunts.
#### Rounding to the nearest 10 BINGO BUNDLE
7 Resources
This rounding to the nearest 10 BINGO BUNDLE activity is a really fun way of practising rounding. Skills include rounding numbers numbers from 2-7 digits to the nearest 10, as well as rounding numbers up to 10 million to the nearest 10. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! Activities included in this BUNDLE: ★Rounding 2 digits to the nearest 10 ★Rounding 3 digits to the nearest 10 ★Rounding 4 digits to the nearest 10 ★Rounding 5 digits to the nearest 10 ★Rounding 6 digits to the nearest 10 ★Rounding 7 digits to the nearest 10 ★Rounding to the nearest 10 with numbers up to 10 million Try out a Rounding BINGO for free here. Each product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This rounding maths bingo is useful for revision, prior learning checks, assessment for learning activities, a starter, a plenary or just a fun activity at the end of the lesson. Click here for more rounding activities, including mazes, dominos and treasure hunts.
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This rounding to the nearest ten BINGO activity is a really fun way of practising rounding. Skills include rounding numbers from 10000 to 99999 to the nearest ten. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! Try out a Rounding BINGO for free here. This product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This rounding maths bingo is useful for revision, prior learning checks, assessment for learning activities, a starter, a plenary or just a fun activity at the end of the lesson. Click here for more rounding activities, including mazes, dominos and treasure hunts.
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This rounding to the nearest ten BINGO activity is a really fun way of practising rounding. Skills include rounding numbers from 100000 to 999999 to the nearest ten. Show the questions on a Smart Board or projector. Your students will love it and won’t realise how many questions they are answering! Try out a Rounding BINGO for free here. This product includes: ★ A PowerPoint display with instructions, student choice bingo numbers and 40 questions ★All 40 numbers on screen for students to make their own 4 by 4 grid and pick their own numbers (saves on printing) ★32 unique student cards (4 per page of paper to save on printing) ★ Answers for the teacher to check if students have crossed off the right numbers! This rounding maths bingo is useful for revision, prior learning checks, assessment for learning activities, a starter, a plenary or just a fun activity at the end of the lesson. Click here for more rounding activities, including mazes, dominos and treasure hunts.
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#### Angles in parallel lines: finding a missing angle and giving reasons Treasure hunt
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A fun 10 card treasure hunt for a finding angles in parallel lines lesson. Skills covered are: finding one missing angle in a set of parallel lines and giving reasons. Student record card and solutions included. Use as a starter, assessment for learning activity or a plenary in a lesson on parallel lines or angles. Difficulties include ★ Gold (medium)finding one missing angle in a set of parallel lines and giving reasons Students need to start on any random card and note down the letter on their recording card (first page). They answer the question on that card then find another card that has the answer on (they then note down that letter and so on). Suggestions for use: ★Print them off on A4 and scatter them around the room (students love getting out of their seats!) ★Print the cards off 4 to a page and give them to students individually to do on their desk Click here for more parallel lines resources, including foldables, worksheets, treasure hunts, lessons and tick or trash activities. U6FE
#### Percentage of amounts worksheet
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A set of differentiated worksheets for finding a percentage of an amount. Skills covered are finding percentages of amounts in multiples of 10%, multiples of 5%, multiples of 1%, multiples of 10% where calculations may involve decimals, multiples of 5% where calculations may involve decimals, multiples of 1% where calculations may involve decimals, percentage increase and decrease. Fantastic to use in lessons where you have a lot of different stages of learning (or over a few lessons!). Also great to give to students for homework and they could pick the difficulty they wanted to do! Included is a PDF version and a Powerpoint version where the text within the tables are editable. Difficulties include (for whole number calculations): ★ Blue (easy) – Finding percentage of amounts in multiples of 10% ★ Gold (medium) - Finding percentage of amounts in multiples of 5% ★ Red (Hard) Finding percentage of amounts in multiples of 1% Difficulties include (for decimal calculations): ★ Blue (easy) – Finding percentage of amounts in multiples of 10% ★ Gold (medium) - Finding percentage of amounts in multiples of 5% ★ Red (Hard) Finding percentage of amounts in multiples of 1% ★ Purple (challenging) Finding percentage increase and decrease of amounts using a mixture of all the skills above. There are approximately 60 blue questions, 60 gold questions, 60 red questions and 30 purple questions in these worksheets. Each topic is colour coded by difficulty. Full solutions included in the same format as the questions to make it easier to mark! Also included is a contents page so you can quickly look up what you need! Click here for more percentage resources.
#### Halloween maths: multiplication treasure hunts
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These Halloween maths multiplication differentiated treasure hunts will give students lots of practice of their times tables and multiplication skills. A great seasonal activity to engage students this halloween.Great for an active starter or fun activity for early finishers. Especially if you want students to practise and consolidate multiplication without using a traditional worksheet. Skills include: times tables to 12, multiplying a 1 digit by 2 digit, multiplying 2 digit by 3 digit. 10 treasure hunt cards per difficulty, solutions and student record cards included. Students absolutely love getting out of their seat and they can check their answers independently as they go along! Difficulties include: ★(blue letter box)times tables to 12 ★(gold letter box)multiplying a 1 digit by 2 digit ★(red letter box)multiplying 2 digit by 3 digit Students need to start on any random card and note down the letter on their recording card (first page). They answer the question on that card then find another card that has the answer on (they then note down that letter and so on). Suggestions for use: ★Print them off on A4 and scatter them around the room (students love getting out of their seats!) ★Print all 3 sets off A4 and scatter them around the room and get students to start at different levels. They could move up a level if they complete the previous one ★Print the cards off 4 to a page and give them to students individually to do on their desk See here for more Halloween themed activities.
#### Halloween Maths - colour by number bundle (x and ÷)
2 Resources
This Halloween maths colour by numbers bundle. Will give students lots of practice of their times tables and division. A great seasonal activity to engage students this halloween. Students need to answer the questions and find which colour corresponds to the answer s in order to reveal a halloween picture. Great starter for lessons or to give to early finishers. Fun Halloween clip art border included that students can also colour in. The pictures are: Multiplcation - Pumpkin Division -ghost If you would like more Halloween themed maths activities please click here.
#### Halloween maths - division colour by number
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This Halloween maths colour by number fun activity is sure to give students lots of practice of their division skills.A great seasonal activity to engage students this halloween. Students need to answer the questions and find which colour corresponds to the answer s in order to reveal a halloween picture. This particular picture reveals a ghost. Great starter for lessons or to give to early finishers. Fun Halloween clip art border included that students can also colour in. If you would like more Halloween themed maths activities please click here.
#### Halloween maths: multiplication mazes
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These Halloween maths multiplication mazes will give students lots of practice of their times tables and multiplication skills. A great seasonal activity to engage students this halloween.Great for a starter or early finishers near halloween. Especially if you want students to practise and consolidate multiplication without using a traditional worksheet. 1 maze works on timestables up to 12 and the other works on multiplying 1 digit by 2 digits. This activity is great because students can check their own answers as they go through and most of the time they don’t realise they are answering multiple questions. This product includes 2 unique mazes with solutions and some great halloween clipart. Students work their way through the maze, colouring the correct answers and following the path to the next question. See here for more halloween themed activities.
#### Halloween Maths - Laws of Indices mazes
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These Halloween maths laws of indices mazes will give students lots of practise of simplifying indices that have been multiplied, divided or have brackets. A great seasonal activity to engage students this Halloween. Great for a starter/plentary or early finishers near Halloween. Especially if you want students to practice and consolidate laws of indices without using a traditional worksheet. This activity is great because students can check their own answers as they go through and most of the time they don’t realise they are answering multiple questions. This product includes 2 unique mazes with solutions and some great Halloween clip art. Students work their way through the maze, coloring the correct answers and following the path to the next question. See here for more Halloween maths activities.
9 Resources
#### Thanksgiving Math Color By Number- Exponent Rules / Laws of Exponents
(0)
This Thanksgiving math color by number fun activity is sure to give students lots of practice using the product rule, quotient rule and power rule. A great seasonal activity to engage students this Thanksgiving. All the squares have numbers in them and students need to simplify the expressions and color in all the squares that correspond to their answer in the color given in order to reveal a Thanksgiving picture. This particular picture reveals a Turkey. This product includes a student sheet and an answer sheet for teachers. Great bell ringer activity or to give to early finishers. Fun fall leaf clip art border included that students can also colour in.
#### Thanksgiving Math Color By Number-Greatest Common Factor & Least Common Multiple
(0)
This Thanksgiving math color by number fun activity is sure to give students lots of practice finding the GCF and LCM. A great seasonal activity to engage students this Thanksgiving. All the squares have numbers in them and students need to answer the questions and color in all the squares that correspond to their answer in the color given in order to reveal a Thanksgiving picture. This particular picture reveals a Turkey. This product includes a student sheet and an answer sheet for teachers. Great bell ringer activity or to give to early finishers. Fun fall leaf clip art border included that students can also colour in.
#### Valentine's Day Maths HCF & LCM
(0)
This Valentine’s day themed colour by number will give students lots of practice of finding the Highest Common Factor and Lowest Common Multiple. A great seasonal activity to engage students this Valentine’s. The Picture will reveal an Emoji with Love Heart eyes! Great for display! Use as a plenary, for early finishers or a fun activity. It’s also great if you want students to practice and consolidate finding the HCF and LCM without using a traditional worksheet. All the squares have a number in them and students need to answer the question and colour in all the squares that correspond to their answer, in the colour given, in order to reveal an emoji. There is also a clip art border for students to color in. This product includes: ♥A Color by Number sheet ♥A question sheet with space for workings ♥An Answer sheet for teachers ♥♥ Find other Valentine’s Day Maths resources here ♥♥
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## ››Convert cord [firewood] to quarter [UK, liquid]
cord quarter
How many cord in 1 quarter? The answer is 0.080271825461359.
We assume you are converting between cord [firewood] and quarter [UK, liquid].
You can view more details on each measurement unit:
cord or quarter
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 0.27589582978642 cord, or 3.4370195046732 quarter.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cord [firewood] and quarters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cord to quarter
1 cord to quarter = 12.45767 quarter
2 cord to quarter = 24.91534 quarter
3 cord to quarter = 37.37301 quarter
4 cord to quarter = 49.83068 quarter
5 cord to quarter = 62.28836 quarter
6 cord to quarter = 74.74603 quarter
7 cord to quarter = 87.2037 quarter
8 cord to quarter = 99.66137 quarter
9 cord to quarter = 112.11904 quarter
10 cord to quarter = 124.57671 quarter
## ››Want other units?
You can do the reverse unit conversion from quarter to cord, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Cord
unit of wood equal to about 128 cubic feet
## ››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!
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# Suboptimal codes. For the Z-channel of Problem 7.8, assume that we choose a code at random, where... 1 answer below »
Suboptimal codes. For the Z-channel of Problem 7.8, assume that we choose a code at random, where each codeword is a sequence of fair coin tosses. This will not achieve capacity. Find the maximum rate R such that the probability of error averaged over the randomly generated codes, tends to zero as the block length n tends to infinity.
Problem 7.8:
Z-channel. The Z-channel has binary input and output alphabets and transition probabilities p(y|x) given by the following matrix:
Find the capacity of the Z-channel and the maximizing input probability distribution.
dharmireddy s
A Z channel has binary input and output alphabets with transition probabilities 2-06 (a) Draw the channel transition...
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Related Articles
# Check if the number is balanced
• Last Updated : 07 May, 2021
Given a number N in the form of a string, the task is to check whether the given number is balanced or not.
Balanced Number: A number is said to be balanced if the sum of digits in the first half of it is equal to the sum of the digits in the second half.
Note: All Palindromic numbers are balanced numbers.
Examples:
Input: N = 19091
Output: Balanced
Explanation:
middle element is 0
Sum of left half = 1 + 9 = 10
Sum of right half = 9 + 1 = 10
Hence, the given number is a Balanced number.
Input: N = 133423
Output: Not Balanced
Explanation:
Sum of left half = 1 + 3 + 3 (7)
Sum of right half = 4 + 2 + 3 (9)
Hence, the given number is not Balanced
Approach:
Iterate over half the length of the number from the beginning. Calculate the sum of digits of the first half and the second half simultaneously by adding s[i] and s[number of digits – 1 – i] to leftSum and rightSum respectively. Finally, check if the leftSum and rightSum are equal or not.
Below is the implementation of the above approach.
## C++
`// C++ program to check``// if a number is``// Balanced or not` `#include ``using` `namespace` `std;` `// Function to check whether N is``// Balanced Number or not``void` `BalancedNumber(string s)``{`` ``int` `Leftsum = 0;`` ``int` `Rightsum = 0;` ` ``// Calculating the Leftsum`` ``// and rightSum simultaneously`` ``for` `(``int` `i = 0; i < s.size() / 2; i++) {` ` ``// Typecasting each character`` ``// to integer and adding the`` ``// digit to respective sums`` ``Leftsum += ``int``(s[i] - ``'0'``);`` ``Rightsum += ``int``(s[s.size() - 1 - i]`` ``- ``'0'``);`` ``}` ` ``if` `(Leftsum == Rightsum)`` ``cout << ``"Balanced"` `<< endl;`` ``else`` ``cout << ``"Not Balanced"` `<< endl;``}` `// Driver Code``int` `main()``{`` ``string s = ``"12321"``;` ` ``// Function call`` ``BalancedNumber(s);` ` ``return` `0;``}`
## Java
`// Java program to check if a number``// is Balanced or not``import` `java.io.*;` `class` `GFG{`` ` `// Function to check whether N is``// Balanced Number or not``private` `static` `void` `BalancedNumber(String s)``{`` ``int` `Leftsum = ``0``;`` ``int` `Rightsum = ``0``;`` ` ` ``// Calculating the Leftsum`` ``// and rightSum simultaneously`` ``for``(``int` `i = ``0``; i < s.length() / ``2``; i++)`` ``{`` ` ` ``// Typecasting each character`` ``// to integer and adding the`` ``// digit to respective sums`` ``Leftsum += (``int``)(s.charAt(i) - ``'0'``);`` ``Rightsum += (``int``)(s.charAt(`` ``s.length() - ``1` `- i) - ``'0'``);`` ``}`` ` ` ``if` `(Leftsum == Rightsum)`` ``System.out.println(``"Balanced"``);`` ``else`` ``System.out.println(``"Not Balanced"``);``}` `// Driver Code``public` `static` `void` `main (String[] args)``{`` ``String s = ``"12321"``;`` ` ` ``// Function call`` ``BalancedNumber(s);``}``}` `// This code is contributed by jithin`
## Python3
`# Python3 program to check``# if a number is``# Balanced or not` `# Function to check whether N is``# Balanced Number or not``def` `BalancedNumber(s):` ` ``Leftsum ``=` `0`` ``Rightsum ``=` `0` ` ``# Calculating the Leftsum`` ``# and rightSum simultaneously`` ``for` `i ``in` `range``(``0``, ``int``(``len``(s) ``/` `2``)):` ` ``# Typecasting each character`` ``# to integer and adding the`` ``# digit to respective sums`` ``Leftsum ``=` `Leftsum ``+` `int``(s[i])`` ``Rightsum ``=` `(Rightsum ``+`` ``int``(s[``len``(s) ``-` `1` `-` `i]))` ` ``if` `(Leftsum ``=``=` `Rightsum):`` ``print``(``"Balanced"``, end ``=` `'\n'``)`` ``else``:`` ``print``(``"Not Balanced"``, end ``=` `'\n'``)` `# Driver Code``s ``=` `"12321"` `# Function call``BalancedNumber(s)` `# This code is contributed by PratikBasu`
## C#
`// C# program to check``// if a number is``// Balanced or not``using` `System;``class` `GFG{`` ` `// Function to check whether N is``// Balanced Number or not``static` `void` `BalancedNumber(``string` `s)``{`` ``int` `Leftsum = 0;`` ``int` `Rightsum = 0;` ` ``// Calculating the Leftsum`` ``// and rightSum simultaneously`` ``for` `(``int` `i = 0; i < s.Length / 2; i++)`` ``{`` ``// Typecasting each character`` ``// to integer and adding the`` ``// digit to respective sums`` ``Leftsum += (``int``)(Char.GetNumericValue(s[i]) -`` ``Char.GetNumericValue(``'0'``));`` ``Rightsum += (``int``)(Char.GetNumericValue(s[s.Length -`` ``1 - i]) -`` ``Char.GetNumericValue(``'0'``));`` ``}` ` ``if` `(Leftsum == Rightsum)`` ``Console.WriteLine(``"Balanced"``);`` ``else`` ``Console.WriteLine(``"Not Balanced"``);``} ` `// Driver code``static` `void` `Main()``{`` ``string` `s = ``"12321"``;` ` ``// Function call`` ``BalancedNumber(s);``}``}` `// This code is contributed by divyeshrabadiya07`
## Javascript
``
Output:
`Balanced`
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
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Question
# 1) A constant current of 0.800 A is run through the electrolytic cell in order to...
1) A constant current of 0.800 A is run through the electrolytic cell in order to produce oxygen (gas) at the anode:
2 H2O → O2(g) + 4 e- + 4 H+
The amount (in grams) of O2 obtained after 15.2 min of electrolysis is [X] g
2) Electrolysis of a solution of Tl3+ produces deposition of elemental thallium, Tl, at the cathode. The time (in min.) needed for a constant current of 1.20 A to deposit 0.500 g of Tl(s) is [X] min.
3) An electrochemical cell is described by the following line notation:
Zn(s) ǀ Zn2+(aq) (0.0420 M) ǀǀ Tl3+(aq) (9.06×10-2 M), Tl+(aq) (0.0400 M) ǀ Pt(s).
Given: E°(Zn2+/Zn) = - 0.763 V; E°(Tl3+/Tl+) = + 1.25 V.
The potential on the cathode half-cell, E+ = V; the potential on the anode half-cell, E- = V; and the overall cell potential, Ecell = V.
4) Consider electrolysis of molten copper bromide, CuBr2. What is the product of this process obtained (a) at the cathode? (b) at the anode?
1. a) Cathode: Cu(s);
(b) Anode: Br2(l)
2. (a) Cathode: Br2(l);
(b) Anode: Cu(s)
3. (a) Cathode: Cu2+;
(b) Anode: Br-
4. (a) Cathode: Br-;
(b) Anode: Cu2+
5. (a) Cathode: Cu(s);
(b) Anode: O2(g)
6. (a) Cathode: H2(g);
(b) Anode: Cu2+
1)
O2- ----> 1/2 O2 + 2e-
for 1 mol of O2, 4 mol of electrons are required
charge passed, Q = I*t
= 0.800 A * (15.2*60)s
= 729.6 C
1 mol of electron = 96480 C
Q = 729.6 C
= 729.6 / (96480 ) mol electrons
= 7.562*10^-3 mol electrons
mol of O2 formed = (7.562*10^-3) / 4 = 1.89*10^-3 mol of O2
molar mass of O2 = 32 g/mol
mass of O2 = mol of O2 * molar mass of O2
=1.89*10^-3 * 32
= 0.060 g
i am allowed to answer only 1 question at a time. Please ask other as different question
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# Number 8003624
Number 8,003,624 spell 🔊, write in words: eight million, three thousand, six hundred and twenty-four , approximately 8.0 million. Ordinal number 8003624th is said 🔊 and write: eight million, three thousand, six hundred and twenty-fourth. The meaning of the number 8003624 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 8003624. What is 8003624 in computer science, numerology, codes and images, writing and naming in other languages
## What is 8,003,624 in other units
The decimal (Arabic) number 8003624 converted to a Roman number is (M)(M)(M)(M)(M)(M)(M)(M)MMMDCXXIV. Roman and decimal number conversions.
#### Weight conversion
8003624 kilograms (kg) = 17644789.5 pounds (lbs)
8003624 pounds (lbs) = 3630420.0 kilograms (kg)
#### Length conversion
8003624 kilometers (km) equals to 4973220 miles (mi).
8003624 miles (mi) equals to 12880589 kilometers (km).
8003624 meters (m) equals to 26258290 feet (ft).
8003624 feet (ft) equals 2439535 meters (m).
8003624 centimeters (cm) equals to 3151033.1 inches (in).
8003624 inches (in) equals to 20329205.0 centimeters (cm).
#### Temperature conversion
8003624° Fahrenheit (°F) equals to 4446440° Celsius (°C)
8003624° Celsius (°C) equals to 14406555.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
8003624 seconds equals to 3 months, 1 week, 1 day, 15 hours, 13 minutes, 44 seconds
8003624 minutes equals to 1 decade, 6 years, 6 months, 2 weeks, 1 hour, 44 minutes
### Codes and images of the number 8003624
Number 8003624 morse code: ---.. ----- ----- ...-- -.... ..--- ....-
Sign language for number 8003624:
Number 8003624 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 8003624
### Multiplications
#### Multiplication table of 8003624
8003624 multiplied by two equals 16007248 (8003624 x 2 = 16007248).
8003624 multiplied by three equals 24010872 (8003624 x 3 = 24010872).
8003624 multiplied by four equals 32014496 (8003624 x 4 = 32014496).
8003624 multiplied by five equals 40018120 (8003624 x 5 = 40018120).
8003624 multiplied by six equals 48021744 (8003624 x 6 = 48021744).
8003624 multiplied by seven equals 56025368 (8003624 x 7 = 56025368).
8003624 multiplied by eight equals 64028992 (8003624 x 8 = 64028992).
8003624 multiplied by nine equals 72032616 (8003624 x 9 = 72032616).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 8003624
Half of 8003624 is 4001812 (8003624 / 2 = 4001812).
One third of 8003624 is 2667874,6667 (8003624 / 3 = 2667874,6667 = 2667874 2/3).
One quarter of 8003624 is 2000906 (8003624 / 4 = 2000906).
One fifth of 8003624 is 1600724,8 (8003624 / 5 = 1600724,8 = 1600724 4/5).
One sixth of 8003624 is 1333937,3333 (8003624 / 6 = 1333937,3333 = 1333937 1/3).
One seventh of 8003624 is 1143374,8571 (8003624 / 7 = 1143374,8571 = 1143374 6/7).
One eighth of 8003624 is 1000453 (8003624 / 8 = 1000453).
One ninth of 8003624 is 889291,5556 (8003624 / 9 = 889291,5556 = 889291 5/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
8003624
#### Is Prime?
The number 8003624 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 8003624 are 2 * 2 * 2 * 1000453
The factors of 8003624 are 1 , 2 , 4 , 8 , 1000453 , 2000906 , 4001812 , 8003624
Total factors 8.
Sum of factors 15006810 (7003186).
#### Powers
The second power of 80036242 is 64.057.997.133.376.
The third power of 80036243 is 512.696.123.248.619.356.160.
#### Roots
The square root √8003624 is 2829,067691.
The cube root of 38003624 is 200,030195.
#### Logarithms
The natural logarithm of No. ln 8003624 = loge 8003624 = 15,895405.
The logarithm to base 10 of No. log10 8003624 = 6,903287.
The Napierian logarithm of No. log1/e 8003624 = -15,895405.
### Trigonometric functions
The cosine of 8003624 is -0,438522.
The sine of 8003624 is 0,89872.
The tangent of 8003624 is -2,049431.
### Properties of the number 8003624
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 8003624 in Computer Science
Code typeCode value
8003624 Number of bytes7.6MB
Unix timeUnix time 8003624 is equal to Friday April 3, 1970, 3:13:44 p.m. GMT
IPv4, IPv6Number 8003624 internet address in dotted format v4 0.122.32.40, v6 ::7a:2028
8003624 Decimal = 11110100010000000101000 Binary
8003624 Decimal = 120001121220112 Ternary
8003624 Decimal = 36420050 Octal
8003624 Decimal = 7A2028 Hexadecimal (0x7a2028 hex)
8003624 BASE64ODAwMzYyNA==
8003624 SHA187ddf30b984032abfaf5e958ff22bb32f37f5d9b
8003624 SHA22433769dffba561178dc998c25a3cfa93cb92880985866cf7c5baf44c0
8003624 SHA25611e243c83b80ee023111261432a2766333dacec600f1d875c3d8142592b4c595
8003624 SHA384858dc5929200a1ee4f478c08732b0a3449a7db34cc4bb18e9786b4826c9936a4536a6df9a4a4402539a0b7046fd1eb49
More SHA codes related to the number 8003624 ...
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## Numerology 8003624
### Character frequency in the number 8003624
Character (importance) frequency for numerology.
Character: Frequency: 8 1 0 2 3 1 6 1 2 1 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 8003624, the numbers 8+0+0+3+6+2+4 = 2+3 = 5 are added and the meaning of the number 5 is sought.
## № 8,003,624 in other languages
How to say or write the number eight million, three thousand, six hundred and twenty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 8.003.624) ocho millones tres mil seiscientos veinticuatro German: 🔊 (Nummer 8.003.624) acht Millionen dreitausendsechshundertvierundzwanzig French: 🔊 (nombre 8 003 624) huit millions trois mille six cent vingt-quatre Portuguese: 🔊 (número 8 003 624) oito milhões e três mil, seiscentos e vinte e quatro Hindi: 🔊 (संख्या 8 003 624) अस्सी लाख, तीन हज़ार, छः सौ, चौबिस Chinese: 🔊 (数 8 003 624) 八百万三千六百二十四 Arabian: 🔊 (عدد 8,003,624) ثمانية ملايين و ثلاثة آلاف و ستمائة و أربعة و عشرون Czech: 🔊 (číslo 8 003 624) osm milionů tři tisíce šestset dvacet čtyři Korean: 🔊 (번호 8,003,624) 팔백만 삼천육백이십사 Danish: 🔊 (nummer 8 003 624) otte millioner tretusinde og sekshundrede og fireogtyve Dutch: 🔊 (nummer 8 003 624) acht miljoen drieduizendzeshonderdvierentwintig Japanese: 🔊 (数 8,003,624) 八百万三千六百二十四 Indonesian: 🔊 (jumlah 8.003.624) delapan juta tiga ribu enam ratus dua puluh empat Italian: 🔊 (numero 8 003 624) otto milioni e tremilaseicentoventiquattro Norwegian: 🔊 (nummer 8 003 624) åtte million, tre tusen, seks hundre og tjue-fire Polish: 🔊 (liczba 8 003 624) osiem milionów trzy tysiące sześćset dwadzieścia cztery Russian: 🔊 (номер 8 003 624) восемь миллионов три тысячи шестьсот двадцать четыре Turkish: 🔊 (numara 8,003,624) sekizmilyonüçbinaltıyüzyirmidört Thai: 🔊 (จำนวน 8 003 624) แปดล้านสามพันหกร้อยยี่สิบสี่ Ukrainian: 🔊 (номер 8 003 624) вiсiм мiльйонiв три тисячi шiстсот двадцять чотири Vietnamese: 🔊 (con số 8.003.624) tám triệu ba nghìn sáu trăm hai mươi bốn Other languages ...
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## Comment
If you know something interesting about the number 8003624 or any other natural number (positive integer), please write to us here or on Facebook.
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CC-MAIN-2023-14
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longest
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https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-3-section-3-2-length-exercise-page-141/16
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## Elementary Technical Mathematics
For this problem, we use unit conversions to find our answer: $1mm=1mm\times\frac{1m}{1000 mm}\times\frac{100cm}{1m}=\frac{100}{1000} cm=\frac{1}{10} cm=0.1 cm$
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| 3.765625
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CC-MAIN-2018-51
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latest
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https://www.sawaal.com/ratio-and-proportion-questions-and-answers/a-child-has-three-different-kinds-of-chocolates-costing-nbsprs2-rs5-rs10-nbsphe-spends-total-rs-120-_6845
| 1,537,581,045,000,000,000
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crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00111.warc.gz
| 860,593,234
| 15,849
|
8
Q:
A child has three different kinds of chocolates costing Rs.2, Rs.5, Rs.10. He spends total Rs. 120 on the chocolates. what is the minimum possible number of chocolates he can buy, if there must be atleast one chocolate of each kind?
A) 22 B) 19 C) 17 D) 15
Explanation:
Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.
Thus, 2 x 5 + 5 x 2 =Rs.20
Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.
Thus minumum number of chocolates = 5 + 2 + 10 = 17
Q:
One year ago the ratio between Maneela’s and Shanthi’s salary was 3 : 4. The ratios of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Maneela, now is?
A) Rs. 1600 B) Rs. 1700 C) Rs. 1800 D) Rs. 1900
Answer & Explanation Answer: A) Rs. 1600
Explanation:
Let the salaries of Maneela and Shanthi one year before be M1, S1 & now be M2, S2 respectively.
Then, from the given data,
M1/S1 = 3/4 .....(1)
M1/M2 = 4/5 .....(2)
S1/S2 = 2/3 .....(3)
and M2 + S2 = 4160 .....(4)
Solving all these eqtns, we get M2 = Rs. 1600.
18 3402
Q:
The ratio of Pens and Pencils in a shop is 3 : 2 respectively. The average number of Pens and Pencils is 180. What is the number of Pencils in the shop?
A) 444 B) 344 C) 244 D) 144
Explanation:
Given ratio of pens and pencils = 3 :2
Number of Pens = 3x
Number of Pencils = 2x
Average number of pencils & Pens = 180
5x = 360
=> x = 72
Hence, the number of pencils = 2x = 72 x 2 = 144.
18 2940
Q:
The ratio of boys and girls in a school is 9:5.If the total number of students in the school is 1050.Then number of boys is
Let the ratio be 'R'
Total number of students = 1050
Then,
9R + 5R = 1050
14R = 1050
=> R = 75
Hence, the number of boys = 9R = 9 x 75 = 675
3706
Q:
3 : 12 :: 5 : ?
A) 17 B) 30 C) 26 D) 32
Explanation:
22 3028
Q:
The ratio of the incomes of Pavan and Amar is 4 : 3 and the ratio of their expenditures are 3:2. If each person saves Rs. 1889, then find the income of Pavan?
A) 6548 B) 5667 C) 7556 D) 8457
Explanation:
Let ratio of the incomes of Pavan and Amar be 4x and 3x
and Ratio of their expenditures be 3y and 2y
4x - 3y = 1889 ......... I
and
3x - 2y = 1889 ...........II
I and II
y = 1889
and x = 1889
Pavan's income = 7556
15 3397
Q:
Maneela lent Rs. 8000 partly at the rate of 5% and partly at the rate of 6% per annum simple interest. The total interest she get after 2 years is Rs. 820, then in which ratio will Rs. 8000 is to be divided?
A) 7:1 B) 13:5 C) 15:7 D) 2:7
Explanation:
Maneela lent Rs. 8000 in two parts,
15 2776
Q:
The ratio of male and female in a city is 7 : 8 respectively and percentage of children among male and female is 25 and 20 respectively. If number of adult females is 156800, what is the total population of the city?
A) 4,12,480 B) 3,67,500 C) 5,44,700 D) 2,98,948
Explanation:
Let the total population be 'p'
Given ratio of male and female in a city is 7 : 8
In that percentage of children among male and female is 25% and 20%
=> Adults male and female % = 75% & 80%
But given adult females is = 156800
=> 80%(8p/15) = 156800
=> 80 x 8p/15 x 100 = 156800
=> p = 156800 x 15 x 100/80 x 8
=> p = 367500
Therefore, the total population of the city = p = 367500
29 4743
Q:
Two numbers are in the ratio 3 : 7. If 6 be added to each of them, then they are in the ratio 5 : 9. Find the numbers ?
A) 11 & 17 B) 7 & 17 C) 9 & 21 D) 13 & 23
Answer & Explanation Answer: C) 9 & 21
Explanation:
Let the two numbers be x and y
Given x : y = 3 : 7 .....(1)
Now, x+6 : y+6 = 5 : 9 .....(2)
From (1), x = 3y/7
From (2), 5y - 9x = 24
=> 5y - 9(3y/7) = 24
=> y = 21
=> From(1), x = 9
Hence, the two numbers be 9 and 21
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CC-MAIN-2018-39
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https://www.hindustantimes.com/trending/brain-teaser-you-re-a-genius-if-you-can-find-the-height-of-this-glass-in-10-seconds-101709622372096.html
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You’re a genius if you can find the height of this glass in 10 seconds | Trending - Hindustan Times
Brain Teaser: You’re a genius if you can find the height of this glass in 10 seconds
ByTrisha Sengupta
Mar 05, 2024 01:11 PM IST
While some people easily calculated the height of this glass, a few shared that they struggled with the brain teaser.
Brain teasers are fun to solve, even if some of them turn out to be real head-scratchers. Adding to that list is this maths problem that may leave you baffled. Shared on X, the brain teaser involves finding the height of a glass.
Also Read: Viral brain teaser: Can you find the relationship between P and T in this mind-bending puzzle?
“Can you find the height of the glass?” reads the caption posted along with an image. The picture shows two stacks of glasses and one single glass. The heights of two glass stacks are mentioned in the image and the challenge is to find the height of the single glass.
Do you think you can solve this maths problem?
The post was shared a day ago on X. Since then, the tweet has accumulated more than 8.5 lakh views and counting. The share has also collected nearly 1,200 likes. From sharing explanations to posting how they couldn’t solve the puzzle, people posted varied reactions.
“X + 4d = 34. X + d = 19. d= 5. X = 19 - 5. The height of the glass = 14,” posted an X user.
“14cm. I copied from someone in the comments section,” joked another.
“Simple maths. Two equations with two variables. X + 4y = 34. x+ y = 19. Now substitute x = 19 - y into the top equation: 19-y + 4y = 34, and solve for y. 3y = 34-19 = 15, y = 5. Now put it back into the bottom equation: x + 5 = 19. x = 19-5 = 14. So the glass is 14 cm tall,” added a third.
“14cm. Each new glass stacked adds an additional 5 cm height,” suggested a fourth.
Also Read: Brain Teaser: ‘99% get it wrong.’ Can you find out how many eggs are left?
While a fifth added “Sure give me a ruler,” a sixth wrote, “Didn’t get, can someone explain this?"
A few also wrote “14” or “15” in the comments section as their answer to the brain teaser.
What are your thoughts on this brain teaser? Did you manage to calculate the correct height of the glass?
Unveiling 'Elections 2024: The Big Picture', a fresh segment in HT's talk show 'The Interview with Kumkum Chadha', where leaders across the political spectrum discuss the upcoming general elections. Watch Now!
Get Latest Updates on Trending News Viral Video, Photos from India and around the world
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CC-MAIN-2024-18
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https://www.onlinemath4all.com/total_surface_area_example.html
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## TOTAL SURFACE AREA EXAMPLE
Total surface area examples :
Here we are going to see some example problems to understand how to find the total surface area of the any shapes.
Example 1 :
A toy is in the form of right circular cylinder with a hemisphere at the one end and a cone at the other end.The base radius measures 3.5 cm , height of the cylindrical portion is 10 cm. and conical part measures 12 cm.Find the total surface-area of the toy(use π = 3.14)
Solution :
Radius of the hemisphere = 3.5 cm
C.S.A of the hemisphere = 2πr²square units
= 2(3.5)²π
= (24.5)π cm²
Radius of the Cylinder r = 3.5 cm
Height of the Cylinder h = 10 cm
C.S.A of the Cylinder = 2πrh square units
= 2 π x .5 x 10
= 70π cm²
Radius of the cone r= 3.5 cm
Height of the cone h = 12 cm
Slant height of the cone l = √(r² + h²)
l = √(3.5² + 12²)
l = √(12.25+144)
l = √(156.25)
l = 12.5 cm
C.S.A of the Cone = πrl square units
= π(3.5)(12.5)
= (43.75)π cm²
T.S.A of the toy = CSA of hemisphere + (CSA of cylinder) + (CSA of cone)
= (24.5 π) + (70 π) + (43.75 π)
= 138.25 x 3.14
Hence, total surface area of the toy = 434.11 cm²
Example 2 :
Find the total surface area of cylinder whose height is 8 cm and radius is 4 cm.
Solution :
Radius of the Cylinder = 4 cm
Height of the cylinder = 8 cm
Required total surface area of the cylinder = 2πr(h+r)
= 2 π (4) (8+4)
= 2 π (4) (12)
= 96 π
Hence, total surface area of the Cylinder = 96π cm²
Example 3
Find the total surface area of cylinder whose height is 16 cm and radius is 7 cm.
Solution:
Radius of the Cylinder = 7 cm
Height of the cylinder = 16 cm
Required Total surface area of the cylinder = 2πr(h+r)
= 2 (22/7) (7) (16+7)
= 2 (22) (23)
= 1012 cm²
Hence, Total surface area of the Cylinder = 1012 cm²
Student who are practicing questions on total surface area can go through the steps of the above questions on total surface area to have better understanding.
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https://www.researchandmarkets.com/reports/2210712/business_decision_analysis_an_active_learning
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+353-1-416-8900REST OF WORLD
+44-20-3973-8888REST OF WORLD
1-917-300-0470EAST COAST U.S
1-800-526-8630U.S. (TOLL FREE)
PRINTER FRIENDLY
# Business Decision Analysis. An Active Learning Approach. Edition No. 1. Open Learning Foundation
• ID: 2210712
• Book
• April 1999
• 644 Pages
• John Wiley and Sons Ltd
Business Decision Analysis is part of a major new national programme of texts and modules designed for undergraduate students on Business Studies degree courses. It provides 150 hours of high quality study to be used in a supported learning environment.
The module provides a comprehensive introduction to the quantitative analysis and solution of business problems and covers some of the key topics in the field, including an introduction to model building for business decision analysis, linear programming, regression analysis, time-series analysis and simulation techniques. Business Decision Analysis contains numerous activities and exercises to develop an understanding of the subject, including many utilizing Microsoft Excel in version 5.0 or later (not supplied with this publication). The module provides the most effective teaching and learning resource available at this level.
Note: Product cover images may vary from those shown
Part I: An Introduction to Business Decision Analysis:.
1. What is Business Decision Analysis?.
2. Model-Building in Business Decision Analysis.
3. The Components of a Mathematical Model.
4. Deterministic and Stochastic Models.
5. Single-attribute and Multi-attribute Problems.
6. Sensitivity Analysis and Model Building.
Part II: Decision Analysis:.
7. Decision Trees and Payoff Matrices.
8. Decision-Making under Conditions of Uncertainty.
9. Decision-Making under Conditions of Risk.
10. Multi-Stage Decision Problems.
11. Revising Probabilities.
12. Extensions.
Part III: Linear Programming:.
13. Formulating a Linear Programming Problem.
14. Solving Linear Programming Problems Using a Graphical Method.
15. Sensitivity Analysis of Solutions.
16. Computer Solution of Linear Programming Problems.
17. The Transportation Problem.
18. The Assignment Problem.
19. Linear Programming - Limitations and Extensions.
Part IV: Regression Analysis:.
20. Functional Relationships.
21. Bi-Variate Causal Models.
22. The Technique of Regression Analysis.
23. Regression Models and Predictive Accuracy.
24. The Analysis of Residuals.
25. Confidence Intervals and Regression Analysis.
26. The Multivariate Model.
27. The Performance of the Multivariate Model.
28. Refining the Multiple Regression Model.
29. Extending Regression Analysis.
Part V: Time Series Analysis:.
30. Time Series: an Overview.
31. Decomposition of a Time Series.
32. Non-Centred Moving Averages and Forecasting Error.
33. Exponential Smoothing.
34. Introduction to ARIMA.
Part VI: Simulation:.
35. What is Simulation?.
36. The Technique of Simulation.
37. Refining the Simulation Model.
38. Waiting Lines and Scheduling Problems.
39. Inventory Problems.
40. Waiting Lines: The Time Element.
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https://quomodocumque.wordpress.com/tag/approximation/
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## How to compute arctangent if you live in the 18th century
Michael Lugo wrote a great post, following an idea of Andrew Gelman, about what would have happened if Pythagoras had known linear regresson. Punchline: he would have found a linear formula for the hypotenuse with an R^2 of 0.9995, and would surely not have seen any need to pursue the matter any further!
I thought this was mostly just a joke, until the mail brought me a copy of the very interesting A Wealth of Numbers from Princeton University Press, an anthology of popular writing about math stretching from the 16th century to the present.
From Hugh Worthington’s 1780 textbook, The Resolution of Triangles:
THE THIRD CASE is, the sides being given, to find the angles, and the rule is as follows. “Half the longer of the two legs added to the hypotenuse, is always in proportion to 86, as the shorter leg is to its opposite angle.”
In modern language: given a right triangle with legs a and b, and hypotenuse 1, how do you find the angle x adjacent to a? Nowadays we would just say “x = arctan b/a.” But this kind of computation was presumably not so easy in 1780. Instead, Worthington offers the approximation
b/x = (a/2 + 1) / 86
which (after converting to radians, as good manners requires) gives
x = (86*pi/180) b / (a/2 + 1)
Of course, when the hypotenuse is set to 1, we have b = sin x and a = cos x. So the approximation is
x = (86*pi/180) (sin x) / (cos x / 2 + 1).
This turns out to be a pretty awesome approximation!
How do you think they came up with this?
Tagged , ,
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## Linear Objective Function MCQs: Questions and Answers
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User lyx - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:37:35Z http://mathoverflow.net/feeds/user/19463 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81626/is-strong-approximation-difficult/81676#81676 Answer by LYX for Is strong approximation difficult? LYX 2011-11-23T01:35:04Z 2011-11-23T01:35:04Z <p>I am quite surprised that it seems to be so poorly known, but an elementary proof of Strong Approximation for $SL_n$ over a Dedekind domain can already be found in Bourbaki (Algebre Commutative, VII, $\S$2, n.4). Essentially, the idea is to deduce it from the Chinese Remainder Theorem, by means of elementary matrices, as hinted in Ralph's answer. Actually, I would be curious to know a bit more about the history of this case of Strong Approximation: who and when did prove it first?</p> http://mathoverflow.net/questions/81626/is-strong-approximation-difficult/81676#81676 Comment by LYX LYX 2011-11-23T02:42:12Z 2011-11-23T02:42:12Z Bourbaki's Algebre Commutative, VII, §2, n.4 has the title: "The Approximation Theorem for Dedekind domains". They take a Dedekind domain $A$ with fraction field $K$ and define the ring of restricted adeles $\bf A$ as the restricted product of completions of $K$ over all (non-zero) primes of $A$. Proposition 4 states: $SL(n,K)$ is dense in $SL(n,{\bf A})$.
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# physics
posted by on .
A robot probe drops a camera off the rim of
a 437 m high cliff on Mars, where the free-fall
acceleration is 3.7 m/s
2
.
Find the velocity with which it hits the
ground.
• physics - ,
vf^2=2*a*h
• physics - ,
tried that...
vf^2=2*3.7*437
vf^2=3233.8
vf=56.86651036
but my teacher says its the wrong answer....
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# functions: find domain and range of f(x) = x^2 - 49
#### gierhame
##### New member
How do you find the domain and range of the function of x[sup:1khd9uk1]2[/sup:1khd9uk1]-49?
I don't really understand how to do this.
#### gierhame
##### New member
Re: functions
I know that i it has something to do with the number 7;
could you explain the process maybe with another problem. I just don't know how to find it with an equation.
#### Mrspi
##### Senior Member
Re: functions
gierhame said:
I know that i it has something to do with the number 7;
could you explain the process maybe with another problem. I just don't know how to find it with an equation.
I'm a bit confused... you mention an "equation" but I don't see one here.
Do you mean
f(x) = x[sup:2c6u8cwo]2[/sup:2c6u8cwo] - 49
Or do you mean something else.
The domain of a function consists of all values for the independent variable (x, in this case) for which the function definition is meaningful.
If
f(x) = x[sup:2c6u8cwo]2[/sup:2c6u8cwo] - 49
f(x) is defined for ANY real number value of x. No matter what real number you pick for x, x[sup:2c6u8cwo]2[/sup:2c6u8cwo] - 49 has a real number value.
Staff member
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Your Federal Quarterly Tax Payments are due April 15th
# Introduction to Climate change Study Cell by k966Xd
VIEWS: 1 PAGES: 46
• pg 1
Training Course on Facing the Challenges of Climate Change: Issues, Impacts and
Organized by International Training Network (ITN) Centre, BUET
Analysis and Modeling of
Climate Change
A.K.M. Saiful Islam
Associate Professor, IWFM
Coordinator, Climate Change Study Cell
Bangladesh University of Engineer and Technology (BUET)
Presentation Outline
• Overview of the Climate System
• Modeling of Climate Change
• General Circulation Model (GCM)
• IPCC SRES Scenarios
• Regional Climate Model (RCM)
• Climatic Modeling at BUET
Climate Models
• Climate models are computer-based simulations that use
mathematical formulas to re-create the chemical and
physical processes that drive Earth’s climate. To “run” a
model, scientists divide the planet into a 3-dimensional grid,
apply the basic equations, and evaluate the results.
• Atmospheric models calculate winds, heat transfer,
radiation, relative humidity, and surface hydrology within
each grid and evaluate interactions with neighboring points.
Climate models use quantitative methods to simulate the
interactions of the atmosphere, oceans, land surface, and
ice.
General Circulation Model (GCM)
• General Circulation Models (GCMs) are a class of computer-
driven models for weather forecasting, understanding climate
and projecting climate change, where they are commonly
called Global Climate Models.
• Three dimensional GCM's discretise the equations for fluid
motion and energy transfer and integrate these forward in
time. They also contain parameterizations for processes -
such as convection - that occur on scales too small to be
resolved directly.
• Atmospheric GCMs (AGCMs) model the atmosphere and
impose sea surface temperatures. Coupled atmosphere-
ocean GCMs (AOGCMs, e.g. HadCM3, EdGCM, GFDL CM2.X,
ARPEGE-Climate) combine the two models.
GCM typical horizontal resolution of between 250 and 600 km, 10 to 20 vertical
layers in the atmosphere and sometimes as many as 30 layers in the oceans.
Heart of Climate Model
Complexity of GCM
Hardware Behind the Climate Model
• Geophysical Fluid Dynamics Laboratory
Special Report on Emissions
Scenarios (SRES)
• The Special Report on Emissions Scenarios (SRES)
was a report prepared by the Intergovernmental Panel on
Climate Change (IPCC) for the Third Assessment Report
(TAR) in 2001, on future emission scenarios to be used for
driving global circulation models to develop climate
change scenarios.
• It was used to replace the IS92 scenarios used for the
IPCC Second Assessment Report of 1995. The SRES
Scenarios were also used for the Fourth Assessment
Report (AR4) in 2007.
SERS Emission Scenarios
• A1 - a future world of very rapid economic growth, global
population that peaks in mid-century and declines
thereafter, and the rapid introduction of new and more
efficient technologies. Three sub groups: fossil intensive
(A1FI), non-fossil energy sources (A1T), or a balance
across all sources (A1B).
• A2 - A very heterogeneous world. The underlying theme
is that of strengthening regional cultural identities, with
an emphasis on family values and local traditions, high
population growth, and less concern for rapid economic
development.
• B1 - a convergent world with the same global population,
that peaks in mid-century and declines thereafter, as in
the A1 storyline.
• B2 - a world in which the emphasis is on local solutions
to economic, social and environmental sustainability.
A1
• The A1 scenarios are of a more integrated world. The A1 family of
scenarios is characterized by:
– Rapid economic growth.
– A global population that reaches 9 billion in 2050 and then
– The quick spread of new and efficient technologies.
– A convergent world - income and way of life converge between
regions. Extensive social and cultural interactions worldwide.
• There are subsets to the A1 family based on their technological
emphasis:
– A1FI - An emphasis on fossil-fuels.
– A1B - A balanced emphasis on all energy sources.
– A1T - Emphasis on non-fossil energy sources.
A2
• The A2 scenarios are of a more divided world. The A2
family of scenarios is characterized by:
– A world of independently operating, self-reliant nations.
– Continuously increasing population.
– Regionally oriented economic development.
– Slower and more fragmented technological changes and
improvements to per capita income.
B1
• The B1 scenarios are of a world more integrated, and
more ecologically friendly. The B1 scenarios are
characterized by:
– Rapid economic growth as in A1, but with rapid changes towards
a service and information economy.
– Population rising to 9 billion in 2050 and then declining as in A1.
– Reductions in material intensity and the introduction of clean
and resource efficient technologies.
– An emphasis on global solutions to economic, social and
environmental stability.
B2
• The B2 scenarios are of a world more divided, but more
ecologically friendly. The B2 scenarios are characterized
by:
– Continuously increasing population, but at a slower rate than in
A2.
– Emphasis on local rather than global solutions to economic,
social and environmental stability.
– Intermediate levels of economic development.
– Less rapid and more fragmented technological change than in
A1 and B1
GCM output described in the 2007 IPCC Fourth
Assessment Report (SRES scenarios), multilayer mean
Models Scenarios Variables
BCC:CM1 1PTO2X specific humidity
BCCR:BCM2 1PTO4X
CCCMA:CGCM3_1-T47 20C3M precipitation flux
CCCMA:CGCM3_1-T63 COMMIT air pressure at sea level
CNRM:CM3 PICTL
CONS:ECHO-G SRA1B net upward shortwave flux in air
CSIRO:MK3 SRA2 air temperature
GFDL:CM2 SRB1
GFDL:CM2_1
air temperature daily max
INM:CM3 air temperature daily min
IPSL:CM4 eastward wind
LASG:FGOALS-G1_0
MPIM:ECHAM5 northward wind
MRI:CGCM2_3_2
NASA:GISS-AOM
NASA:GISS-EH
NASA:GISS-ER
NCAR:CCSM3
NCAR:PCM
NIES:MIROC3_2-HI
NIES:MIROC3_2-MED
List of GCM – Page 1
• BCC-CM1
– AgencyBeijing Climate Center, National Climate
S.Road, Zhongguancun Str., Beijing 100081, China
• BCCR
– Bjerknes Centre for Climate Research (BCCR), Univ.
of Bergen, Norway
• CGCM3
– Canadian Centre for Climate Modelling and Analysis
(CCCma)
• CNRM-CM3
– Centre National de Recherches Meteorologiques,
Meteo France, France
List of GCM– Page 2
• CONS-ECHO-G
– Meteorological Institute of the University of Bonn
(Germany), Institute of KMA (Korea), and Model and
Data Group.
• CSIRO, Australia
• INMCM3.0
– Institute of Numerical Mathematics, Russian Academy
of Science, Russia.
• GFDL
– Geophysical Fluid Dynamics Laboratory, NOAA
• NASA-GISS-AOM
– NASA Goddard Institute for Space Studies
(NASA/GISS), USA
List of GCM – Page 3
• MRI-CGCM2_3_2
– Meteorological Research Institute, Japan
Meteorological Agency, Japan
• NCAR-PCM
– National Center for Atmospheric Research (NCAR),
NASA, and NOAA
• Model NIES-MIROC3_2-MED
– CCSR/NIES/FRCGC, Japan
– Hadley Centre for Climate Prediction and Research,
Met Office, United Kingdom
Arctic Sea Ice Prediction using
community climate system model
Arctic Sea Ice in Arctic Sea Ice in
2000 2040
Prediction of Global Warming
• Figure shows the distribution of warming during the late 21st
century predicted by the HadCM3 climate model. The average
warming predicted by this model is 3.0 °C.
Prediction of Temperature increase
Prediction of Sea level rise
Regional details of Climate Change
Regional Climate modeling
• An RCM is a tool to add small-scale detailed information of
future climate change to the large-scale projections of a
GCM. RCMs are full climate models and as such are
physically based and represent most or all of the processes,
interactions and feedbacks between the climate system
components that are represented in GCMs.
• They take coarse resolution information from a GCM and
then develop temporally and spatially fine-scale information
consistent with this using their higher resolution
representation of the climate system.
• The typical resolution of an RCM is about 50 km in the
horizontal and GCMs are typically 500~300 km
RCM can simulate cyclones and
hurricanes
Regional Climate change modeling in
• PRECIS regional climate
modeling is now running
in Climate change study
cell at IWFM,BUET.
• Uses LBC data from
• LBC data available for
baseline, A2, B2, A1B
scenarios up to 2100.
• Predictions for every
hour. Needs more than
100 GB free space.
Domain used in PRECIS experiment
Topography of Experiment Domain
Simulation Domain = 88 x 88
Resolution = 0.44 degree
Predicted Change of Mean
Temperature (0C) using A1B
Baseline = 2000
2050 2090
Predicting Maximum Temperature
using A2 Scenarios
[Output of PRECIS model using SRES A2 scenario]
Predicting Minimum Temperature
using A2 Scenarios
[Output of PRECIS model using SRES A2 scenario]
Change of Mean Rainfall (mm/d)
using A1B Scenarios
Baseline = 2000
2050 2090
Predicting Rainfall using A2
Scenarios
[Output of PRECIS model using SRES A2 scenario]
Change of mean climatic variables of
Temperate (0C) Rainfall (mm/d)
Monthly Average Rainfall (mm/d)
Month 1990 2000 2010 2020 2030 2040 2050 2060 2070 2080 2090
January 2.61 0.34 0.03 0.03 0.42 0.99 1.24 0.21 0.12 1.66 1.02
February 0.61 0.55 1.38 1.01 1.24 1.88 0.45 1.10 0.53 1.61 0.76
March 2.42 1.02 4.82 3.04 1.87 3.07 0.99 3.62 2.84 1.27 3.59
April 5.84 1.38 11.46 5.99 2.82 7.84 11.41 6.60 8.39 8.74 3.66
May 10.03 5.59 10.36 6.42 11.92 18.16 33.47 16.53 29.47 11.29 11.96
June 17.06 7.90 14.79 13.59 10.84 21.48 12.87 12.93 7.24 10.04 11.70
July 7.20 9.07 7.97 8.13 7.32 11.26 5.62 10.26 10.31 6.33 9.98
August 7.39 5.46 5.11 3.92 9.79 6.67 7.46 13.60 10.65 9.13 9.59
September 4.49 6.71 5.47 7.83 7.51 8.82 10.29 10.80 10.52 8.18 7.48
October 5.68 1.48 4.16 2.76 6.16 3.11 1.89 3.94 2.55 8.84 7.58
November 0.14 0.16 0.41 0.91 0.03 0.73 0.08 1.91 0.27 1.23 0.51
December 0.14 0.06 0.10 0.26 0.06 0.18 1.09 0.04 0.13 0.32 0.03
Monthly Average Temperature (0C)
Month 1990 2000 2010 2020 2030 2040 2050 2060 2070 2080 2090
January 14.74 15.08 14.63 15.94 15.66 17.66 19.52 16.49 17.68 21.55 20.88
February 14.27 21.18 20.18 22.36 20.61 20.65 23.14 25.37 24.50 23.00 23.32
March 24.25 26.34 25.68 25.66 28.82 26.70 29.23 29.04 29.71 28.53 28.84
April 27.95 32.36 29.10 31.28 34.07 31.96 31.29 32.64 32.81 31.53 34.52
May 29.51 32.11 32.16 33.17 31.97 32.37 29.31 32.00 32.59 33.88 35.62
June 29.18 31.42 30.66 31.44 30.82 31.56 31.94 31.18 37.24 34.80 35.07
July 28.59 28.23 28.88 28.99 29.35 30.28 30.58 30.45 31.03 31.76 30.44
August 28.19 28.24 29.06 29.65 28.62 30.34 30.26 29.31 30.12 29.93 30.09
September 28.02 27.29 28.65 28.11 28.58 30.72 29.07 29.79 30.72 29.01 29.87
October 25.24 25.21 27.10 27.29 26.14 28.48 28.22 29.25 29.72 27.82 29.09
November 19.44 20.20 21.03 20.52 21.06 23.21 22.64 22.04 23.76 25.52 26.30
December 14.48 17.37 17.86 18.53 16.24 18.85 19.99 18.26 19.36 20.90 20.80
Trends of Temperature of
Max. Temp. = 0.63 0C/100 year Min. Temp. = 1.37 0C/100 year
Trends of Maximum Temperature Trends of Minimum Temperature
31.4 22
31.2 y = 0.0063x + 17.855 21.8 y = 0.0137x - 6.0268
31 21.6
30.8 21.4
30.6 21.2
30.4 21
30.2 20.8
30 20.6
29.8 20.4
29.6 20.2
29.4 20
1948
1951
1954
1957
1960
1963
1966
1969
1972
1975
1978
1981
1984
1987
1990
1993
1996
1999
2002
2005
2008
1948
1951
1954
1957
1960
1963
1966
1969
1972
1975
1978
1981
1984
1987
1990
1993
1996
1999
2002
2005
2008
Spatial Distribution of Trends of
Temperature (1947-2007)
Maximum Temperature Minimum Temperature
Maximum increase: 0.0581 at Shitakunda Maximum increase: 0.0404 at Bogra
Minimum increase: -0.026 at Rangpur Minimum increase: -0.023 at Tangail
Conclusions
Analysis of the historic data (1948-2007) shows that
daily maximum and minimum temperature has been
increased with a rate of 0.63 0C and 1.37 0C per 100
years respectively.
PRECIS simulation for Bangladesh using A1B climate
change scenarios showed that mean temperature will be
increased at a constant rate 40C per 100 year from the
base line year 2000.
On the other hand, mean rainfall will be increased by
4mm/d in 2050 and then decreased by 2.5mm/d in
2100 from base line year 2000.
Recommendations
• In future, Climate change predictions will be
generated in more finer spatial scale(~25km).
• PRECIS model will be simulated with other
Boundary condition data such as ECHAM5 using
A1B scenarios.
• Results will be compared with other regional
climate models such as RegCM3 etc.
Climate Change Study Cell, BUET
http://teacher.buet.ac.bd/diriwfm/climate/
Thank you
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# If 4<(7-x)/3, which of the following must be true?
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Intern
Joined: 23 Mar 2014
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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04 Sep 2015, 21:06
mn2010 wrote:
If 4<(7-x)/3, which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am confused about statement II ????
According to II, x can be 2 or -10. But according to the given question x<-5. Hence II can not be true always.
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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23 Sep 2015, 08:27
got this question correct, but selected wrong option as statement D is partially correct .
Manager
Joined: 10 May 2014
Posts: 135
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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02 Jan 2016, 14:15
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
Hi Bunuel,
Statements I and III are perfectly clear. But let me ask you a question about statement II to clear all my doubts, if you don´t mind.
Question stem states that x < -5 and then asks "if this is true, then what else must be true?"
Statement II gives us 2 options. Case A: x > -1 OR Case B: x < -5. Since the question stem already stated that x < -5, then Case A cannot be true (since x cannot be less than -5 and bigger than -1 at the same time) and Case B must be true. Therefore, Statement II must be true.
Is this reasoning correct?
Thank you so much!
Math Expert
Joined: 02 Sep 2009
Posts: 61258
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
### Show Tags
03 Jan 2016, 10:16
minwoswoh wrote:
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
Hi Bunuel,
Statements I and III are perfectly clear. But let me ask you a question about statement II to clear all my doubts, if you don´t mind.
Question stem states that x < -5 and then asks "if this is true, then what else must be true?"
Statement II gives us 2 options. Case A: x > -1 OR Case B: x < -5. Since the question stem already stated that x < -5, then Case A cannot be true (since x cannot be less than -5 and bigger than -1 at the same time) and Case B must be true. Therefore, Statement II must be true.
Is this reasoning correct?
Thank you so much!
Yes. Basically we are given that x<-5 and then asked whether |x+3|>2 is true. We know that if x < -5, then |x+3|>2 must be true.
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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20 Mar 2016, 13:54
try to plug in any number less than -5 and check because that's the condition given in question.
x < -5 means x can be -6,-7,-8 ....
|x+3|>2
when x= -6 => |-6+3|>2 true
when x=-100 => |-100+3|>2 true
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Joined: 13 Mar 2011
Posts: 21
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
### Show Tags
01 Apr 2016, 01:17
Bunuel wrote:
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
Yes. Basically we are given that x<-5 and then asked whether |x+3|>2 is true. We know that if x < -5, then |x+3|>2 must be true.
Chiragjordan wrote:
mn2010 wrote:
If 4<(7-x)/3, which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
Hey Everyone I am facing majo issues with this one ..
the second statement specifies X>1 or X<-5
now x can be 100 too hence it will make the inequality insufficient...
I am rooting for B only
Someone i the thread mentioned X<-5 is given right... but what if X is 100 ..
i have every statement in this thread and still believe the answer is B ..
Can anyone tell me where am i doing wrong
P.S => DON'T tell me X<-5 IS GIVEN
Bunuel, Sorry that bother you but, in my humble opinion, it seems that Chiragjordan is correct. II is not true.
A simple check can prove it :
From II we have that Case A: x > -1 OR Case B: x < -5
From Case A let's take x = 0
4<[(7-x)/3] => 4<[(7-0)/3] - is not true => Case A => is not working
Because problem's type is MUST BE not COULD BE => the answer is B ( only III MUST BE true ) .
Bunuel, please, could you confirm ?
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
### Show Tags
11 Apr 2016, 23:31
mn2010 wrote:
If 4<(7-x)/3, which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am confused about statement II ????
Simplify 4<(7-x)/3 and you get x < -5. Any of the statement says it would be true.
(I) 5 < X i.e. X > 5 NOT true.
(II) |x+3|>2 - range for this is -5 < X < 5. True.
(III) -(x+5) i.e. -x-5 as x is less than -5 the values it would take will be -6, -7 and so on. in all these cases -(x+5) will be positive. True
Hence D is the correct answer.
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Posts: 28
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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01 Nov 2016, 20:11
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
In II. |x+3|>2
If we have two cases where x>-1 and x<-5, then how can it MUST BE TRUE? Not 100% clear on this.
Thanks for Looking into this.
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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02 Jan 2017, 21:31
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
Regarding the option II shouldn't the condition x>-1 be satisfied ?
Math Expert
Joined: 02 Sep 2009
Posts: 61258
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
### Show Tags
03 Jan 2017, 01:02
Animatzer wrote:
Bunuel wrote:
mn2010 wrote:
If 4<[(7-x)/3], which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am not confused about statement II ????
Good question, +1.
Note that we are asked to determine which MUST be true, not could be true.
$$4<\frac{7-x}{3}$$ --> $$12<7-x$$ --> $$x<-5$$. So we know that $$x<-5$$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $$x<-5$$.
Basically the question asks: if $$x<-5$$ which of the following is true?
I. $$5<x$$ --> not true as $$x<-5$$.
II. $$|x+3|>2$$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $$x+3>2$$, so when $$x>-1$$ or 2. when $$-x-3>2$$, so when $$x<-5$$. We are given that second range is true ($$x<-5$$), so this inequality holds true.
Or another way: ANY $$x$$ from the range $$x<-5$$ (-5.1, -6, -7, ...) will make $$|x+3|>2$$ true, so as $$x<-5$$, then $$|x+3|>2$$ is always true.
III. $$-(x+5)>0$$ --> $$x<-5$$ --> true.
Hope it's clear.
Regarding the option II shouldn't the condition x>-1 be satisfied ?
Check here: if-4-7-x-3-which-of-the-following-must-be-true-99015.html#p853541
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Posts: 132
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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23 Mar 2017, 07:02
let's simpify from the beginning: 12<7-x
x<-5
let's test 1: doesn't work C,E out
let's test 2: |-6+3|>2 works fine
let's test 3: -(-6+5)= 1, hence, positive
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Posts: 10105
Location: Pune, India
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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05 Apr 2017, 09:24
mn2010 wrote:
If 4<(7-x)/3, which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am confused about statement II ????
Responding to a pm:
Quote:
In this question, I found that option two has two ranges. x > - or x < -5. So why this becomes sufficient? Don't we need only one clear answer to prove the sufficiency? If I have three ranges and one of them is good to prove the sufficiency, then also can we discard the other two?
You are given that x < -5. You obtain this because you are given that 4<(7-x)/3.
So x can take values such as -5.4, -6, -100, -54637 etc
Which of the following MUST BE TRUE?
II. |x+3|>2
Gives x < -5 OR x > -1
For every value that x can take, is this statement true? Yes. For every value that x can take it is "'less than 5" or "more than -1".
All our values are actually "less than -5". We need x to satisfy either "x < -5" or "x > -1".
_________________
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Math Expert
Joined: 02 Sep 2009
Posts: 61258
Re: If 4<(7-x)/3, which of the following must be true? [#permalink]
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29 May 2017, 21:50
mn2010 wrote:
If 4<(7-x)/3, which of the following must be true?
I. 5<x
II. |x+3|>2
III. -(x+5) is positive
A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
I am confused about statement II ????
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-4-7-x-3-w ... 68681.html
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-4-7-x-3-w ... 68681.html
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Re: If 4<(7-x)/3, which of the following must be true? [#permalink] 29 May 2017, 21:50
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## Are there 364 or 365 days in a year?
In the Julian calendar, the average (mean) length of a year is 365.25 days. In a non-leap year, there are 365 days, in a leap year there are 366 days. A leap year occurs every fourth year, or leap year, during which a leap day is intercalated into the month of February. The name “Leap Day” is applied to the added day.
## How many days are in a year 2021?
A calendar year consists of 52 weeks, 365 days in total. 2021 marks the beginning of a new decade. 2021 began on Friday, January 1 and will end on Friday, December 31, 2021.
## Are there exactly 365 days in a year?
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Remember that there are 8760 hours in total in the year 2021. So if you’re full-time you will be working, or at least at work about 22.84% of the time.
## What is the 100th day of 2021?
And since the 100th day of 2021 falls on April 10, it’s the perfect time to start emerging from our winter hibernation and enjoy some spring activities, while recognizing all that we’ve accomplished in this first 100 days of the year.
## How many work days were there in 2021?
There are a total of 261 working days in the 2021 calendar year.
## Who decided 365 days a year?
To solve this problem the Egyptians invented a schematized civil year of 365 days divided into three seasons, each of which consisted of four months of 30 days each. To complete the year, five intercalary days were added at its end, so that the 12 months were equal to 360 days plus five extra days.
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## Why does 1 year have 365 days?
A year is the amount of time it takes a planet to orbit its star one time. It takes Earth approximately 365 days and 6 hours to orbit the Sun. It takes Earth approximately 24 hours — 1 day — to rotate on its axis. So, our year is not an exact number of days.
## How much is a year in months?
Answer: There are 12 months in a year. There are 4 months in a year that have 30 days.
## What is the actual year of the earth?
The current year by the Gregorian calendar, AD 2021, is 12021 HE in the Holocene calendar. The HE scheme was first proposed by Cesare Emiliani in 1993 (11993 HE).
## Who decided 12 months in a year?
Julius Caesar’s astronomers explained the need for 12 months in a year and the addition of a leap year to synchronize with the seasons. At the time, there were only ten months in the calendar, while there are just over 12 lunar cycles in a year.
## How many hours exactly is a day?
Modern timekeeping defines a day as the sum of 24 hours —but that is not entirely correct. The Earth’s rotation is not constant, so in terms of solar time, most days are a little longer or shorter than that. The Moon is—very gradually—slowing the Earth’s rotation because of friction produced by tides.
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# Bee algorithm
## 1 Algorithm
The Bee Algorithm was mathematically first described relatively recently. It is one of the representatives of a large family of algorithms that allow modeling swarm intelligence. This article will provide an example of the application of the Bee Algorithm to search for the global extremum of the function. The two-dimensional Schwefel function, having a large number of local minima, and the Rosenbrock function, whose global minimum lies in a narrow, parabolic valley, were chosen as the target functions.
### 1.1 Intuition
A colony of honey bees can spread over long distances (more than 10 km) and in several directions, while using a huge number of food sources. A colony can only be successful by deploying its foragers to good fields. The main idea is that flower fields that provide large amounts of nectar or pollen that are easy to collect with less energy consumption should be visited by more bees, whereas areas with less nectar or pollen should receive fewer bees.
The search for food begins with the sending of scout bees in search of honey flower fields. Scout bees search randomly through their journey from one patch to another. Also throughout the harvest season, the colony continues its research, keeping a percentage of the entire population as bee scouts.
When the bees return to the hive, those who found a source which is above a certain threshold (a combination of some constituents, such as sugar percentage regarding the source) deposit their nectar or pollen and go to the dance floor to perform their waggle dance. This mysterious dance is essential for colony communication and contains three vital pieces of information about flower spots: direction, distance, and source quality.
The nectar search process is described in more detail here.
### 1.2 Mathematical interpretation
And now imagine that the location of the global extremum is the site where the most nectar, and this site is the only one, that is, in other places there is nectar, but less. And bees do not live on a plane, where it is enough to know two coordinates to determine the location of sites, but in a multidimensional space, where each coordinate represents one parameter of a function that needs to be optimized. The amount of nectar found is the value of the target function at this point.
The list below shows the pseudocode for a simple Bee Algorithm.
1. Initialize the set of parameters: number of scout bees - n, number of elite bees - e, number of selected regions out of n points - m, number of recruited around elite regions - nep, number of recruited around other selected (m-e) regions - nsp, and stopping criteria.
2. Every bee evaluates the value of target function
3. While (stopping criteria not met): //Forming new population
1. Elite bees (e) that have better fitness are selected and saved for the next population
2. Select sites for neighbourhood search (m-e)
3. Recruit bees around selected sites and evaluate fitness. More bees will be recruited around elite points(nep) and fewer bees will be recruited around the remaining selected points(nsp).
4. Select the bee with the highest fitness from each site.
5. Assign remaining bees (n-m-e) to search randomly and evaluate their fitness.
4. End While
### 1.3 Examples
Two standard functions problems were selected to test the bee algorithm. Code implementation of The Bee Algorithm in Python is described here
The following parameters were set for this test:
• population n = 300
• number of elite bees e = 5
• selected sites m = 15
• bees round elite points nep = 30
• bees around selected points nsp = 10
• stopping criteria: max_iteration = 2000
A random point is selected from the definition area to initialize the algorithm.
#### 1.3.1 SCHWEFEL FUNCTION
The Schwefel function is complex, with many local minima. The plot shows the two-dimensional form of the function.
The function is usually evaluated on the hypercube x_i \in [-500, 500] for all i = 1, ..., d.
f(x_1 \cdots x_d) = 418.9829 \cdot d -\sum_{i=1}^d (x_i sin(\sqrt{|x_i|}))
The function has one global minimum:
f(x_1 \cdots x_d) = 0, \quad x_i = 420.9687
The plot below shows drop of the objective function averaged over 100 runs of the algorithm can be observed in the following graph.
#### 1.3.2 ROSENBROCK FUNCTION
The Rosenbrock function, also referred to as the Valley or Banana function, is a popular test problem for gradient-based optimization algorithms. It is shown in the plot below in its two-dimensional form.
The function is unimodal, and the global minimum lies in a narrow, parabolic valley. However, even though this valley is easy to find, convergence to the minimum is difficult.
The function is usually evaluated on the hypercube x_i \in [-5, 10] for all i = 1, ..., d, although it may be restricted to the hypercube x_i \in [-2.048, 2.048] for all i = 1, ..., d.
f(x_1 \cdots x_d) = \sum_{i=1}^{d-1} (100(x_i^2 - x_{i+1})^2 + (1-x_i)^2) \\-2.048 \leq x_i \leq 2.048
The function global minimum:
f(x_1 \cdots x_d) = 0, \quad x_i = 1
To test the algorithm the four-dimensional Rosenbrock function was chosen. The fall of the objective function averaged over 100 runs of the algorithm can be observed in the following graph.
Moreover, the number of iterations required to find the point at which the value of the target function differs from the optimal one by no more than 0.2%(new stopping criteria) has also been tested for some dimensions of the Rosenbrock function. One can see the results at the chart below.
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Re: Reposted, Reformatted Re: "mapping" functions over lists, again!
• To: mathgroup at smc.vnet.net
• Subject: [mg96343] Re: Reposted, Reformatted Re: "mapping" functions over lists, again!
• From: dh <dh at metrohm.com>
• Date: Thu, 12 Feb 2009 06:33:56 -0500 (EST)
• References: <gmu8ik\$geb\$1@smc.vnet.net>
```
Hi Paul,
assume you have some function fun of 2 variables (e.g.fun[r_,c]=r+ Im[c]
) then we can map this by:
Map[{#[[1]], fun @@ #} &, shortTestdata, {2}]
hope this helps, Daniel
Paul Ellsmore wrote:
> This is a repost of an earlier post, as there were problems with the =
> email
> formatting for some people. I have composed this one in Notepad, and cut =
> and
> pasted to Outlook. Hope it works better.
>
> Thanks to all who have already given me some advice, but I still haven't
> quite got it yet!
>
> My data is in the form:
>
> shortTestdata = {{{40., 28.06776 + 1.208548*I}, {43.094, 28.05721 +
> 1.298519*I}, {46.428, 28.05527 + 1.400228*I}, {50.019, 28.05509 +
> 1.508759*I},
> {53.888, 28.05729 + 1.624517*I}, {58.057, 28.05651 + 1.75026*I}}, =
> {{40.,
> 7.42169 + 0.2198589*I}, {43.094, 7.408397 + 0.2343525*I},
> {46.428, 7.403769 + 0.2521353*I}, {50.019, 7.401313 + 0.2715986*I},
> {53.888, 7.400478 + 0.2920617*I}, {58.057, 7.39994 + 0.3145005*I}},
> {{40., 1685.526 + 0.04809976*I}, {43.094, 1694.691 - 0.09133625*I},
> {46.428, 1698.265 - 0.02731824*I}, {50.019, 1699.761 - 0.0491538*I},
> {53.888, 1700.523 - 0.2179222*I}, {58.057, 1701.162 - 0.2423136*I}},
> {{40., 1808.702 - 0.006286621*I}, {43.094, 1808.524 - 0.1140757*I},
> {46.428, 1808.534 - 0.02445889*I}, {50.019, 1808.443 - 0.1061664*I},
> {53.888, 1808.481 - 0.1762974*I}, {58.057, 1808.631 - 0.2894506*I}}}
>
> This is a list of lists, the lowest level lists containing pairs of =
> {real,
> complex}. The individual lists are not all the same length, and the =
> total
> number of lists can vary, and I need to preserve the list structure.
>
> I want to "map" functions across all the lists, to convert the data =
> pairs to
> {real, f(real,complex)}.
>
> One suggestion was to use a Rule in Cases:
>
> Cases[shortTestdata, {r_Real, c_Complex} :> {r, Re[c]}], but when =
> applied to
> the data above it Flattens my list structure:
>
> In: realpart = Cases[shortTestdata, {r_Real, c_Complex} :> {r, Re[c]}, =
> 2]
>
> Out: {{40., 28.06776}, {43.094, 28.05721}, {46.428, 28.05527}, {50.019,
> 28.05509}, {53.888, 28.05729}, {58.057, 28.05651}, {40., 7.42169}, =
> {43.094,
> 7.408397},
> {46.428, 7.403769}, {50.019, 7.401313}, {53.888, 7.400478}, {58.057,
> 7.39994}, {40., 1685.526}, {43.094, 1694.691}, {46.428, 1698.265}, =
> {50.019,
> 1699.761},
> {53.888, 1700.523}, {58.057, 1701.162}, {40., 1808.702}, {43.094,
> 1808.524}, {46.428, 1808.534}, {50.019, 1808.443}, {53.888, 1808.481},
> {58.057, 1808.631}}
>
> I can ressurect the list structure by checking the length of every list =
> in
> the data, and using these lengths in a Partition statement, but I'd =
> rather
> not lose the list structure in the first place. Is there a way to do =
> that?
>
> I am sure I could use Map in some way:
>
> realpart=Map[fxn, shortTestdata,2]
>
> but I have no real idea how to set up fxn to do what I want. I have =
> tried:
> In: Map[Cases[#_,{r_Real,c_Complex}->{r,Re[c]}&,shortTestdata,2] but I =
> get a
> list of empty lists out:
> Out:{{},{},{},{}}
>
> Another suggestion was to use First[#], Last[#], so:
>
> In: realpart=Map[{First[#],Re[Last[#]]}&,shortTestdata,2] but this =
> takes
> first and last of the level 2 list:
> Out: {{{40., 28.06776}, {58.057, 28.05651}}, {{40., 7.42169}, {58.057,
> 7.39994}}, {{40., 1685.526}, {58.057, 1701.162}}, {{40., 1808.702}, =
> {58.057,
> 1808.631}}}
>
> If I Map at level 3 I get an error message. Why doesn't this work? =
> Surely at
> Level 3, each element is a list of length 2, so First would be the real =
> and
> Last would be the complex?
>
> My basic problem is that I don't know how to structure a fxn to be =
> Mapped
> over my lists, so that it applies different transformations to each =
> element
> of my data pairs.
>
> So I think the most succinct way of expressing my problem is, what form =
> does
> fxn take if I want to Map it across my lists of {real,complex} so that =
> it
> returns {fxn1[real],fxn2[complex]} or even {real,fxn[complex]}?
>
> Apologies if this is either a trivial question, or a nonsense question.
> Ultimately, I think I can make Cases and Partition work for me, but I =
> feel
> sure there is a more elegant way, if only I understood the Mathematica
> syntax better.
>
> Thanks again,
>
> Cheers,
>
> Paul.
>
>
>
> Dr. Paul A. Ellsmore
>
> Nanion Limited
> Oxford Centre for Innovation
> Mill Street
> Oxford
> United Kingdom
> OX2 0JX
>
> Tel: +44 (0) 1865 811175
> Fax: +44 (0) 1865 248594
>
>
>
```
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0
# What is 3.4 percent in decimal form?
Wiki User
2010-12-27 18:08:33
3.4% in decimal form is .034.
Remember - to change a percentage to decimal move the decimal point TWO (2) places to left. To change a decimal to a percentage - move the decimal point TWO (2) places to the right.
Wiki User
2010-12-27 18:08:33
Study guides
20 cards
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See all cards
3.8
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## "Dynamic Compound Convex" Questions
Eternl Knight
Posts: 58
Joined: Sun Jan 22, 2006 4:31 am
### "Dynamic Compound Convex" Questions
I am working on implementing a volumetric approach to the height-field collision shape so that fast moving objects are detected as colliding with the terrain even if they pass through thin shell of triangles making up it's surface.
I am making a new shape class for this as the existing height-field will work just fine when using continuous collision detection algorithms. This is a problem primarily with the Discrete Collision Detection code path.
Anyhow, the idea is relatively simple (and will hopefully tie easily into a "water/bouyancy" class later). Instead of generating just the surface triangles, I take the points for the triangle and a copy of these points translated "down" by a configurable margin and make a btConvexHullShape for the results. So in effect there would be a triangular "column" in the new height-field object per surface triangle in the current height-field object.
The problem is that this is not easy to "hack" into the current Bullet architecture. There are really only two ways of handling multiple "collision shapes" at runtime based on the bounds of the colliding object. There is the current method used by the height-field (i.e. generates the triangles for collision based on the bounds, callback derived from btConcaveShape) or the btCompoundShape which requires that all shapes be added to it before processing the collision.
Hence a naïve method would be to simply create a btConvexHullShape at the start (making the terrain non-modifiable and negating the primary benefit of it being stored as a 2D height-field in the first place) or to create a new "compound shape" that enables the creation of child shapes on the fly as required.
The question coming from all this is to ask people more knowledgable than me as to whether this should be done by creating a new compund shape or by altering the existing one to allow dynamic creation/return of convex shapes as required by bounding box?
--EK
Erwin Coumans
Posts: 4232
Joined: Sun Jun 26, 2005 6:43 pm
Location: California, USA
Contact:
### Re: "Dynamic Compound Convex" Questions
Eternl Knight wrote: The problem is that this is not easy to "hack" into the current Bullet architecture. There are really only two ways of handling multiple "collision shapes" at runtime based on the bounds of the colliding object. There is the current method used by the height-field (i.e. generates the triangles for collision based on the bounds, callback derived from btConcaveShape) or the btCompoundShape which requires that all shapes be added to it before processing the collision.
For now, you could hack it in the btConvexConcaveCollisionAlgorithm. There is a place, where a temporary btTriangleShape is constructed.
Code: Select all
``````void btConvexTriangleCallback::processTriangle(btVector3* triangle,int partId, int triangleIndex)
``````
Instead of such triangle, create a temporarily btConvexHullShape with the points. This would be the best place to start. You can copy/paste the collision algorithm, and register it specially for a heightfield interactions, in the btDefaultCollisionConfiguration.
I will make modifications in the future that makes this easier.
Hope this helps,
Erwin
Erin Catto
Posts: 324
Joined: Fri Jul 01, 2005 5:29 am
Location: Irvine
Contact:
### Re: "Dynamic Compound Convex" Questions
If you make those columns tall, you might get objects wedged between them with only horizontal contact normals. The object will then ping-pong back-and-forth falling into the abyss.
Perhaps if you detect deep penetration, you could just generate sloppy contact points to push the object up and out.
Erwin Coumans
Posts: 4232
Joined: Sun Jun 26, 2005 6:43 pm
Location: California, USA
Contact:
### Re: "Dynamic Compound Convex" Questions
Erin Catto wrote:If you make those columns tall, you might get objects wedged between them with only horizontal contact normals. The object will then ping-pong back-and-forth falling into the abyss.
Well, you could use overlapping truncated pyramids instead of colums, by enlarging the bottom. Then the resulting normal would always point upwards (as long as you make the pyramid large enough.
Perhaps if you detect deep penetration, you could just generate sloppy contact points to push the object up and out.
Yes, some heuristic would be doable. It would be good to get Eternl Knight up and running.
Thanks,
Erwin
Eternl Knight
Posts: 58
Joined: Sun Jan 22, 2006 4:31 am
### Re: "Dynamic Compound Convex" Questions
Erin Catto wrote:If you make those columns tall, you might get objects wedged between them with only horizontal contact normals. The object will then ping-pong back-and-forth falling into the abyss.
I have thought of this and is something I have discussed with Erwin (briefly) in the beginning. For any face but the top triangle you need a specialized contact normal pointing in the direction of the height-field's "up" direction. How to do this is something I am not sure of.
Erwin Coumans wrote:Well, you could use overlapping truncated pyramids instead of colums, by enlarging the bottom. Then the resulting normal would always point upwards (as long as you make the pyramid large enough.
The problem with this solution is that in areas of steep changes in height, the truncated pyramids may have faces "above" the actual terrain - causing spurious contacts where there should be none. Hence the single column design. I have a feeling that changing the direction of the contact normal is easier than creating/checking the pyramids for intersection with the terrain surface.
--EK
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Using only those 2 fuses and the lighter, how would you measure a period of exactly 45 minutes?
When you first think about this puzzle (it may even be called a riddle), it may seem impossible to find a solution. But, there must be an answer since someone is asking this question. What do we have and what do we know? We have 2 fuses, a lighter, and we know each fuse takes one hour to burn. It seems like all we can do is measure one hour, since we can’t make any other assumptions. So, this means we must really try to think creatively, because there is something here that we are missing.
Thinking “outside the box”
What if we start the flame from the middle of a fuse? Would that allow us to measure half an hour? Actually, no it would not, because we can not assume that half of the length of the fuse would burn in half an hour – it may just be that 1/10th of the length of the fuse take 50 minutes to burn and the other 9/10ths of the fuse takes 10 minutes to burn. So, starting the flame in the middle of the fuse is not a valid option.
What would happen if we burn a fuse from both sides? That sounds interesting. Well, what would happen is that the fuse would burn out in just 30 minutes – this is because the fuse would be burning from both sides and the flames would burn until they meet each other and extinguish after exactly 30 minutes. Well, that sounds very useful – because we can now measure 30 minutes!
We made an extra assumption here – that the fuses burn at a uniform rate
We want to make a little side note here: we are assuming that the burn rate is uniform – meaning that burning 1/4th of the fuse from one end will take exactly the same amount of time as burning 1/4th of the fuse from the other end of the fuse. It could potentially take 1 minute to burn the first 9/10th of the rope and 59 minutes to burn just the last 1/10th of the rope if the burn rate were not uniform. This was not an assumption stated in the problem, but it is important, and if you ever do encounter this question in an interview it is an assumption you should probably make whether the interviewer states it or not (some interviewers are bad enough to just ask this question without really understanding) because there really would be no solution to this problem if we did not assume the burn rate was uniform. Anyways, that is our little side note, please carry on and read the entire solution below.
Now that we can measure 30 minutes, how could we measure 15 minutes more to get 45 minutes total? Well, can we use the idea of burning a fuse from both sides to measure that extra 15 minutes? That sounds like it has potential – what if we burn fuse # 1 from both ends, and we burn fuse #2 from only one end. Then, after 30 minutes has passed, we can burn the other end of fuse #2. Fuse #2 would finish burning in 15 minutes because it has already has 30 minutes worth of time burned from it, but it is also burning from both ends – so that cuts the burning time in half. And 30 + 15 would give us 45 minutes – so we finally have an answer!
2 Ropes 45 Minutes
You may also see a variation of this puzzle being asked with ropes instead of fuses. But, keep in mind that the answer is exactly the same as the one we gave above whether you are dealing with ropes or fuses.
Subscribe to our newsletter for more free interview questions.
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# How to imagine 52 factorial
I remember being fascinated by a description of eternity in "The Shepard Boy," from the Brothers Grimm:
In lower pomerania is the Diamond mountain, which is two miles high, two miles wide, and two miles deep. Every hundred years a little bird comes and sharpens its beak on it, and when the whole mountain is worn away by this, then the first second of eternity will be over.
Similarly, Scott Czepiel has a great essay on imagine the immensity of 52!, or 80658175170943878571660636856403766975289505440883277824000000000000, which is the number of ways an ordinary deck of cards can be shuffled:
This number is beyond astronomically large. I say beyond astronomically large because most numbers that we already consider to be astronomically large are mere infinitesimal fractions of this number. So, just how large is it? Let's try to wrap our puny human brains around the magnitude of this number with a fun little theoretical exercise. Start a timer that will count down the number of seconds from 52! to 0. We're going to see how much fun we can have before the timer counts down all the way.
Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you've emptied the ocean.
Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven't even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won't do it. There are still more than 5.385e67 seconds remaining. You're just about a third of the way done.
To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you've filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you've leveled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt. Exercise for the reader: at what point exactly would the timer reach zero?
Michael Stevens of Vsauce made a good YouTube video visualization of Czepiel's essay (above).
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### Dirichlet-multinomial model for Skittle proportions
Last year I came across a blog post describing how the author collected count data from 468 packs of Skittles. (It's a great blog post, definitely worth reading.) For each bag, the author counted up how many of each color were present.
Using data from the Github repo for the project, we can fit a Bayesian model to see what the underlying proportions of each color might be.
In [1]:
%matplotlib inline
from matplotlib import pyplot as plt
import seaborn as sns
import numpy as np
import pandas as pd
import warnings
warnings.filterwarnings("ignore")
skittles_palette = ["red", "orange", "yellow", "green", "purple"]
df = pd.read_csv("../data/skittles.txt", sep="\t").drop("Uncounted", axis=1, errors="ignore")
Out[1]:
Strawberry Orange Lemon Apple Grape
0 10 15 11 7 18
1 5 12 17 15 10
2 16 11 15 11 9
3 15 8 13 16 7
4 11 14 20 8 7
5 10 11 11 17 11
6 9 17 6 17 14
7 14 12 13 11 12
8 11 11 14 10 17
9 12 6 18 11 12
In [2]:
df.shape
Out[2]:
(468, 5)
We can also normalize each row by its sum (the total number of skittles in that bag) to show the proportion of each color in the bag:
In [3]:
counts = df.sum(axis="columns")
normalized = df.div(counts, axis="rows")
Out[3]:
Strawberry Orange Lemon Apple Grape
0 0.163934 0.245902 0.180328 0.114754 0.295082
1 0.084746 0.203390 0.288136 0.254237 0.169492
2 0.258065 0.177419 0.241935 0.177419 0.145161
3 0.254237 0.135593 0.220339 0.271186 0.118644
4 0.183333 0.233333 0.333333 0.133333 0.116667
### Exploratory analysis¶
Now that we have the data, we can ask a few simple questions just with descriptive statistics and visualizations. Here's an attempt at a quick summary of the whole data set all in one picture:
In [4]:
def summary_plot():
fig, ax = plt.subplots(figsize=(10, 20))
# plot the mean for each color
for i, cumulative_mean in enumerate(df.mean(axis="rows").cumsum().values):
color = df.columns[i]
label = f"mean={df[color].mean():0.2f}"
plt.axvline(cumulative_mean, c=skittles_palette[i], alpha=0.5, lw=2, label=label)
# draw stacked bars
_sorted_index = df.sum(axis=1).argsort()[::-1]
df.loc[_sorted_index].plot(kind="barh", stacked=True, ax=ax, color=skittles_palette)
# plot the overall mean
_overall_mean = df.sum(axis="columns").mean()
_jitter = 0.2
plt.axvline(
_overall_mean + _jitter,
c=skittles_palette[-1],
alpha=0.5,
ls="--",
label=f"mean total={_overall_mean:0.2f}"
)
ax.set_yticks([])
ax.xaxis.tick_top()
plt.tick_params(axis="x", top=True)
sns.despine()
plt.legend(loc="upper right")
plt.show()
summary_plot()
How many skittles are usually in a bag?
In [5]:
sns.distplot(counts, bins=31)
sns.despine()
plt.xlim(50, 70)
plt.show()
How many of each color are usually in a bag?
In [6]:
def plot_counts(axs):
ax = axs[0]
sns.boxplot(
data=pd.melt(df, var_name="color", value_name="count"),
x="color",
y="count",
palette=skittles_palette,
ax=ax
)
sns.despine()
ax.set_title("counts per bag")
ax = axs[1]
for i, color in enumerate(df.columns):
sns.distplot(df[color], color=skittles_palette[i], ax=ax, hist_kws={"alpha": 0.1})
sns.despine()
ax.set_xlabel("count")
fig, axs = plt.subplots(ncols=2, figsize=(14, 4))
plot_counts(axs)
plt.show()
In [7]:
def plot_proportions(axs):
ax = axs[0]
# counts
sns.boxplot(
data=normalized.melt(var_name="color", value_name="proportion"),
x="color",
y="proportion",
palette=skittles_palette,
ax=axs[0]
)
ax.set_ylim(0, 0.4)
sns.despine(ax=ax)
ax.set_title("proportions per bag")
ax = axs[1]
for i, color in enumerate(normalized.columns):
sns.distplot(normalized[color], color=skittles_palette[i], ax=ax, hist_kws={"alpha": 0.1})
sns.despine()
ax.set_xlim(0, 0.4)
ax.set_xlabel("proportion")
fig, axs = plt.subplots(ncols=2, figsize=(14, 4))
plot_proportions(axs)
plt.show()
These graphs give a descriptive feel for the data, but we can't yet draw more general conclusions. That's a job for statistical inference.
### Inference: what are the true proportions for each color?¶
We can assume that the manufacturer is motivated to control the average number of Skittles per bag for cost and packing reasons. What is less clear is whether they exert any intentionality over the proportion of colors.
One thing we might want to look into: are the proportions for each color in a bag supposed to be the same or are any flavors purposely favored or disfavored? And if the differences are intentional, what are the intended proportions for each color?
Eyeballing the plot above, we might hypothesize that the machines are trying to get about 20% of each color and that fluctuations are due to machinery imprecision. For example, it is possible that it is solely due to random fluctuations that we see slightly more lemon and and slightly less apple. The means give us our maximum likelihood estimate (MLE) for the "true" proportions:
In [8]:
normalized.mean(axis="rows")
Out[8]:
Strawberry 0.201242
Orange 0.198415
Lemon 0.204962
Apple 0.191098
Grape 0.204283
dtype: float64
The MLE converges on the true population values given infinite trials, but how do we know how different the real population was given we have observed only $N=468$ trials?
The frequentist attempt to answer the question might use a null hypothesis significance test (NHST) where the null hypothesis is that the proportions are the same. For this scenario the standard test would be a one-way ANOVA, which is sort of like a Student's t-test generalized to more than two groups:
In [9]:
import statsmodels.api as sm
from statsmodels.formula.api import ols
melted_proportions = normalized.melt(var_name="color", value_name="proportion")
model = ols("proportion ~ color", data=melted_proportions).fit()
sm.stats.anova_lm(model, typ="II")
Out[9]:
sum_sq df F PR(>F)
color 0.059094 4.0 4.967424 0.000548
Residual 6.944487 2335.0 NaN NaN
This p-value of $0.0005$ is very small, telling us that if the null hypothesis were true then we would be very unlikely to see data with differences in means (as captured by the F statistic) as extreme or more extreme than what we actually saw in our sample. Since $p < 0.05$ or some other small-ish number we would reject the null hypothesis and conclude that the intended proportions of each color are probably not the same.
• It doesn't really tell us what we want to know. The test only tells us that not all of the proportions are exactly the same. It doesn't tell us which ones are different and by how much. If you wanted to know about individual differences you would have to do a bunch of pairwise comparisons which is a recipe for falsely finding significance by random chance.
• It doesn't quantify our remaining uncertainty about the estimates. When we take the mean of each color, that MLE is a point estimate: one single number that represents the whole group. We know that this estimate is subject to randomness, so how much plausibility do we allocate to higher or lower values?
Both of these drawbacks are typical motivations for fitting a Bayesian model that more fully characterizes the data generating process.
### Choosing the Dirichlet-multinomial¶
In order to build a Bayesian model, we need to come up with a believable data generating process, which is a story of how our data came to be. Here's one for the Skittles data:
• The factory decides on a proportion of Skittles $\mathbf{p} \in \mathbb R^5$ where each $p_j \geq 0$ and they sum up to 1.
• The factory makes a large number of Skittles in these proportions and then mixes them all up into a big hopper.
• For each bag $i$, the Skittles factory selects a number of Skittles $n$ to put in that bag.
• Then the factory draws $n$ Skittles from the big hopper and puts them into the bag, and it turns out that a count $k_j$ of each color is observed in the bag.
This corresponds almost exactly to the Dirichlet-multinomial model, where the Dirichlet is the prior distribution over proportions and the multinomial is the distribution for counts of each group. (This is the generalization of the Beta-Binomial model which does the same but for only two groups, e.g. heads versus tails on flips of a biased coin.) In particular, having proportional amounts of Skittles in a hopper ready to dispense corresponds directly to the Pólya urn interpretation with an extemely large $K$.
The Dirichlet gives us a distribution over simplexes (vectors that add up to one), and the multinomial gives us a distribution over an enumerated set of discrete outcomes. Together, we have a model for the proportions which we can condition on the observed data. Based on the question we're asking above, the parameter we are most interested in is the $p$ parameter of the multinomial which is sampled from the Dirichlet prior - or the idealized proportions for each color before they are probabilistically sampled into actual counts.
Here's how we could set this up in PyMC3:
In [10]:
import pymc3 as pm
import arviz as az
color_names = df.columns
with pm.Model() as model:
# make the Dirichlet an uninformative prior
alpha = np.ones(len(color_names))
# choose true proportions
p = pm.Dirichlet("p", a=alpha)
# choose a bag size
n = pm.DiscreteUniform("n", lower=40, upper=100, observed=df.sum(axis=1).values)
# choose how many of each color to put in that bag adding up to n based on proportions p
k = pm.Multinomial("k", n=n, p=p, observed=df)
trace = pm.sample(2500)
fit = az.from_pymc3(trace=trace, coords={"p_dim_0": color_names})
Auto-assigning NUTS sampler...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [p]
Sampling 4 chains, 0 divergences: 100%|██████████| 12000/12000 [00:03<00:00, 3632.16draws/s]
Setting the $\alpha$ parameter of the Dirichlet to ones makes it a uniform prior over simplexes, so it's uninformative here. In any case there's probably plenty of data to overwhelm whatever reasonable prior we would set.
#### Diagnostics¶
There were no divergences during sampling, so that is good. But before we draw any conclusions from the model, we should check to make sure the sampling went well. Having sampled 4 chains of 2,500 iterations each, we can use arviz to look at the summary:
In [11]:
summary_df = az.summary(fit).reset_index().assign(color_name=df.columns).set_index("color_name")
summary_df
Out[11]:
index mean sd hpd_3% hpd_97% mcse_mean mcse_sd ess_mean ess_sd ess_bulk ess_tail r_hat
color_name
Strawberry p[0] 0.201 0.002 0.197 0.206 0.0 0.0 14607.0 14586.0 14605.0 7130.0 1.0
Orange p[1] 0.198 0.002 0.194 0.203 0.0 0.0 14988.0 14968.0 15029.0 8035.0 1.0
Lemon p[2] 0.205 0.002 0.200 0.210 0.0 0.0 14070.0 14070.0 14045.0 7831.0 1.0
Apple p[3] 0.191 0.002 0.186 0.195 0.0 0.0 14890.0 14890.0 14878.0 8025.0 1.0
Grape p[4] 0.204 0.002 0.200 0.209 0.0 0.0 15598.0 15598.0 15600.0 6927.0 1.0
Here, we can see the summary stats for our estimates of each parameter. Especially of note:
• The Gelman-Rubin $\hat R$ statistic (r_hat) is less than 1.01 for each parameter indicating that we don't have signifcant divergence between chains.
• The tail effective sample size (ess_tail) is still in the high thousands meaning we have probably explored the parameter space pretty well.
We can also look at the traceplot for the MCMC sampling:
In [12]:
az.plot_trace(fit)
plt.show()
On the left, we see that each of the 4 chains generally agreed with all the others for each of the parameters. On the right, we see nice "fuzzy caterpillars" indicating that chains didn't get stuck in any proposals.
#### Getting our answers from the trace¶
Let's look at what our MCMC chains say about these parameters. We'll look at the 89% credible interval after McElreath, who in suggests 89% in "Statistical Rethinking" because (1) unlike 95% or other neat numbers we are used to seeing, 89% is a little weird so it reminds us that it's an arbitrary choice, and (2) it's prime making it slightly special, so why not.
In [13]:
az.plot_forest(fit, credible_interval=0.89)
plt.show()
This plot nicely shows the uncertainty around each maximum a posteriori (MAP) estimate with four lines per parameter, one per MCMC chain. They are all aligned on the same x axis this time, so we can clearly see which ones overlap and which ones do not.
As we can see, our credible interval for apple does not overlap with lemon or grape at all, which means we don't believe that we could have observed the data we collected if the underlying proportions were actually the same.
### A different data story¶
In the Dirichlet-multinomial model, we imagined a giant hopper where all the colors were mixed up and then sampled into a noisy bag size. We tried to learn $p$ while throwing away the distribution over bag sizes as a nuisance parameter.
The thing is, we might not be right about this — what if all the colors are kept separate, and the way the bag is filled is just 5 independent attempts to put roughly $μ_i$ in the bag from each color $i$ plus or minus machine error $\sigma$?
The nice thing about Bayesian models is that we can try to model this story directly:
In [15]:
with pm.Model() as model2:
# error, centered at zero with a large sigma to make this prior pretty uniformative
sd = pm.HalfNormal("σ", sigma=5.0)
# mu vector with 5 entries that we are trying to fit; use a Normal as the natural
# representation of an expected value with random errors
mu = pm.Normal(f"μ", mu=10, sigma=2, shape=df.shape[1])
for i, color in enumerate(df.columns):
# condition each color's mu with the observed counts for that color
n_i = pm.Normal(f"n_{color}", mu[i], sd, observed=df[color])
trace2 = pm.sample(2_500)
fit2 = az.from_pymc3(trace=trace2, coords={"μ_dim_0": color_names})
Auto-assigning NUTS sampler...
Multiprocess sampling (4 chains in 4 jobs)
NUTS: [μ, σ]
Sampling 4 chains, 0 divergences: 100%|██████████| 12000/12000 [00:04<00:00, 2964.88draws/s]
In [16]:
az.summary(fit2)
Out[16]:
mean sd hpd_3% hpd_97% mcse_mean mcse_sd ess_mean ess_sd ess_bulk ess_tail r_hat
μ[0] 11.919 0.152 11.632 12.201 0.001 0.001 12653.0 12651.0 12680.0 8350.0 1.0
μ[1] 11.739 0.151 11.455 12.022 0.001 0.001 11041.0 11041.0 11030.0 8261.0 1.0
μ[2] 12.141 0.150 11.857 12.416 0.001 0.001 12125.0 12120.0 12126.0 7961.0 1.0
μ[3] 11.320 0.152 11.037 11.608 0.001 0.001 13276.0 13276.0 13297.0 8659.0 1.0
μ[4] 12.100 0.150 11.824 12.398 0.001 0.001 13161.0 13158.0 13169.0 8024.0 1.0
σ 3.257 0.048 3.166 3.348 0.000 0.000 13040.0 13037.0 13025.0 8588.0 1.0
In [17]:
az.plot_trace(fit2)
plt.show()
In [20]:
az.plot_forest(fit2, var_names=["μ"], credible_interval=0.89)
plt.show()
Nicely, this result is pretty similar to the above model even though it's describing a much different story. It confirms that apple has a substantially different underlying probability.
### Conclusion¶
Most notably, we can conclude that apple really is less present than all the other colors by an average of one Skittle per bag.
In actual inference, we might want to compare these models using LOO or WAIC. But that's probably enough for Skittle modeling.
Any comments or suggestions? Let me know.
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https://community.powerbi.com/t5/Desktop/New-Measure-similar-to-SumIf/td-p/1263689
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Frequent Visitor
## New Measure similar to SumIf
I'm new to Power BI so this one would be easy I think. I'd like to show a calculation on "Total Commit" and "Total Pipeline" in our forecast. Data is organised like below
Region Opportunity Forecast Stage AU Opportunity 1 \$ 1,000 COMMIT AU Opportunity 2 \$ 2,000 COMMIT NZ Opportunity 3 \$ 3,000 FUNNEL AU Opportunity 4 \$ 4,000 COVER NZ Opportunity 5 \$ 5,000 COMMIT NZ Opportunity 6 \$ 6,000 FUNNEL
End Result
Total Commit = \$8,000 (ie stage = Commit)
Total Pipeline = \$21,000 (All stages)
I want to show in Matrix and Bar visual the 2 values together.
Total Commit Total Pipleine AU \$ 3,000 \$ 7,000 NZ \$ 5,000 \$ 14,000 \$ 8,000 \$ 21,000
1 ACCEPTED SOLUTION
Super User
Have two measures like
Total Commit = calculate(sum(Table[Forecast]), stage[Stage] ="COMMIT")
Total Pipeline =calculate(sum(Table[Forecast]), stage[Stage] <> "COMMIT")
Or have new column like
Stage Type =if(stage[Stage] ="COMMIT","COMMIT","Pipeline")
you can stage Type as column or legend (in Bar).
https://www.burningsuit.co.uk/blog/2019/04/7-secrets-of-the-matrix-visual/
Dashboard of My Blogs !! Connect on Linkedin
Want To Learn Power BI
Learn Power BI Beginners !! Advance Power BI Concepts !! Power BI For Tableau User !! Learn Power BI in Hindi !!
Proud to be a Super User!
!! Subscribe to my youtube Channel !!
2 REPLIES 2
Super User
Have two measures like
Total Commit = calculate(sum(Table[Forecast]), stage[Stage] ="COMMIT")
Total Pipeline =calculate(sum(Table[Forecast]), stage[Stage] <> "COMMIT")
Or have new column like
Stage Type =if(stage[Stage] ="COMMIT","COMMIT","Pipeline")
you can stage Type as column or legend (in Bar).
https://www.burningsuit.co.uk/blog/2019/04/7-secrets-of-the-matrix-visual/
Dashboard of My Blogs !! Connect on Linkedin
Want To Learn Power BI
Learn Power BI Beginners !! Advance Power BI Concepts !! Power BI For Tableau User !! Learn Power BI in Hindi !!
Proud to be a Super User!
!! Subscribe to my youtube Channel !!
Frequent Visitor
Thank you!
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# 7th Grade x y input output tables
I don't understand them please some one help me!!!!ASAP!!!
1. Ms. Sue
http://www.mathsisfun.com/sets/function.html
2. Anonymous
help me too
3. Amber Age 12
Okay take this problem for example :
x 8x y
0 8(0) 0
1 8(1) 8
2 8(2) 16
3 8(3) 24
You multiply the numbers in parenthesis
by 8.
## Similar Questions
1. ### programming
Can You Please help me understand how to do this and what i have to do to solve the problem.... Please!! Consider the following selection statement where X is an integer test score between 0 and 100. input X if (0 <= X and X < …
write a function rule for the input-output table input X 0 1 2 3 output Y 5 4 3 2
3. ### 7th grade input/output patterns
I need to see a completed example of input/output patterns at the 7th grade level. Thanks,
4. ### programming
Consider the following selection statement where X is an integer test score between 0 and 100. input X if (0<=X and X<49) Output "you fail" Else if (50<=X and X <70) Output "your grade is" X Output " you did OK" Else if …
I have a input and output machine . The input number is 5 and the output number 4. the second input number is 10 and the output is 5. I am having a hard time finding the operation. Can some please help thanks
6. ### 4th grade math input output tables
input x 1 2 4 5 6 8 9 output Y 7 9 13 _ _ _ _
7. ### 7th Grade x y input output tables
i need help with x and y input output tables for 7th grade i don't understand what 5x means or what to do! Math hw problems: 1. y=4x+3 x=1,2,5,10,20 2.d=3.5t t=1,2,5,10,20 Please help ASAP!!!!!
8. ### 7 th grade math
What is the rule for this function? input 1 output 3 input 2 output 5 Input 4 output 9 Input 5 output 11 Input 10 output 21 Input 12 output 25 Please help now thanks
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Mathbox for Scott Fenton < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > brbigcup Structured version Visualization version GIF version
Theorem brbigcup 33352
Description: Binary relation over Bigcup . (Contributed by Scott Fenton, 11-Apr-2012.)
Hypothesis
Ref Expression
brbigcup.1 𝐵 ∈ V
Assertion
Ref Expression
brbigcup (𝐴 Bigcup 𝐵 𝐴 = 𝐵)
Proof of Theorem brbigcup
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 relbigcup 33351 . . 3 Rel Bigcup
21brrelex1i 5601 . 2 (𝐴 Bigcup 𝐵𝐴 ∈ V)
3 brbigcup.1 . . . 4 𝐵 ∈ V
4 eleq1 2898 . . . 4 ( 𝐴 = 𝐵 → ( 𝐴 ∈ V ↔ 𝐵 ∈ V))
53, 4mpbiri 260 . . 3 ( 𝐴 = 𝐵 𝐴 ∈ V)
6 uniexb 7478 . . 3 (𝐴 ∈ V ↔ 𝐴 ∈ V)
75, 6sylibr 236 . 2 ( 𝐴 = 𝐵𝐴 ∈ V)
8 breq1 5060 . . 3 (𝑥 = 𝐴 → (𝑥 Bigcup 𝐵𝐴 Bigcup 𝐵))
9 unieq 4838 . . . 4 (𝑥 = 𝐴 𝑥 = 𝐴)
109eqeq1d 2821 . . 3 (𝑥 = 𝐴 → ( 𝑥 = 𝐵 𝐴 = 𝐵))
11 vex 3496 . . . . 5 𝑥 ∈ V
12 df-bigcup 33312 . . . . 5 Bigcup = ((V × V) ∖ ran ((V ⊗ E ) △ (( E ∘ E ) ⊗ V)))
13 brxp 5594 . . . . . 6 (𝑥(V × V)𝐵 ↔ (𝑥 ∈ V ∧ 𝐵 ∈ V))
1411, 3, 13mpbir2an 709 . . . . 5 𝑥(V × V)𝐵
15 epel 5462 . . . . . . 7 (𝑦 E 𝑧𝑦𝑧)
1615rexbii 3245 . . . . . 6 (∃𝑧𝑥 𝑦 E 𝑧 ↔ ∃𝑧𝑥 𝑦𝑧)
17 vex 3496 . . . . . . 7 𝑦 ∈ V
1817, 11coep 32980 . . . . . 6 (𝑦( E ∘ E )𝑥 ↔ ∃𝑧𝑥 𝑦 E 𝑧)
19 eluni2 4834 . . . . . 6 (𝑦 𝑥 ↔ ∃𝑧𝑥 𝑦𝑧)
2016, 18, 193bitr4ri 306 . . . . 5 (𝑦 𝑥𝑦( E ∘ E )𝑥)
2111, 3, 12, 14, 20brtxpsd3 33350 . . . 4 (𝑥 Bigcup 𝐵𝐵 = 𝑥)
22 eqcom 2826 . . . 4 (𝐵 = 𝑥 𝑥 = 𝐵)
2321, 22bitri 277 . . 3 (𝑥 Bigcup 𝐵 𝑥 = 𝐵)
248, 10, 23vtoclbg 3567 . 2 (𝐴 ∈ V → (𝐴 Bigcup 𝐵 𝐴 = 𝐵))
252, 7, 24pm5.21nii 382 1 (𝐴 Bigcup 𝐵 𝐴 = 𝐵)
Colors of variables: wff setvar class Syntax hints: ↔ wb 208 = wceq 1531 ∈ wcel 2108 ∃wrex 3137 Vcvv 3493 ∪ cuni 4830 class class class wbr 5057 E cep 5457 × cxp 5546 ∘ ccom 5552 Bigcup cbigcup 33288 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1905 ax-6 1964 ax-7 2009 ax-8 2110 ax-9 2118 ax-10 2139 ax-11 2154 ax-12 2170 ax-ext 2791 ax-sep 5194 ax-nul 5201 ax-pow 5257 ax-pr 5320 ax-un 7453 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1084 df-tru 1534 df-ex 1775 df-nf 1779 df-sb 2064 df-mo 2616 df-eu 2648 df-clab 2798 df-cleq 2812 df-clel 2891 df-nfc 2961 df-ne 3015 df-ral 3141 df-rex 3142 df-rab 3145 df-v 3495 df-sbc 3771 df-dif 3937 df-un 3939 df-in 3941 df-ss 3950 df-symdif 4217 df-nul 4290 df-if 4466 df-pw 4539 df-sn 4560 df-pr 4562 df-op 4566 df-uni 4831 df-br 5058 df-opab 5120 df-mpt 5138 df-id 5453 df-eprel 5458 df-xp 5554 df-rel 5555 df-cnv 5556 df-co 5557 df-dm 5558 df-rn 5559 df-res 5560 df-iota 6307 df-fun 6350 df-fn 6351 df-f 6352 df-fo 6354 df-fv 6356 df-1st 7681 df-2nd 7682 df-txp 33308 df-bigcup 33312 This theorem is referenced by: dfbigcup2 33353 fvbigcup 33356 ellimits 33364 brapply 33392 dfrdg4 33405
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# Solving linear equations
Are you struggling with Solving linear equations? In this post, we will show you how to do it step-by-step. We can solve math word problems.
## Solve linear equations
We will also provide some tips for Solving linear equations quickly and efficiently There are many ways to solve slope, but one of the most common and effective methods is to use the slope formula. This formula involves finding the rise and run of a line, and then plugging those values into the formula. The rise is the vertical distance between two points on the line, while the run is the horizontal distance between those same two points. Once you have those values, you can plug them into the slope formula and solve for the slope.
If you're struggling to remember all of the trigonometric identities, there's now an app for that! With a trig identities solver, all you need to do is input the equation you're trying to solve and it will give you the answer. No more memorizing dozens of identities!
In other words, they are outside of the real number line. However, imaginary numbers are not really imaginary. They are just numbers that cannot be expressed using the real number line. There are many ways to solve imaginary numbers. One way is to square the number. This will usually give you a real number. Another way is to use the quadratic equation. This will usually give you two answers, one of
Linear equations are equations that involve only linear functions. To solve a linear equation, first determine what the variable is and what the equation is saying about that variable. Then use algebraic methods to solve for the variable.
A new app called "Math Cheat" is available for download on smartphones. The app promises to provide users with solutions to math problems in a matter of seconds. The app has been criticized for being a "cheat" tool, as it takes away the challenge and effort required to solve math problems on one's own. However, many students and parents see the app as a valuable resource that can save time and help students succeed in math class.
To solve an equation or inequality, you need to find the value or values of the variable or variables that make the equation or inequality true. To do this, you can use algebraic methods, graphical methods, or numerical methods. Algebraic methods involve using algebraic rules to manipulate the equation or inequality until you can solve it. Graphical methods involve using a graph to visualize the equation or inequality and find the solution. Numerical methods involve using a calculator or computer to
## We will support you with math difficulties
I'm an adjunct instructor in a community college. I teach courses in electronics, automation, and industrial engineering. So, yeah, a bit of math involved. A student brought the app to my attention several years ago. I was a little skeptical of how well it would be able to recognize various equations, etc., especially handwritten problems. So, I installed the app and gave it a try. I was immediately impressed by the optical OCR and the way solutions were presented. It's outstanding.
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Symbolic algebra solver Math explanation Equation solver calculator with steps College math cheat sheet Basic algebra help Solving domain and range
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# Inelastic Collision Energy loss
1. Oct 23, 2014
### camorin
1. The problem statement, all variables and given/known data
Two particles are in a perfectly inelastic collision with no external force acting on them. Particle 1 has m1 and v1>0. Particle 2 has mass m2 and v2<0. Find the loss in kinetic energy after the collision. Express your answer in
reduced mass as well as relative velocity.
2. Relevant equations
KE=1/2mv^2
Momentum conservation.
Reduced mass=m1*m2/(m2-m1)
3. The attempt at a solution
So far all I've managed to come up with is the final velocity which is (m1v1+m2v2_/(m1+m2)
I'm not sure where to go form here, I've tried writing out the KE after the collision by squaring the final velocity but I cant find a place to go from here.
2. Oct 23, 2014
### Simon Bridge
What is the total kinetic energy before the collision?
What is the total kinetic energy after the collision?
Do you know how to relate the velocities and masses to the relative velocity and reduced mass?
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# Low cost launch technology
#### PhilKravitz
Why is there no low cost space launch technology? The cost to buy the energy (in the form of electricity) required is very low (don't have the number on me).
Related General Engineering News on Phys.org
#### russ_watters
Mentor
How do you define "low cost" and what does electricity have to do with it? The problem of launching an object into orbit is pretty straightforward and to do better than chemical rockets would require something pretty exotic.
#### PhilKravitz
How do you define "low cost" and what does electricity have to do with it? The problem of launching an object into orbit is pretty straightforward and to do better than chemical rockets would require something pretty exotic.
If one buy the joules per kilogram required to reach orbit from the electric company in the form of electricity it is well let see
1 kilo to say 1e4 m/sec using 1/2*m*v^2 is 5E7 joules
so 1 kilowatt hour at 20 cents from con edison is 3.6e6 joules for 20 cents. So 5E7 joules would cost about $3. Of course there are always losses in any system but even with only 10% eff. that would be$30 per kilo to orbit as far as energy costs are concerned. But rockets cost roughly $10,000 per kilo. I do like the beamed power solutions I have been seeing articles about recently. Beam power to the "rocket" in the form of microwaves or as laser light. #### russ_watters Mentor The fundamental problem with chemical rockets is that you aren't just lifting the payload but are also lifting the fuel. Finding a technology that keeps the fuel on the ground hasn't been very easy. #### Bob S If one buy the joules per kilogram required to reach orbit from the electric company in the form of electricity it is well let see 1 kilo to say 1e4 m/sec using 1/2*m*v^2 is 5E7 joules so 1 kilowatt hour at 20 cents from con edison is 3.6e6 joules for 20 cents. So 5E7 joules would cost about$3. Of course there are always losses in any system but even with only 10% eff. that would be $30 per kilo to orbit as far as energy costs are concerned. But rockets cost roughly$10,000 per kilo.
I do like the beamed power solutions I have been seeing articles about recently. Beam power to the "rocket" in the form of microwaves or as laser light.
There is a lot more than just energy required to launch a rocket. The force required to accelerate a rocket is based on momentum transfer from the rocket to the exhaust gases. Momentum transfer requires mass (photons are the exception*). Because mass is transferred to the exhaust from the rocket, the rocket loses mass, so the rocket has to carry this mass in addition to the energy needed to accelerate the expelled mass. Also, the acceleration rate has to be limited to a few g's (10?) in order not to crush the astronauts inside. See the rocket equations in
http://www.braeunig.us/space/propuls.htm
Bob S
*photons are an extremely poor source of momentum transfer, in terms of thrust per unit energy.
Gold Member
Not to mention that power on the grid is generated by giant machines. For a rocket, you have to compact all that required energy and mass into a tiny space, which is generally super expensive.
#### Bob S
I do like the beamed power solutions I have been seeing articles about recently. Beam power to the "rocket" in the form of microwaves or as laser light.
Kinetic energy may be written (Newtoninan classical mechanics)
Eclass = ½mv2 = ½(mv)2/m = p2/(2m)
So the momentum p is, for a given classical energy Eclass
p = sqrt(2mEclass)
So classically, photons have no mass and therefore can transfer no momentum.
Relativistically, however
p = Etotal/c
where Etotal = [STRIKE]Eclass + mc2[/STRIKE] = EPlanck, and c = speed of light.
So there is a very small momentum transfer with photons.
Bob S
Last edited:
#### mheslep
Gold Member
There must have been some prior looks at some kind of pre-launch acceleration tower for heavy lift vehicles, to save on launch fuel? It appears that a 3g (?) acceleration along, say, a 1000M 'elevator' tower takes the vehicle to ~250 m/s = sqrt(2 * 3g * height) before it needs to burn its own fuel. This would be the heavy lift version of what is already done with air launched rockets such as Pegasus.
Edit: Listing problems/comments here as they occur to me:
• A launch failure might destroy not only the vehicle but the significant investment in the vehicle pre-launch elevator.
• Drag incurred at low altitude (250 m/s) forces the vehicle to give back some energy gained by using the elevator. (Meaning the launch needs to move from the Cape to Pike's Peak :tongue:)
• Fuel savings: ~2kg of LOX per metric ton of payload pre-accelerated to 250 m/s [using a savings of 15 MJ combustion energy per kg of LOX to reach 13 kJ of kinetic energy per kg of payload at launch velocity]. So on the shuttle for instance at 2000 mt launch weight, this would save 4 mt of LOX, not counting mechanical structure savings, etc.
• Mechanical complexity of the elevator. Lifting a 2000 mt vehicle slowly is achievable with current technology. Lifting and accelerating something to 150 m/s is achievable with current technology. Doing both things together might be quite difficult - rapidly changing large moments along the tower, etc
I suppose what I'm looking for here are the right terms for which to google, as nothing pops up under the terms I have used above.
Last edited:
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# Homework Help: Physics - Question - Stress, Area & Factor of Safety?
1. Nov 8, 2009
### Danyolb
Diagram: http://img11.imageshack.us/img11/2442/boltlw.jpg [Broken]
Question: The material for the bolt shown in the angled joint has an ultimate tensile stress of 600MPa and a shear strength of 250MPa. The diameter of the bolt is 10mm. Determine the factor of safety. Assume F=9KN.
Stress = Applied Force/ cross-sectional area
Factor of safety = ultimate tensile strength / actual tensile stress
Last edited by a moderator: May 4, 2017
2. Nov 8, 2009
### PhanthomJay
The bolt is subject to both tensile and shear stresses. You have to determine each. What have you tried so far?
3. Nov 16, 2009
### Matty G
I'm also stuck on virtually the same question as Danyolb's (different values though), so can anyone help in showing the full workings of the following please:
Diagram: http://img11.imageshack.us/img11/2442/boltlw.jpg [Broken]
Question: The material for the bolt shown in the angled joint has an ultimate tensile strength of 600MPa and a shear strength of 250MPa. The diameter of the bolt is 10mm. Assume K=9kN. Determine (1) Factor of safety in tension, (2) Factor of safety in shear & (3) State which factor of safety is safer of the two.
Thanks.
Last edited by a moderator: May 4, 2017
4. Nov 16, 2009
### PhanthomJay
Matty, you are going to have show some attempt before we can help. You should first determine the vector components of the force acting along the axis of the bolt and perpendicular to its axis.
Last edited by a moderator: May 4, 2017
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Preschool
Math
Lesson Plan
Magic 2D Shapes
Lesson Plan
Magic 2D Shapes
In this mini-lesson, your ELs will learn all about 2D shapes using hands on practice! This can be used as a stand-alone or support lesson for the Shape It Up lesson plan.
Preschool
Math
Lesson Plan
Color the Shapes!
Worksheet
Color the Shapes!
Ready for some shape practice? This sheet is jam-packed with shape learning and coloring fun!
Preschool
Math
Worksheet
Shape Mix-Up
Worksheet
Shape Mix-Up
Kids cut out pictures to put silly, mixed-up shapes back together on this prekindergarten math worksheet.
Preschool
Math
Worksheet
It's a Shape Zoo!
Lesson Plan
It's a Shape Zoo!
Preschool students will love getting creative as they make shape animals while learning all about 2D shapes in this fun lesson! It can be used as a stand alone or support lesson plan for the Shapes and Shadows lesson plan.
Preschool
Math
Lesson Plan
Match the Shapes
Worksheet
Match the Shapes
Oh, what a mix-up! With this worksheet, your child matches mixed-up shapes together.
Preschool
Math
Worksheet
How Do You Know the Shape?
Lesson Plan
How Do You Know the Shape?
In this fun music and text inspired lesson plan your students will love making their own shape pictures as they learn all about 2D shapes! It can be used on its own or as support for the lesson Fingerpaint and Fun with Shapes.
Preschool
Math
Lesson Plan
Stamping Patterns
Lesson Plan
Stamping Patterns
Your students will love learning about patterns and positional language as they create their very own piece of art! It can be used alone or as a support lesson for the Pasta Patterning lesson plan.
Preschool
Math
Lesson Plan
Tracing Shapes: Diamonds
Worksheet
Tracing Shapes: Diamonds
Your preschooler will love this tracing page, all about the diamond shape. He'll get to color in the picture when he's finished!
Preschool
Math
Worksheet
Where Does It Go?
Lesson Plan
Where Does It Go?
In this fun hands on lesson, students will be introduced to and get practice using positional language!
Preschool
Math
Lesson Plan
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# Homework Help: I need help very confused and frustrated
1. Apr 25, 2005
### Jayhawk1
I asked this question earlier and I didn't get any significant help, atleast beyond what I already knew. I do not know how to calculate this, nor do my classmates who I have seen also posed this question to the forum. Please help!! I know the concept, but I can not calculate it. I have worked on it for a few days now, and it is due in about an hour.
2. Apr 25, 2005
### HallsofIvy
You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.
3. Apr 25, 2005
### OlderDan
Diameter of brass plug as a function of the temperature change is
$$D_b = D_{b0}(1+10*10^{-6}/C^o*{\Delta}T)$$
Diameter of iron ring as a function of the temperature change is
$$D_i = D_{i0}(1+12*10^{-6}/C^o*{\Delta}T)$$
Set the two diameters equal and solve for the temperature change. Add the change to the starting temperature.
4. Apr 25, 2005
### OlderDan
The diameter of the inside of the ring actually gets larger. If it were a disk, all the linear dimensions would increase. The diameter of the "missing" part of the disk that makes it a ring will expand in proportion to the outer diameter.
5. Apr 25, 2005
### Jayhawk1
Thank you but...
Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.
6. Apr 25, 2005
### OlderDan
$$D_b = 8.755cm(1+10*10^{-6}/C^o*{\Delta}T)$$
$$D_i = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)$$
$$8.755cm(1+10*10^{-6}/C^o*{\Delta}T) = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)$$
$$8.755cm-8.745cm = 8.745cm(12*10^{-6}/C^o*{\Delta}T) - 8.755cm(10*10^{-6}/C^o*{\Delta}T)$$
$$(8.755cm-8.745cm)*10^6C^o = (8.745cm*12 - 8.755cm*10){\Delta}T$$
Can you finish this?
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This site is supported by donations to The OEIS Foundation.
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A058512 Coefficients of replicable function number 15a. 1
1, 0, 5, -2, 0, 10, -1, 0, 25, 2, 0, 50, 1, 0, 100, 4, 0, 170, -6, 0, 305, -2, 0, 500, 2, 0, 825, 0, 0, 1300, 10, 0, 2040, -14, 0, 3100, -5, 0, 4700, 8, 0, 6950, 4, 0, 10225, 20, 0, 14800, -28, 0, 21285, -10, 0, 30200, 14, 0, 42625, 4, 0, 59500, 39, 0, 82610, -56, 0, 113690, -20, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET -1,3 LINKS G. C. Greubel, Table of n, a(n) for n = -1..1000 D. Ford, J. McKay and S. P. Norton, More on replicable functions, Commun. Algebra 22, No. 13, 5175-5193 (1994). FORMULA Expansion of A + 5*q^2/A, where A = q*(eta(q^3)/eta(q^15))^2, in powers of q. - G. C. Greubel, Jun 21 2018 EXAMPLE T15a = 1/q + 5*q - 2*q^2 + 10*q^4 - q^5 + 25*q^7 + 2*q^8 + 50*q^10 + q^11 + ... MATHEMATICA eta[q_] := q^(1/24)*QPochhammer[q]; e15D := q^(1/3)*(eta[q]/eta[q^5])^2; a[n_]:= SeriesCoefficient[(e15D /. {q -> q^3}) + 5*q^2/(e15D /. {q -> q^3}), {q, 0, n}]; Table[a[n], {n, 0, 50}] (* G. C. Greubel, Feb 14 2018 *) PROG (PARI) {a(n) = my(A); if( n<-1, 0, n++; A = x * O(x^n); A = (eta(x^3 + A) / eta(x^15 + A))^2; polcoeff( A + 5*x^2 / A, n))}; /* Michael Somos, Feb 18 2018 */ CROSSREFS Cf. A000521, A007240, A014708, A007241, A007267, A045478, etc. Sequence in context: A257406 A269980 A319231 * A111560 A324609 A211991 Adjacent sequences: A058509 A058510 A058511 * A058513 A058514 A058515 KEYWORD sign AUTHOR N. J. A. Sloane, Nov 27 2000 EXTENSIONS Terms a(24) onward added by G. C. Greubel, Feb 14 2018 STATUS approved
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Steam Nozzles and Turbines Online Quiz - ObjectiveBooks
# Steam Nozzles and Turbines Online Quiz
1. The pressure at which the steam leaves the nozzle is known as back pressure.
Correct
Incorrect
2. The turbine, in which the general direction of the steam flow is parallel to the turbine axis, is called axial flow turbines
True
False
3. The blade velocity coefficient is ratio of relative velocity of steam at outlet tip of the blade to the relative velocity of steam at inlet tip of the blade.
True
False
4. The critical pressure ratio for initially dry saturated steam is more as compared to initially wet steam.
Yes
No
5. In a De-Laval impulse turbine, the nozzle is kept very close to the blades.
Yes
No
6. The ratio of the work delivered at the turbine shaft to the heat supplied is called overall thermal efficiency of turbine.
True
False
7. The velocity of steam, in reaction turbines, is increased in the fixed blades as well as in moving blades.
True
False
8. When the cross-section of a nozzle first increases from its entrance to throat, and then decreases from its throat to exit, it is not a convergent-divergent nozzle.
Yes
No
9. In a convergent divergent nozzle, the discharge depends upon the initial conditions of steam and the area of nozzle at throat.
Correct
Incorrect
10. The efficiency ratio is the ratio of total useful heat drop to the total isentropic heat drop.
Agree
Disagree
11. In a single stage impulse turbine, the velocity of steam approaching nozzles is negligible.
True
False
12. The ratio of the cumulative heat drop to the isentropic heat drop is called reheat factor.
Yes
No
13. The Rankine efficiency depends upon total useful heat drop and total isentropic heat drop.
Correct
Incorrect
14. The discharge through a nozzle is maximum for a certain value of exit pressure. This pressure is known as critical pressure.
Agree
Disagree
15. The pressure of steam, in reaction turbines, is reduced in the fixed blades as well as in moving blades.
Correct
Incorrect
16. The ratio of the energy supplied to the blades per kg of steam to the total energy supplied per stage per kg of steam is called mechanical efficiency.
Yes
No
17. In a nozzle, the effect of super-saturation is to increase the dryness fraction of steam.
Yes
No
18. In pressure compounding of an impulse turbine, the total pressure drop of the steam does not take place in the first nozzle ring, but is divided equally among all the nozzle rings.
Agree
Disagree
19. During flow through a nozzle, no heat is supplied or rejected by the steam.
Agree
Disagree
20. In velocity compounding of an impulse turbine, the expansion of steam takes place in a nozzle or a set of nozzles from the boiler pressure to condenser pressure.
True
False
Score =
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https://www.justintools.com/unit-conversion/area.php?k1=square-leagues-US-STATUTE&k2=square-links-RAMDEN-ENGINEER
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# AREA Units Conversionsquare-leagues-US-STATUTE to square-links-RAMDEN-ENGINEER
1 Square Leagues US STATUTE
= 250905596.44892 Square Links RAMDEN ENGINEER
Embed this to your website/blog
Category: area
Conversion: Square Leagues US STATUTE to Square Links RAMDEN ENGINEER
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Leagues US STATUTE] symbol/abbrevation: (sq leag [US])
[Square Links RAMDEN ENGINEER] symbol/abbrevation: (sq li [engineer])
How to convert Square Leagues US STATUTE to Square Links RAMDEN ENGINEER (sq leag [US] to sq li [engineer])?
1 sq leag [US] = 250905596.44892 sq li [engineer].
1 x 250905596.44892 sq li [engineer] = 250905596.44892 Square Links RAMDEN ENGINEER.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Leagues US STATUTE (sq leag [US]) is equal to 23309986 square-meters, while 1 Square Links RAMDEN ENGINEER (sq li [engineer]) = 0.092903412 square-meters.
1 Square Leagues US STATUTE to common area units
1 sq leag [US] = 23309986 square meters (m2, sq m)
1 sq leag [US] = 233099860000 square centimeters (cm2, sq cm)
1 sq leag [US] = 23.309986 square kilometers (km2, sq km)
1 sq leag [US] = 250906709.14825 square feet (ft2, sq ft)
1 sq leag [US] = 36130550561.101 square inches (in2, sq in)
1 sq leag [US] = 27878511.235418 square yards (yd2, sq yd)
1 sq leag [US] = 9.0000359113618 square miles (mi2, sq mi)
1 sq leag [US] = 3.6130550561101E+16 square mils (sq mil)
1 sq leag [US] = 2330.9986 hectares (ha)
1 sq leag [US] = 5760.0178904138 acres (ac)
Square Leagues US STATUTEto Square Links RAMDEN ENGINEER (table conversion)
1 sq leag [US] = 250905596.44892 sq li [engineer]
2 sq leag [US] = 501811192.89785 sq li [engineer]
3 sq leag [US] = 752716789.34677 sq li [engineer]
4 sq leag [US] = 1003622385.7957 sq li [engineer]
5 sq leag [US] = 1254527982.2446 sq li [engineer]
6 sq leag [US] = 1505433578.6935 sq li [engineer]
7 sq leag [US] = 1756339175.1425 sq li [engineer]
8 sq leag [US] = 2007244771.5914 sq li [engineer]
9 sq leag [US] = 2258150368.0403 sq li [engineer]
10 sq leag [US] = 2509055964.4892 sq li [engineer]
20 sq leag [US] = 5018111928.9785 sq li [engineer]
30 sq leag [US] = 7527167893.4677 sq li [engineer]
40 sq leag [US] = 10036223857.957 sq li [engineer]
50 sq leag [US] = 12545279822.446 sq li [engineer]
60 sq leag [US] = 15054335786.935 sq li [engineer]
70 sq leag [US] = 17563391751.425 sq li [engineer]
80 sq leag [US] = 20072447715.914 sq li [engineer]
90 sq leag [US] = 22581503680.403 sq li [engineer]
100 sq leag [US] = 25090559644.892 sq li [engineer]
200 sq leag [US] = 50181119289.785 sq li [engineer]
300 sq leag [US] = 75271678934.677 sq li [engineer]
400 sq leag [US] = 100362238579.57 sq li [engineer]
500 sq leag [US] = 125452798224.46 sq li [engineer]
600 sq leag [US] = 150543357869.35 sq li [engineer]
700 sq leag [US] = 175633917514.25 sq li [engineer]
800 sq leag [US] = 200724477159.14 sq li [engineer]
900 sq leag [US] = 225815036804.03 sq li [engineer]
1000 sq leag [US] = 250905596448.92 sq li [engineer]
2000 sq leag [US] = 501811192897.85 sq li [engineer]
4000 sq leag [US] = 1003622385795.7 sq li [engineer]
5000 sq leag [US] = 1254527982244.6 sq li [engineer]
7500 sq leag [US] = 1881791973366.9 sq li [engineer]
10000 sq leag [US] = 2509055964489.2 sq li [engineer]
25000 sq leag [US] = 6272639911223.1 sq li [engineer]
50000 sq leag [US] = 12545279822446 sq li [engineer]
100000 sq leag [US] = 25090559644892 sq li [engineer]
1000000 sq leag [US] = 2.5090559644892E+14 sq li [engineer]
1000000000 sq leag [US] = 2.5090559644892E+17 sq li [engineer]
(Square Leagues US STATUTE) to (Square Links RAMDEN ENGINEER) conversions
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http://www.ck12.org/book/CK-12-Middle-School-Math-Grade-6/r4/section/8.1/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
8.1: Ratios
Difficulty Level: At Grade Created by: CK-12
Introduction
The Milk Comparison
On the way home from soccer practice, Casey’s mom sends her into the grocery store to get a half gallon of milk. Casey is hungry after practice, so she isn’t paying attention to what kind her Mom has asked her to get. In Casey’s house they drink both whole milk and skim milk.
Casey runs to the dairy section of the grocery store and stops short. She isn’t sure what to get. There are five different kinds of milk. There is whole milk, reduced fat milk, lowfat milk, skim milk and organic milk. There are also different brands to choose from: Hood, Eagle Brand, and Garelic for non-organic milk, and Organic Valley and Nature’s Valley for organic brands.
Casey notices that there are three non-organic brands to the two organic brands. Then she notices that the supermarket has its own brand of non-organic milk as well. That makes four non-organic brands to two organic brands.
Casey is sure that this means something. She has been reading about organic food in school and is interested in organic milk and food. Casey wishes that there were more organic brands than non organic brands. She decides to make a note of this to show her teacher at school.
If Casey wants to document this information as a ratio, how could she do it?
Does simplifying the ratio change anything? What conclusions can Casey figure out by working with these ratios?
This lesson is all about ratios, by the end of the lesson, you will know how to help Casey write her milk ratios. Pay close attention and what you learn will be very helpful!!
What You Will Learn
By the end of this lesson, you will be able to demonstrate the following skills:
• Identify and write different forms of equivalent ratios.
• Write ratios in simplest form.
• Write and compare ratios in decimal form.
• Solve real-world problems involving ratios.
Teaching Time
I. Identify and Write Different Forms of Equivalent Ratios
This lesson focuses on ratios. Ratios are everywhere in everyday life. In fact, we work with ratios so much that we probably don’t even realize that we are working with them. In this lesson, you will learn how to write ratios, simplify ratios and compare ratios, but there is a question that we must answer first.
What is a ratio?
A ratio is a comparison of two quantities. The quantities can be nearly anything; people, cars, dollars... even two groups of things!
Let’s look at a picture.
We can write ratios that compare the boys in this picture, but how?
How do we write a ratio?
A ratio is written in three different ways. It can be written as a fraction, with the word “to” or with a colon.
Let’s take a look at this in action by writing ratios that compare the boys in the picture.
Example
What is the ratio of boys with striped shirts to boys with solid shirts?
There are two boys with striped shirts and two boys with solid shirts.
Let’s write the ratio in three ways.
222:22 to 2\begin{align*}&\frac{2}{2}\\ &2:2\\ &2 \ \text{to}\ 2\end{align*}
Each of these ratios is correct. Notice that we are comparing an individual quality here.
What about comparing a category to the group?
Example
What is the ratio of boys that are holding binders to all of the boys?
There are two boys holding binders and four boys in the group.
Let’s write the ratio three different ways. Notice that the first thing being compared comes first when writing the ratio. Or the first thing becomes the numerator in the fraction form of the ratio.
242:42 to 4\begin{align*}&\frac{2}{4}\\ &2:4\\ &2 \ \text{to}\ 4\end{align*}
Each of these ratios is equivalent, meaning that they are all equal. Each ratio, though written in a different form, is an equivalent ratio.
We can write many different ratios by comparing these figures. Let’s list some and use the word “to” to write our ratio form.
Stars to circles = 3 to 2
Red stars to total stars = 2 to 3
Red stars to blue stars = 2 to 1
Blue stars to red stars = 1 to 2
Blue stars to total stars = 1 to 3
We could continue making this list.
We can also write the same ratios using a colon or a fraction.
Practice on your own. Use the picture to write each ratio three different ways.
1. What is the ratio of orange marbles to green marbles?
2. What is the ratio of yellow marbles to total marbles?
3. What is the ratio of orange marbles to total marbles?
Take a few minutes to check your work with a peer.
II. Write Ratios in Simplest Form
Sometimes, a ratio does not represent a clear comparison. If you look at one of the ratios in the practice problems you just finished you will see what I mean. The ratio of orange marbles to total marbles was 2 to 22.
We can simplify a ratio just as we would a fraction. Let’s look at the ratio 2 to 22 in the fraction form of the ratio.
222\begin{align*}\frac{2}{22}\end{align*}
We simplify a ratio in fraction form in the same way that we would simplify a fraction.
We use the greatest common factor of both the numerator and the denominator. By dividing the numerator and the denominator by the GCF we can simplify the fraction.
The GCF of both 2 and 22 is 2.
2÷222÷2=111\begin{align*}\frac{2 \div 2}{22 \div 2} = \frac{1}{11}\end{align*}
The simplest form of the ratio is 1 to 11. We can write this in three ways 1 to 11, 1:11 and 111\begin{align*}\frac{1}{11}\end{align*}.
When we simplify a ratio in fraction form, we also write an equivalent form of the original ratio.
111=222\begin{align*}\frac{1}{11} = \frac{2}{22}\end{align*}
Simplify these ratios on your own. If the ratio is not written in fraction form, you will need to do that first.
1. 210\begin{align*}\frac{2}{10}\end{align*}
2. 6 to 8
3. 5:20
III. Write and Compare Ratios in Decimal Form
We just finished writing ratios in fraction form and simplifying them. What about decimal form? Fractions and decimals are related, in fact a fraction can be written as a decimal and a decimal can be written as a fraction.
Is it possible to write a ratio as a decimal too?
Yes! Because a ratio can be written as a fraction, it can also be written as a decimal. To do this, you will need to remember how to convert fractions to decimals.
Now we can apply this information to our work with ratios.
Example
Convert 2:4 into a decimal.
First, write it as a ratio in fraction form.
2:4=24\begin{align*}2:4 = \frac{2}{4}\end{align*}
Next, simplify the fraction if possible.
24=12\begin{align*}\frac{2}{4} = \frac{1}{2}\end{align*}
Finally, convert the fraction to a decimal.
2)1.0¯¯¯¯¯¯¯¯¯¯.5\begin{align*} \overset{\quad .5}{2\overline{)1.0 \;}}\end{align*}
Practice converting ratios to decimal form.
1. 4 to 5
2. 520\begin{align*}\frac{5}{20}\end{align*}
3. 6 to 10
Take a few minutes to check your work with a partner.
Real Life Example Completed
The Milk Comparison
Remember Casey and her milk comparison? Well, now you know all that you need to know about ratios to help her with her comparison. Here is the problem once again.
On the way home from soccer practice, Casey’s mom sends her into the grocery store to get a half gallon of milk. Casey is hungry after practice, so she isn’t paying attention to what kind her Mom has asked her to get. In Casey’s house they drink both whole milk and skim milk.
Casey runs to the dairy section of the grocery store and stops short. She isn’t sure what to get. There are five different kinds of milk. There is whole milk, reduced fat milk, lowfat milk, skim milk and organic milk. There are also different brands to choose from: Hood, Eagle Brand, and Garelic for non-organic milk, and Organic Valley and Nature’s Valley for organic brands.
Casey notices that there are three non-organic brands to the two organic brands. Then she notices that the supermarket has its own brand of non-organic milk as well. That makes four non-organic brands to two organic brands.
Casey is sure that this means something. She has been reading about organic food in school and is interested in organic milk and food. Casey wishes that there were more organic brands than non organic brands. She decides to make a note of this to show her teacher at school.
If Casey wants to document this information as a ratio, how could she do it?
Does simplifying the ratio change anything? What conclusions can Casey figure out by working with these ratios?
First, let’s underline the important information. (This has been done for you above.)
A ratio is a comparison. We can write a ratio to compare two quantities in three different ways. In this problem, Casey wants to compare organic and non-organic brands of milk.
She notices that there are four non-organic brands and two organic brands.
Casey can write this comparison three different ways.
4 to 2424:2\begin{align*}4\ \text{to}\ 2 \qquad \frac{4}{2} \qquad 4:2\end{align*}
If Casey simplifies these ratios, what conclusions can she draw?
4 to 2 simplifies to 2 to 1
424:2=21=2:1\begin{align*}\frac{4}{2} &= \frac{2}{1}\\ 4 : 2 &= 2 : 1\end{align*}
Casey concludes that there are twice as many non-organic brands as there are organic. When she shows her teacher, Ms. Gilson challenges Casey to do some research about organic brands of milk to bring to the grocery store manager. Casey rises to the challenge!!
Vocabulary
Here are the vocabulary words that are found in this lesson.
Ratio
a comparison between two quantities; can be written three different ways.
Equivalent
equal
Simplify
to make smaller
Greatest Common Factor
the largest number that will divide into two or more numbers evenly.
Technology Integration
Other Videos:
http://www.mathplayground.com/howto_ratios.html – This is a great video that explains the concept of ratios.
http://www.mathplayground.com/howto_equalratios.html – This is a nice basic video on understanding equal ratios.
Time to Practice
Directions: Use the picture to answer the following questions. Write each ratio three ways.
1. What is the ratio of hens to chicks?
2. What is the ratio of green chicks to yellow chicks?
3. What is the ratio of white chicks to total chicks?
4. What is the ratio of green chicks to total chicks?
5. What is the ratio of yellow chicks to total chicks?
6. What is the ratio of green chicks to white chicks?
7. 2 to 4
8. 3:6
9. 5 to 15
10. 2 to 30
11. 10 to 15
12. 46\begin{align*}\frac{4}{6}\end{align*}
13. 3:9
14. 6:8
15. 28\begin{align*}\frac{2}{8}\end{align*}
16. 416\begin{align*}\frac{4}{16}\end{align*}
17. 10 to 12
18. 7:21
19. 12:24
20. 25 to 75
Directions: Convert the following ratios into decimals.
21. 3 to 4
22. 2 to 4
23. 15\begin{align*}\frac{1}{5}\end{align*}
24. 25 to 100
25. 16 to 32
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# Week 1 & 2 Notes GIS3015
Katie Lebow
FSU
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These notes are somewhat short, as not much was covered during weeks 1 and 2, because of Hurricane Hermine and class cancellations.... However, they define the first bit of material we have covered...
COURSE
Map Analysis
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Gregory Burris
TYPE
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## Popular in Department
This 1 page Class Notes was uploaded by Katie Lebow on Friday September 9, 2016. The Class Notes belongs to GIS3015 at Florida State University taught by Gregory Burris in Summer 2016. Since its upload, it has received 64 views.
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### What is Karma?
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Date Created: 09/09/16
Introduction to Map Analysis/Understanding: • Is the map "clustered?" : Depends on the size of the frame; frame sizes areVERY easy to manipulate, thus making studies easy to change for personal gain (i.e. "Adjusting" the numbers to represent what a specific company/person wants to prove) • Mercader Projections: Distortion occurs because of the flattening of the Earth (a sphere) to have a 2- dimensional picture (For example: Mexico and Greenland are approximately the same size by land area, despite the visual representation on most maps) • Declination Diagrams: A way to determine which "North" is being used in a map (Magnetic North, Grid North, True North) * Compass North (Magnetic North) is in a different direction based on location * Cell phone/GPS North (Grid North) is completely different from Magnetic North • Vocabulary to Know: Spatial: The way that objects/areas/topographic elements relate to one another in a map (how far from each other they are, if they're near water, etc.) ‣ Spatial Data: The digital representation of topography ‣ Spatial Units: The organizational representations used in map making to make them easier to read (countries, states, counties, etc.) Features: Specific earth objects/land uses/land coverings in maps, such as buildings, towns, rivers, trees, etc. Distance: Can be measured in length (such as miles or kilometers) or in time (such as hours) Direction: Demonstrated via compass points (N, S, E, W), radians (. ), degrees (0-360) Cartography: The science of map making (regardless of map format or type) GIS: Geographic Information Systems: Digital systems used to relate spatial data in an easy to understand way (using colors, legends, borders, etc.)
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# Distance between Tangier (TNG) and Al Hoceima (AHU)
Flight distance from Tangier to Al Hoceima (Tangier Ibn Battouta Airport – Cherif Al Idrissi Airport) is 123 miles / 198 kilometers / 107 nautical miles. Estimated flight time is 43 minutes.
Driving distance from Tangier (TNG) to Al Hoceima (AHU) is 188 miles / 302 kilometers and travel time by car is about 4 hours 49 minutes.
## Map of flight path and driving directions from Tangier to Al Hoceima.
Shortest flight path between Tangier Ibn Battouta Airport (TNG) and Cherif Al Idrissi Airport (AHU).
## How far is Al Hoceima from Tangier?
There are several ways to calculate distances between Tangier and Al Hoceima. Here are two common methods:
Vincenty's formula (applied above)
• 123.159 miles
• 198.205 kilometers
• 107.022 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 122.935 miles
• 197.845 kilometers
• 106.828 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Tangier Ibn Battouta Airport
City: Tangier
Country: Morocco
IATA Code: TNG
ICAO Code: GMTT
Coordinates: 35°43′36″N, 5°55′0″W
B Cherif Al Idrissi Airport
City: Al Hoceima
Country: Morocco
IATA Code: AHU
ICAO Code: GMTA
Coordinates: 35°10′37″N, 3°50′22″W
## Time difference and current local times
There is no time difference between Tangier and Al Hoceima.
+01
+01
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 43 kg (95 pounds).
## Frequent Flyer Miles Calculator
Tangier (TNG) → Al Hoceima (AHU).
Distance:
123
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
123
Round trip?
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Marks a rectangular box with an open top is to have a
• Test Prep
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• 100% (6) 6 out of 6 people found this document helpful
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7.[10marks]A rectangular box with an open top is to have a volume of 48 m3.The length of the baseis to be twice the width.The material for the base costs \$8 per square metre.Material for the sideswill cost \$6 per square metre.Find the cost of producing the cheapest such box.5
8.[8marks]Use a Riemann sum with a regular partition to determine(x+ 3)2dx.6R
9.[7marks]LetRdenote the region bounded between the curvesy=x2andy=px.Find thevolume of the solid obtained by rotatingRabout the linex= 4.-2246-2-112346
10.[5marks]Use a substitution to evaluate1xpx+ 3dx.11.[7marks]Use integration by parts to evaluate2R1tan°1px°1dx.6R7
8
13.[6marks]Show that the two-variable functionu= sin (x°at) + ln (x+at)satis°es thewave equationutt=a2uxx(whereais a constant).14.[8marks]Find all critical points and determine the relative extrema (if any) off(x; y) =x3°3xy+ 3y2°9y.Use the Second Derivative Test to justify your conclusions.9
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# Number 1568376
### Properties of number 1568376
Cross Sum:
Factorization:
2 * 2 * 2 * 3 * 3 * 3 * 53 * 137
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
17ee78
Base 32:
1frjo
sin(1568376)
-0.96367889830264
cos(1568376)
0.26706362718686
tan(1568376)
-3.6084243610919
ln(1568376)
14.265551247064
lg(1568376)
6.1954501879081
sqrt(1568376)
1252.3481943932
Square(1568376)
### Number Look Up
Look Up
1568376 which is pronounced (one million five hundred sixty-eight thousand three hundred seventy-six) is a unique number. The cross sum of 1568376 is 36. If you factorisate the number 1568376 you will get these result 2 * 2 * 2 * 3 * 3 * 3 * 53 * 137. 1568376 has 64 divisors ( 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 53, 54, 72, 106, 108, 137, 159, 212, 216, 274, 318, 411, 424, 477, 548, 636, 822, 954, 1096, 1233, 1272, 1431, 1644, 1908, 2466, 2862, 3288, 3699, 3816, 4932, 5724, 7261, 7398, 9864, 11448, 14522, 14796, 21783, 29044, 29592, 43566, 58088, 65349, 87132, 130698, 174264, 196047, 261396, 392094, 522792, 784188, 1568376 ) whith a sum of 4471200. The figure 1568376 is not a prime number. The number 1568376 is not a fibonacci number. 1568376 is not a Bell Number. The figure 1568376 is not a Catalan Number. The convertion of 1568376 to base 2 (Binary) is 101111110111001111000. The convertion of 1568376 to base 3 (Ternary) is 2221200102000. The convertion of 1568376 to base 4 (Quaternary) is 11332321320. The convertion of 1568376 to base 5 (Quintal) is 400142001. The convertion of 1568376 to base 8 (Octal) is 5767170. The convertion of 1568376 to base 16 (Hexadecimal) is 17ee78. The convertion of 1568376 to base 32 is 1frjo. The sine of the figure 1568376 is -0.96367889830264. The cosine of the figure 1568376 is 0.26706362718686. The tangent of the number 1568376 is -3.6084243610919. The root of 1568376 is 1252.3481943932.
If you square 1568376 you will get the following result 2459803277376. The natural logarithm of 1568376 is 14.265551247064 and the decimal logarithm is 6.1954501879081. You should now know that 1568376 is unique figure!
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THIS IS DOCUMENTATION FOR AN OBSOLETE PRODUCT.
SEE THE DOCUMENTATION CENTER FOR THE LATEST INFORMATION.
FindInstance
FindInstance[expr, vars]finds an instance of vars that makes the statement expr be True. FindInstance[expr, vars, dom]finds an instance over the domain dom. Common choices of dom are Complexes, Reals, Integers and Booleans. FindInstance[expr, vars, dom, n]finds n instances.
• FindInstance[expr, {x1, x2, ...}] gives results in the same form as Solve: {{x1->val1, x2->val2, ...}} if an instance exists, and {} if it does not.
• expr can contain equations, inequalities, domain specifications and quantifiers, in the same form as in Reduce.
• With exact symbolic input, FindInstance gives exact results.
• Even if two inputs define the same mathematical set, FindInstance may still pick different instances to return.
• The instances returned by FindInstance typically correspond to special or interesting points in the set.
• FindInstance[expr, vars] assumes by default that quantities appearing algebraically in inequalities are real, while all other quantities are complex.
• FindInstance[expr, vars, Reals] assumes that not only vars but also all function values in expr are real. FindInstance[expr&&varsReals, vars] assumes only that the vars are real.
• FindInstance may be able to find instances even if Reduce cannot give a complete reduction.
• Every time you run FindInstance with a given input, it will return the same output.
• Different settings for the option RandomSeed->s may yield different collections of instances.
• FindInstance[expr, vars, dom, n] will return a shorter list if the total number of instances is less than n.
Find a solution instance of a system of equations:
Find a real solution instance of a system of equations and inequalities:
Find an integer solution instance:
Find Boolean values of variables that satisfy a formula:
Find several instances:
Find a solution instance of a system of equations:
Out[1]=
Find a real solution instance of a system of equations and inequalities:
Out[1]=
Find an integer solution instance:
Out[1]=
Find Boolean values of variables that satisfy a formula:
Out[1]=
Find several instances:
Out[1]=
Scope (41)
A linear system:
A univariate polynomial equation:
A multivariate polynomial equation:
Systems of polynomial equations and inequations:
This gives three solution instances:
If there are no solutions FindInstance returns an empty list:
If there are fewer solutions than the requested number, FindInstance returns all solutions:
Quantified polynomial system:
An algebraic system:
Transcendental equations:
In this case there is no solution:
A solution in terms of transcendental Root objects:
A system of transcendental equations:
A linear system:
A univariate polynomial equation:
A univariate polynomial inequality:
A multivariate polynomial equation:
A multivariate polynomial inequality:
Systems of polynomial equations and inequalities:
Get four solution instances:
If there are no solutions FindInstance returns an empty list:
If there are fewer solutions than the requested number, FindInstance returns all solutions:
A quantified polynomial system:
An algebraic system:
Piecewise equations:
Piecewise inequalities:
Transcendental equations:
A solution in terms of transcendental Root objects:
Transcendental inequalities:
Transcendental systems:
A linear system of equations:
A linear system of equations and inequalities:
Find more than one solution:
A univariate polynomial equation:
A univariate polynomial inequality:
A Thue equation:
If there are fewer solutions than the requested number, FindInstance returns all solutions:
A sum of squares equation:
The Pythagorean equation:
A bounded system of equations and inequalities:
A high-degree system with no solution:
Transcendental Diophantine systems:
A polynomial system of congruences:
A linear system:
A univariate polynomial equation:
A multivariate polynomial equation:
Find seven instances:
A system of polynomial equations and inequations:
A quantified polynomial system:
Mixed real and complex variables:
Find a real value of and a complex value of for which is real and less than :
An inequality involving Abs[z]:
Options (3)
Find a solution over the integers modulo 9:
Find three solutions:
Finding instances often involves random choice from large solution sets:
By default, FindInstance chooses the same solutions each time:
With a different RandomSeed, FindInstance may give different solutions:
Finding an exact solution to this problem is hard due to high degrees of algebraic numbers:
With a finite WorkingPrecision, FindInstance is able to find an approximate solution:
Applications (6)
Find a point in the intersection of two regions:
Find a counterexample to a geometric conjecture:
Prove the conjecture using stronger assumptions:
Prove that a statement is a tautology:
This can be proven with TautologyQ as well:
Show that a statement is not a tautology; get a counterexample:
This can be done with SatisfiabilityInstances as well:
Find a Pythagorean triple:
Find Pythagorean triples when they exist:
Two instances are now found when :
Show that there are no 2×2 magic squares with all numbers unequal:
Solution instances satisfy the input system:
Use RootReduce to prove that algebraic numbers satisfy equations:
When there are no solutions, FindInstance returns an empty list:
If there are fewer solutions than the requested number, FindInstance returns all solutions:
To get a complete description of the solution set use Reduce:
To get a generic solution of a system of complex equations use Solve:
Solving a sum of squares representation problem:
Use SquaresR to find the number of solutions to sum of squares problems:
Solving a sum of powers representation problem:
Use PowersRepresentations to enumerate all solutions:
Find instances satisfying a Boolean statement:
Use SatisfiabilityInstances to obtain solutions represented as Boolean vectors
Integer solutions for a Thue equation:
New in 5
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Show without a calculator : $e^{-\gamma}<\omega$
constantsinequality
Hi I think this question is new :
Problem :
Show that :
$$e^{-\gamma}<\omega$$
Where we have the Euler's number and constant and omega constant wich is the value taking at $$x=1$$ of the Lambert's function .
I recall two interesting fact on The Lambert function :
$$W_{0}(x)=\sum_{n=1}^{\infty}{\frac{(-n)^{n-1}}{n!}}x^{n}$$
And the Euler-Mascheroni constant can be determined via the Harmonic numbers .
Question :
Can we hope to find a proof by hand without calculator as if Euler meet Lambert at the 18 century ?
Thanks !
The Omega constant $$\Omega = W_0(1)$$ is the unique solution for real $$x$$ of $$xe^x = 1,$$ therefore it is strictly less than the unique positive solution $$a$$ of the equation $$p(a) = 1,$$ where $$p(x) = x + x^2 + x^3/2 + x^4/6$$ (a strictly increasing function of $$x$$ for $$x \geqslant 0$$). We will show that $$p(4/7) > 1,$$ whence $$a < 4/7,$$ whence: $$\Omega < 4/7.$$ Proof: $$p\left(\frac47\right) = \frac47 + \frac{16}{49} + \frac{32}{343}\cdot\frac{25}{21} = \frac{44}{49} + \frac{800}{7203} = \frac{44}{49} + \frac19\left(1 - \frac3{7203}\right) > \frac{44}{49} + \frac19\left(1 - \frac4{49}\right) = 1.$$
Because $$\Omega = e^{-\Omega},$$ the required inequality is equivalent to $$e^{-\gamma} < e^{-\Omega},$$ whence (as has been pointed out in the comments) it is equivalent to $$\Omega < \gamma.$$ It is therefore enough to prove that: $$\gamma > 4/7.$$
The Wikipedia page for Euler's constant, referring to Duane W. DeTemple, "A Quicker Convergence to Euler's Constant", The American Mathematical Monthly 100 (5), pp. 468–470, states that for each positive integer $$n,$$ $$$$\label{4337532:eq:1}\tag{1} \frac1{24(n+1)^2} < \gamma_n\left(\frac12\right) - \gamma < \frac1{24n^2},$$$$ where for all $$\alpha > -n,$$ $$\gamma_n(\alpha) = 1 + \frac12 + \frac13 + \cdots + \frac1n - \log_e(n + \alpha).$$ We only need take $$n = 2$$ in \eqref{4337532:eq:1}, obtaining: $$\gamma > \frac{143}{96} - \log_e\left(\frac52\right).$$ Temporarily taking for granted this inequality (whose hairy proof is exiled to a separate section below): $$$$\label{4337532:eq:2}\tag{2} \log_e\left(\frac52\right) < \frac{11}{12},$$$$ we conclude $$\gamma > \frac{143}{96} - \frac{11}{12} = \frac{55}{96} > \frac47,$$ whence $$\gamma > \Omega$$ (as explained above). $$\square$$
$$\begin{gather*} \log_e\left(\frac52\right) = \log_e\left(\frac{10}4\right) = \log_e\frac{1 + 3/7}{1-3/7} \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7} + \frac{3^9}{9\cdot7^9} + \frac{3^{11}}{11\cdot7^{11}} + \cdots\right) \\ < 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7}\left( 1 + \frac{3^2}{7^2} + \frac{3^4}{7^4} + \cdots\right)\right) \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7}\cdot\frac{49}{40}\right) = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{5\cdot7^6\cdot8}\right) \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5}\left(1 + \frac9{56}\right)\right) = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5}\cdot\frac{65}{56}\right) \\ = \frac67\left(1 + \frac3{49} + \frac{3^4}{5\cdot7^4}\cdot\frac{65}{56}\right) = \frac67\left(1 + \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8}\right). \end{gather*}$$ So we have to prove $$\begin{gather*} 1 + \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8} < \frac{77}{72}, \text{ i.e., }\ \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8} < \frac5{72}, \text{ i.e., }\ \frac{216}{49} + \frac{3^6\cdot13}{7^5} < 5,\\ \text{ i.e., }\ \frac{20}{49} + \frac{3^6\cdot13}{7^5} < 1, \text{ i.e., }\ \frac{3^6\cdot13}{7^5} < \frac{29}{7^2}, \text{ i.e., }\ 729\cdot13 < 343\cdot29, \end{gather*}$$ i.e., $$9477 < 9947$$ - which is true. $$\square$$
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Babyboy
Joined: 11 Jun 2003
Posts: 499
Posted: 08 Mar 2004 07:44:04 pm Post subject: here are some code snippets from final fantasy, which is coming along sloooowly, as i have other things to do first is the 2 player code, so far it finds the enemy's name, and stores it seperate from yours, and can detect when the other calc hits the "attack" button. Code: ```1->LSTAT(41) Goto P0 End If K=93 Then Goto P1 End End Lbl P0 For(Z,5,9 While Z GetCalc(LSTAT(40) If LSTAT(40)=92 Then 0->M For(Z,5,7 While Z Output(1,1,"ATTACKED" If LSTAT(7)=0 Then 54->H 24->L Pt-On(H,L Pt-On(H+1,L Pt-On(H,L-1 Pt-On(H+1,L-1 Pt-On(H+1,L-1 Pt-On(H+1,L-2 Pt-On(H+1,L-1 Pt-On(H+2,L-2 Pt-On(H+2,L-3 Pt-On(H+3,L-2 Pt-On(H+4,L-3 Line(H+6,L-1,H+1,L-6 Line(H+9,L+10,H+3,L-4 Line(H+10,L-9,H+4,L-3 Pt-On(H+10,L-10 M+1->M If M=2 or M=3:Then:H+1->H:L+1->L:End End End LSTAT(2)-L1(5)->LSTAT(2 ClrHome Output(1,1,"GOT HIT FOR",L1(5) Goto P2 End If LSTAT(40)=93 Then Output(4,5,"ATTACKED BY MAGIC" Pause End Lbl P2 ClrHome ClrDraw Text(1,30,Str0,"'S STATS" Text(9,5,"HP-",LSTAT(2),"/",LSTAT(1)," ATTACK-",LSTAT(5) Text(18,5,"MP-",LSTAT(4),"/",LSTAT(3)," DEFENCE-",LSTAT(4) Text(30,25,Str1,"'S STATS Text(40,5,"HP-",L1(1),"/",L1(2)," ATTACK-",L1(5) Text(48,5,"MP-",L1(3),"/",L1(4)," DEFENCE-",L1(6) Pause Goto P1 Lbl P1 GetCalc(L1(1) Output(4,5,"YOUR TURN Pause ClrHome For(Z,5,7 RecallPic 1 While Z getKey->LSTAT(40) If LSTAT(40)=92 Then Output(1,1,"ATTACK" Pause Goto P1 End If LSTAT(40)=93 Then Output(1,1,"MAGIC" Goto P1 End End``` And here is the main charecter as a teen, my art style follows Akira Toryama sensei's art Code: ```ClrDraw Pt-On(7,45 Pt-On(6,45 Pt-On(5,44 Pt-On(4,45 Pt-On(15,44 Pt-On(14,45 Pt-On(13,44 Pt-On(12,45 Pt-On(11,45 Pt-On(11,44 Pt-On(12,43 Pt-On(6,43 Pt-On(7,44 Pt-On(4,45 Pt-On(3,44 Pt-On(16,57 Pt-On(16,56 Pt-On(17,55 Pt-On(1,55 Pt-On(0,55 Pt-On(0,54 Pt-On(9,60 Pt-On(10,59 Pt-On(10,58 Pt-On(9,57 Pt-On(8,57 Pt-On(8,59 Pt-On(8,58 Pt-On(7,56 Pt-On(6,56 Pt-On(6,52 Pt-On(7,52 Pt-On(9,50 Pt-On(8,51 Pt-On(9,52 Pt-On(8,53 Pt-On(8,54 Pt-On(10,53 Pt-On(10,54 Pt-On(11,56 Pt-On(12,56 Pt-On(11,52 Pt-On(12,52 Line(18,55,18,53 Line(17,52,11,46 Line(11,46,7,46 Line(7,46,0,53 Line(16,50,16,44 Line(2,44,2,50 Line(3,51,3,55 Line(2,54,7,59 Line(11,59,16,54 Line(15,54,15,51 Line(10,48,8,48 Line(13,61,10,61 Line(13,61,16,58 Line(8,61,5,61 Line(5,61,2,58 Line(2,58,2,56 Line(5,55,5,53 Line(13,55,13,53``` what i would like though is some code for a program that puts lists into strings, then strings to graphscreen, i know there was a topic on it some time ago, i cant find it thoughLast edited by Guest on 08 Mar 2004 07:49:27 pm; edited 1 time in total
Babyboy
Joined: 11 Jun 2003
Posts: 499
Posted: 10 Mar 2004 03:09:18 pm Post subject: guys! i made this graphical engine that uses PXL-TEST and only 4 if statements! ill post code when i get home (1.5 hours) Note- if this is common knowledge again, dont laugh!
DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
Posted: 10 Mar 2004 05:13:09 pm Post subject: Quote:Code: ```ClrDraw Pt-On(7,45 Pt-On(6,45 Pt-On(5,44 Pt-On(4,45 Pt-On(15,44 Pt-On(14,45 Pt-On(13,44 Pt-On(12,45 Pt-On(11,45 Pt-On(11,44 Pt-On(12,43 Pt-On(6,43 Pt-On(7,44 Pt-On(4,45 Pt-On(3,44 Pt-On(16,57 Pt-On(16,56 Pt-On(17,55 Pt-On(1,55 Pt-On(0,55 Pt-On(0,54 Pt-On(9,60 Pt-On(10,59 Pt-On(10,58 Pt-On(9,57 Pt-On(8,57 Pt-On(8,59 Pt-On(8,58 Pt-On(7,56 Pt-On(6,56 Pt-On(6,52 Pt-On(7,52 Pt-On(9,50 Pt-On(8,51 Pt-On(9,52 Pt-On(8,53 Pt-On(8,54 Pt-On(10,53 Pt-On(10,54 Pt-On(11,56 Pt-On(12,56 Pt-On(11,52 Pt-On(12,52 Line(18,55,18,53 Line(17,52,11,46 Line(11,46,7,46 Line(7,46,0,53 Line(16,50,16,44 Line(2,44,2,50 Line(3,51,3,55 Line(2,54,7,59 Line(11,59,16,54 Line(15,54,15,51 Line(10,48,8,48 Line(13,61,10,61 Line(13,61,16,58 Line(8,61,5,61 Line(5,61,2,58 Line(2,58,2,56 Line(5,55,5,53 Line(13,55,13,53``` Put those in a list and do Code: ```For(I,1,whatever Pt-On(L1(I),L2(I End For(I,1,whatever2 Line(L3(I),L4(I),L5(I),L6(I End```
Babyboy
Joined: 11 Jun 2003
Posts: 499
Posted: 10 Mar 2004 07:24:17 pm Post subject: yea but lists are big, the most ive found to get in it are 999, that isnt enough for most screens, here is that code Code: ```50->A 50->B While Z(does not = )1000 getKey->K If K=24 Then B-1->B pxl-Test(A,B->C If C=1 Then B+1->B Pxl-On(A,B End End If K=26 Then B+1->B pxl-Test(A,B->C If C=1 Then B-1->B Pxl-On(A,B End End If K=25 Then A-1->A pxl-Test(A,B->C If C=1 Then A+1->A Pxl-On(A,B End End If K=34 Then A+1->A pxl-Test(A,B->C If C=1 Then A-1->A Pxl-On(A,B End End Pxl-On(A,B End``` has this been done before in an RPG, i think i have seen it talked about before, but have never seen actual code till mine
nether fish
Newbie
Joined: 04 Feb 2004
Posts: 5
Posted: 10 Mar 2004 08:52:10 pm Post subject: Quote:If K=24 Then B-1->B pxl-Test(A,B->C If C=1 Then B+1->B Pxl-On(A,B End End you can optimize it a lil more by doing something like this... Quote:B-1(K=24)->B pxl-Test(A,B->C B+1(-C=1)->B Pxl-On(A,B End End that should work... but i havent tested it.
DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
Posted: 11 Mar 2004 05:08:23 pm Post subject: If one list isn't enough for a screen, maybe you should use a pic for that screen.
Babyboy
Joined: 11 Jun 2003
Posts: 499
Posted: 11 Mar 2004 08:22:24 pm Post subject: there is only 10 pics, and alot of screens
DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
Posted: 12 Mar 2004 05:00:30 pm Post subject: Beg to differ... there are actually 255 pics. And using asm you can store pics to programs and it will only take about 800 bytes per screen, close to a pic.
aka Tianon
Know-It-All
Joined: 02 Jun 2003
Posts: 1874
Posted: 12 Mar 2004 05:51:06 pm Post subject: using ASM you could (theoretically) store a pic to Str1 if you really wanted... you could just store the pic to a prgm file (same syntax...)
X1011
10100111001
Active Member
Joined: 14 Nov 2003
Posts: 657
Posted: 12 Mar 2004 07:32:09 pm Post subject: You could have an ASM program with compressed pic data in it that would display them.
Newbie
Joined: 29 Jan 2004
Posts: 36
Posted: 12 Mar 2004 08:52:09 pm Post subject: To optemize your code, you replace While Z(does not = ) 1000 with Repeat Z=1000 Anyway, IF statements are icky in my view, especially if within a loop. I would use something like this, putting IFs outside the loop.: Repeat Z=1000 Repeat K getkey->K End If K=24 (some code) If K=?? (some code) End
Babyboy
Joined: 11 Jun 2003
Posts: 499
Posted: 12 Mar 2004 10:49:51 pm Post subject: here is the modified code for that, with built in code for nit going off the screen Code: ```50->A 50->B Pxl-On(A,B Pxl-On(A+1,B Pxl-On(A+2,B-1 Pxl-On(A+2,B-1 Pxl-On(A+2,B+1 Pxl-On(A-2,B Pxl-On(A,B+1 Pxl-On(A,B-1 Repeat Z=1000 getKey->K If K=24 and B-3>0 Then pxl-Test(A,B-3->C If C=0 and B-3÷93 Then Pxl-Off(A,B Pxl-Off(A+1,B Pxl-Off(A+2,B-1 Pxl-Off(A+2,B+1 Pxl-Off(A-2,B Pxl-Off(A,B-1 Pxl-Off(A,B+1 B-3->B End End If K=26 and B+3<94 Then pxl-Test(A,B+3->C If C=0 and B+3÷92 Then Pxl-Off(A,B Pxl-Off(A+1,B Pxl-Off(A+2,B-1 Pxl-Off(A+2,B+1 Pxl-Off(A-2,B Pxl-Off(A,B-1 Pxl-Off(A,B+1 B+3->B End End If K=25 and A-3>1 Then pxl-Test(A-3,B->C If C=0 and A÷63 Then Pxl-Off(A,B Pxl-Off(A+1,B Pxl-Off(A+2,B-1 Pxl-Off(A+2,B+1 Pxl-Off(A-2,B Pxl-Off(A,B-1 Pxl-Off(A,B+1 A-3->A End End If K=34 and A+3<60 Then pxl-Test(A+3,B->C If C=0 Then Pxl-Off(A,B Pxl-Off(A+1,B Pxl-Off(A+2,B-1 Pxl-Off(A+2,B+1 Pxl-Off(A-2,B Pxl-Off(A,B-1 Pxl-Off(A,B+1 A+3->A End End Pxl-On(A,B Pxl-On(A+1,B Pxl-On(A+2,B-1 Pxl-On(A+2,B+1 Pxl-On(A-2,B Pxl-On(A,B+1 Pxl-On(A,B-1 End```
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The growth of individual grains of a solid phase during cooling liquid melt has been studied. During cooling, the compositions of the liquid and solid phases and the equilibrium conditions change. Therefore, each subsequent layer, "freezing" on the grain, will have a slightly different composition. This paper proposes a method for calculating the structure of the grain changes with increasing distance from the center. Corresponding mathematical model has been created. It is based on the following assumptions. Growing grains are considered as spherical. The temperature alignment in the system and the composition alignment in the liquid phase occur instantly. The alignment of the solid phase composition does not occur. Also, it is assumed that local equilibrium of the liquid phase with the solid phase on the grain surface is observed at any temperature. The characteristics of this local equilibrium can be found out from the corresponding equilibrium state diagram. The balance equation of phase masses and masses of their components at infinitesimal temperature decrease was made. It was assumed that the local equilibrium of the liquid phase and the infinitely thin layer of the solid phase, formed during this decrease in temperature, is observed. Taking to the limits, we have obtained a differential equation describing the investigated process. The solution of this equation has been obtained in the form of the solid phase mass as an integral function of the temperature. The grain composition depending on the distance from its center is obtained in the form of parametric functions expressing the radius of the current grain point and its composition at this point depending on the temperature. A computer program for the calculation of the mathematical model equations have been created. To use the model, it is needed to know the composition of the initial melt, the average density of the solid phase and the equations of the liquidus and solidus lines in the form of functions of concentration vs the temperature. An example of the calculation is presented.
## МАТЕМАТИЧНА МОДЕЛЬ ЗМІНИ СКЛАДУ ЗЕРНА при охолодженні двокомпонентних РОЗПЛАВУ
Розглядається зростання окремих зерен твердої фази при охолодженні рідкого розплаву. У міру охолодження змінюються склади рідкої і твердої фаз і умови рівноваги. Тому кожен наступний шар, «намерзає» на зерно, матиме дещо інший склад. У даній роботі запропоновано метод розрахунку зміни складу зерна в міру віддалення від його центру. Для цього розроблена математична модель, що базується на наступних припущеннях: зростаюче зерно вважається сферичним; вирівнювання температури в системі і вирівнювання складу рідкої фази відбуваються моментально; вирівнювання складу твердої фази не відбувається. При цьому вважали, що при будь-якій температурі дотримується локальне рівновагу рідкої фази і поверхневого шару твердої фази. Характеристики цього локального рівноваги можуть бути визначені з відповідною рівноважної діаграми стану. Було складено рівняння балансу мас фаз і мас їх компонентів при нескінченно малому зниженні температури. Вважали, що при цьому дотримується локальне рівновагу рідкої фази і нескінченно тонкого шару твердої фази, що виділився при цьому зниженні температури. Переходячи до меж, отримали диференціальне рівняння, що описує досліджуваний процес. Рішення цього рівняння було отримано у вигляді інтегральної функції маси затверділого сплаву від температури. Так як маса затверділого сплаву при наших припущеннях однозначно пов'язана з його масою, рішення всієї задачі визначення складу зерна в залежності від відстані до його центру було отримано у вигляді параметричної функції, що виражає радіус поточної точки зерна і його склад у цій точці через температуру. Складено комп'ютерна програма розрахунку за рівняннями математичної моделі. Для використання моделі потрібно знати склад вихідного розплаву, середню щільність твердої фази і рівняння ліній ліквідусу і солідусу у вигляді функцій складу від температури. Представлений приклад розрахунку.
Область наук:
• хімічні науки
• Рік видавництва: 2019
Журнал: Вісник Південно-Уральського державного університету. Серія: Металургія
## Текст наукової роботи на тему «A mathematical model of Changes in the composition of grains during cooling a two-component melt»
?DOI: 10.14529 / met190107
A MATHEMATICAL MODEL OF CHANGES IN THE COMPOSITION OF GRAINS DURING COOLING A TWO-COMPONENT MELT
A.D. Drozin, Ця електронна адреса захищена від спам-ботів. Вам потрібно увімкнути JavaScript, щоб побачити її.,
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A.V. Vorobyov, Ця електронна адреса захищена від спам-ботів. Вам потрібно увімкнути JavaScript, щоб побачити її.,
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South Ural State University, Chelyabinsk, Russian Federation
The growth of individual grains of a solid phase during cooling liquid melt has been studied. During cooling, the compositions of the liquid and solid phases and the equilibrium conditions change. Therefore, each subsequent layer, "freezing" on the grain, will have a slightly different composition. This paper proposes a method for calculating the structure of the grain changes with increasing distance from the center. Corresponding mathematical model has been created. It is based on the following assumptions. Growing grains are considered as spherical. The temperature alignment in the system and the composition alignment in the liquid phase occur instantly. The alignment of the solid phase composition does not occur. Also, it is assumed that local equilibrium of the liquid phase with the solid phase on the grain surface is observed at any temperature. The characteristics of this local equilibrium can be found out from the corresponding equilibrium state diagram. The balance equation of phase masses and masses of their components at infinitesimal temperature decrease was made. It was assumed that the local equilibrium of the liquid phase and the infinitely thin layer of the solid phase, formed during this decrease in temperature, is observed. Taking to the limits, we have obtained a differential equation describing the investigated process. The solution of this equation has been obtained in the form of the solid phase mass as an integral function of the temperature. The grain composition depending on the distance from its center is obtained in the form of parametric functions expressing the radius of the current grain point and its composition at this point depending on the temperature. A computer program for the calculation of the mathematical model equations have been created. To use the model, it is needed to know the composition of the initial melt, the average density of the solid phase and the equations of the liquidus and solidus lines in the form of functions of concentration vs the temperature. An example of the calculation is presented.
Keywords: mathematical model, physical metallurgy, state diagram, liquation.
Introduction
Taking into account the influence of liquation in the crystallization of metal is a longstanding problem of metallurgy [1-7]. One of the approaches to its study is the creation of an adequate mathematical model that takes into account the main factors affecting the process under study [8-10].
Consider the growth of a separate grain during cooling of a multicomponent melt, namely, changing its size and composition. State diagrams show how the equilibrium compositions of the liquid and solid phases change when the melt is cooled. However, such an equilibrium cooling can be achieved only if the phase composition and temperature have time to align with any decrease in temperature. This is only possible in two cases: the temperature decreases infinitely slow or have infinitely large thermal
conductivity and diffusion coefficients of each phase.
Therefore, only a local equilibrium observed at short distances from the phase boundary can be performed. During cooling of the melt from the temperature T to T + AT (AT < 0), a thin layer of the solid phase freezing (from the surrounding liquid phase) on the grain, will be in equilibrium corresponding to the new temperature composition. With the subsequent decrease in the temperature to T + 2AT, a new layer of a new composition corresponding to this new temperature will appear on the grain. With subsequent temperature decreases, the grain will be overgrown with new layers, which compositions correspond to the new temperatures, as shown in Fig. 1. In fact, we will not even deal with separate layers, but with a continuous change in composition as we go away from the grain center.
Fig. 1. Changes in grain composition as it grows
The purpose of this work was to create a model that allows calculation of the change in the composition of the grain as it goes away from its center.
1. Basic assumptions
The growth of individual grains, which are considered to be spherical is studied Accepted that:
1) the temperature alignment in all phases (liquid and solid) is instantaneous;
2) the composition alignment in the liquid phase is instantaneous;
3) the alignment of the solid phase composition does not occur.
2. Derivation of basic equations
Consider the model state diagram (Fig. 2).
Consider a melt A-B with the mass M0 and the fraction of the component B equaled c0. Let the system has been cooled to a temperature T. Let the mass of the solid phase be MS (T) and the number of identical grains in it be N. Let the mass of the liquid phase be ML (T), and
the fraction of the component B in it be cL (T).
Let the system now be further cooled to a temperature T + AT (AT < 0). In this case, the new
part of the solid phase with a mass AMs and
the composition cs (T), where T + AT <f<T,
will be formed. Then the mass of the liquid phase will become ML (T + AT) = ML (T) -AMS. Its
composition will be cL (T + AT).
Let's make the balance equation of the component B at the temperature T + AT:
AMscs (f) + (ML (T) -AMs) cL (T + AT) =
= ML (T) Cl (T) and obtain
, .Cj (T + AT) - Cj (T)
AM, = M, (T) ^ - H ^ f.
S} cL {T + AT) -cs (T)
Divide this equality by AT:
AMo
= ML (T)-
1
AT ^; cL (T + AT) -cs (T)
cl (T + AT) -Cl (T)
AT '
Let's go to the limits as AT 0. Taking into account that if AT 0 then 7 ^ '/'. \\ c get
dMQ
= Ml (T)
1
dcr
dT
In this case
cL (T) - cs (T) dT •
-L = cr and
dT L
1
cl (T) - cs (T) are known values, from equilibrium state diagram.
Because of ML (T) = M0 -Ms (T) then
dMs ~ dT
= (Mo -Ms (T))
cl (T) - cs (T) •
Having solved this differential equation we have obtained
fir, W
Ms = Mo
1 - exp
i
cl (T) - cs (T)
dT
Fig. 2. Model state diagram
J J
c
We have obtained the dependence of the the crystallized metal mass on the temperature. Assuming that the solid phase has the shape of a sphere of radius R and density p, we obtain the ratio
MS = 4% R3pN from which follows R (T) =
3MS (T) 4% pN
l
or
R (T) =
3Mn
4 ^ pN
1 - exp
"J
cs (T) - cL (T)
dT
J J J
Thus, for any T we can calculate the grain radius by this formula, and by the state diagram determine the composition of the corresponding layer. We now have the defined parametrically function
. (R) using the parameter T:
R (T) = С (t).
3Mn
4% pN
1 - exp
J
V TL
cS (T) - Cl (T)
dT
(1)
J J)
3. The calculation algorithm
To calculate, we should know the initial composition of the melt c0, the initial melt mass M0, the density of the solid phase p, the state diagram of the system (the equations of the diagram lines in the form of cL (T), cS (T)).
1. Let's find the the liquidus temperature TL and the solidus temperature TS.
2. Let's divide the interval [TS; TL] into equal parts by points T1, T2, ..., Tn_x, where T0 = TS, Tn = TL.
3. For the each point, calculate R (T), cS (T) by the formula (1).
4. Get a table of the values R (T), cS (T), on which we build a graph of the dependence cS (R). According to the above algorithm, a computer program was written in VBA Excel.
4. Calculation example
Consider a cooling of the tin-bismuth alloy (mas. 30% Bi)). Let's use the Sn-Bi state diagram [11] (Fig. 3).
V
Sn mole% Bi Bi
Fig. 3. Sn-Bi phase diagram
The selected composition corresponds to mole 19.576%., That is, between the points ESn and E. Thus, for the calculation we will need the equations of the liquidus ASnA1E and solidus ASnA2ESn lines in the form of c = ^ (T). To do this, we took these dependencies parabolic, found the coordinates of the points ASn, A1, E, ESn, A1, A2 on the state diagram, and transferred the concentrations into mass fractions. Further, by the points ASn, A1, E the equation of the liquidus line has been determined and by the points ASn, A2, ESn the equation of the solidus line has been determined. We have obtained
ASnA1E: c = -5,7 -10 "5 T2 +1,4883 • 10" 2 T - 0,403 84;
ASnA2ESn: c = 2,25 -10 "6 T2 - 3,18 • 10-3 T + 0,616724.
The calculation results are shown in Fig. 4-6. To ensure that they do not depend on arbitrarily specified values: the melt mass and the number of grains, some of the results are presented in a normalized form. So the current radius of the grain is presented in the form of its ratio to the maximum radius reached by the time of solidification of the entire melt.
In Fig. 4 the data on the the final composition of the grain are presented. Inconstancy of the grain composition is clearly visible. The alignment of the composition can occur only after a fairly long period of time.
g_i_i_i_i_i_i_i_i_i_
O 0.1 02 0.3 0.4 0.5 0.6 0.7 0.6 0.9 1
Fig. 4. The composition of the grain phase as a function of the distance to the grain center
Fig. 5, 6 show a comparison of the obtained results with the data that would be obtained directly from the equilibrium state diagrams, which would be realized if the cooling was infinitely slow (or if the diffusion and the thermal conductivity coefficients of both phases were infinitely large ).
From Fig. 5 it can be seen that if diffusion in the solid phase is absent, the mass of the crystallized metal is less than it follows from the equilibrium state diagram. When the supercooling is 60.0794 (the temperature drops out to 139 ° C), the eutectic precipitates and the "equilibrium" curve rises to 100%. What happens to previously formed growing grains is not considered in this paper.
Fig. 5. The fraction of the solid phase as a function of the corresponding supercooling (dotted line - the "equilibrium" case)
Fig. 6. The average composition of the solid phase in the absence of diffusion in solid phase (solid line) and in accordance with the equilibrium state diagram (dotted line)
Conclusions
The paper considers the limiting (not quite real) case of cooling of a two-component system in the case when the alignment of the solid phase composition does not occur is considered. In this case, there are a number of significant differ-
ences from the usual conclusions resulting from the equilibrium state diagram.
• At any supercooling, the mass of the solid phase is less than it follows from the equilibrium state diagram.
• At any supercooling the total fraction of
the dissolved substance in a solid phase is less than it follows from the equilibrium state diagram.
• The solidus temperature of a melt is decreases.
References
1. Roshchin V.E, Roshchin A.V. Elektro-metallurgiya i metallurgiya stali [Electrometallurgy and Metallurgy of Steel]. Chelyabinsk, South Ural St. Univ. Publ., 2013. 571 p.
2. Kushner V.S., Vereshchaka A.S., Skhirt-lazdze A.G., Negrov D.A., Burgonova O.Yu. Materialovedenie [Materials Technology]. Omsk, OGTU, 2008. 232 p.
3. Rabinovich S.V., Chermenskij V.I., Ogo-rodnikova O.M., Harchuk M.D. [Mathematical Modeling of Nickel Dendritic Liquation in Casting Invar and Superinvar Alloys]. Litejnoe proizvod-stvo [Foundry], 2002 no. 6, pp. 9-12. (In Russ.)
4. Zhukova S.Yu. Distribution of Chemical Elements in Dendrites and Inter-Dendritic Areas of Cast Metal. Metallurgist 2009, vol. 53, iss. 1-2, pp. 32-37. DOI: 10.1007 / s11015-009-9133-4
5. Martyushev N.V. [Effect of Crystallization Conditions on the Structure and Properties of Lead-Containing Bronzes]. Metallurgiya mashi-nostroeniya [Metallurgy of Mechanical Engineering] 2010, no. 4, pp. 32-36. (In Russ.)
6. Koltygin A.V., Belov V.D., Bazhenov V.E. Effect of the Specific Features of Solidification
of an ML10 Magnesium Alloy on the Zirconium Segregation during Melting. Russian Metallurgy (Metally), 2013, vol. 2013, iss. 1, pp. 66-70. DOI: 10.1134 / S0036029513010060
7. Kondratyuk S.E., Stoyanova E.N., Plyah-tur A.A., Parhomchuk Zg.V. [Structure and Segregation of Steel R6M5L Depending on the Crystallization Conditions]. Metallurgiya mashino-stroeniya [Metallurgy of Mechanical Engineering], 2014 року, no. 1, pp. 23-25. (In Russ.)
8. Gamov P.A., Drozin A.D., Dudorov M.V., Roshchin V.E. Model for Nanocrystal Growth in an Amorphous Alloy. Russian Metallurgy (Metally) 2012, no. 11, pp. 1002-1005. DOI: 10.1134 / S0036029512110055
9. Drozin A.D., Roshchin V.E. [Thermodynamics of Nucleation of Heterophase Chemical Reactions Products in Liquid Solutions]. Bulletin of South Ural State University. Mathematics, Physics, Chemistry, 2002 vol. 16, no. 4, pp. 54-66. (In Russ.)
10. Drozin A.D., Dudorov M.V., Roshchin V.E., Gamov P.A., Menihes L.D. [Mathematical Model of Crystal Nuclei Formation in Supercooled Melt of Eutectic Composition]. Bulletin of South Ural State University. Mathematics, Physics, Chemistry 2012, no. 6, pp. 66-77. (In Russ.)
11. Lyakishev N.P. Diagrammy sostoyaniya dvojnyh metallicheskih sistem [State Diagrams of Double Metal Systems]. Moscow, Mashinostroe-nie Publ., 1996. 996 p.
УДК 669.017 + 51 -74 DOI: 10.14529 ^ М 90107
МАТЕМАТИЧНА МОДЕЛЬ ЗМІНИ СКЛАДУ ЗЕРНА при охолодженні двокомпонентних РОЗПЛАВУ
А.Д. Дрозін, С.Р. Найман, Н.Є. Кочетов, А.В. Воробйов, І.А. Моїсеєв
Південно-Уральський державний університет, м Челябінськ, Росія
Розглядається зростання окремих зерен твердої фази при охолодженні рідкого розплаву. У міру охолодження змінюються склади рідкої і твердої фаз і умови рівноваги. Тому кожен наступний шар, «намерзає» на зерно, матиме дещо інший склад. У даній роботі запропоновано метод розрахунку зміни складу зерна в міру віддалення від його центру. Для цього розроблено математичну модель, що базується на наступних припущеннях: зростаюче зерно вважається сферичним; вирівнювання температури в системі і вирівнювання складу рідкої фази відбуваються моментально; вирівнювання складу твердої фази не відбувається. При цьому вважали, що при будь-якій температурі дотримується локальне
рівновагу рідкої фази і поверхневого шару твердої фази. Характеристики цього локального рівноваги можуть бути визначені з відповідною рівноважної діаграми стану. Було складено рівняння балансу мас фаз і мас їх компонентів при нескінченно малому зниженні температури. Вважали, що при цьому дотримується локальне рівновагу рідкої фази і нескінченно тонкого шару твердої фази, що виділився при цьому зниженні температури. Переходячи до меж, отримали диференціальне рівняння, що описує досліджуваний процес. Рішення цього рівняння було отримано у вигляді інтегральної функції маси затверділого сплаву від температури. Так як маса затверділого сплаву при наших припущеннях однозначно пов'язана з його масою, рішення всієї задачі - визначення складу зерна в залежності від відстані до його центру - було отримано у вигляді параметричної функції, що виражає радіус поточної точки зерна і його склад у цій точці через температуру. Складено комп'ютерна програма розрахунку за рівняннями математичної моделі. Для використання моделі потрібно знати склад вихідного розплаву, середню щільність твердої фази і рівняння ліній ліквідусу і солідусу у вигляді функцій складу від температури. Представлений приклад розрахунку.
Ключові слова: математична модель, металознавство, діаграма стану, ізоляція.
література
1. Рощин, В.Є. Електрометалургія і металургія стали / В.Є. Рощин, А.В. Рощин. - Челябінськ, ЮУрГУ, 2013. - 571 с.
2. Матеріалознавство: навч. для студентів вузів / В.С. Кушнер, А.С. Верещака. А.Г. Схірт-ладзе і ін .; під ред. В.С. Кушнера. - Омськ: Изд-во ОмГТУ, 2008. - 232 с.
3. Математичне моделювання дендритних ліквації нікелю в ливарних інварних і суперінварних сплавах / С.В. Рабинович, В.І. Черменський, О.М. Огороднікова, М.Д. Харчук // Ливарне виробництво. - 2002. - № 6. - P. 9-12.
4. Жукова, С.Ю. Розподіл хімічних елементів в дендритах і междендрітних ділянках литого металу / С.Ю. Жукова // Металург. - 2009. - № 1. - С. 46-49.
5. Мартюшев, Н.В. Вплив умов кристалізації на структуру та властивості бронз, що містять свинець /Н.В. Мартюшев // Металургія машинобудування. - 2010. - № 4 - С. 32-36.
6. Колтигін, А.В. Вплив особливостей кристалізації магнієвого сплаву МЛ10 на ликвацию цирконію в процесі плавки / А.В. Колтигін, В.Д. Бєлов, В.Є. Баженов // Метали. - 2013. -№ 1. - С. 78-83. DOI: 10.1134 / S0036029513010060
7. Структура і сегрегація стали Р6М5Л в залежності від умов кристалізації / С.Є. Кондратюк, Е.Н. Стоянова, А.А. Пляхтур, З.В. Пархомчук // Металургія машинобудування. - 2014. - № 1. - С. 23-25.
8. Model for Nanocrystal Growth in an Amorphous Alloy / P.A. Gamov, A.D. Drozin, M. V. Du-dorov, V.E. Roshchin // Russian Metallurgy (Metally). - 2012. - No. 11. - P. 1002-1005. DOI: 10.1134 / S0036029512110055
9. Дрозін, А.Д. Термодинаміка утворення зародків продуктів гетерофазних хімічних реакцій в рідких розчинах / А.Д. Дрозін, В.Є. Рощин // Вісник ЮУрГУ. Серія «Математика, фізика, хімія». - 2002. - Т. 16, № 4. - С. 54-66.
10. Математична модель освіти кристалічних зародків в переохолодженому розплаві евтектичного складу / А.Д. Дрозін, М.В. Дудоров, В.Є. Рощин та ін. // Вісник ЮУрГУ. Серія «Математика, фізика, хімія». - 2012. - № 6. - С. 66-77.
11. Лякишев, Н.П. Діаграми стану подвійних металевих систем / Н.П. Лякишев. -М. : Машинобудування, 1996. - 996 с.
Дрозін Олександр Дмитрович, д-р техн. наук, професор, директор Центру елітного освіти, Південно-Уральський державний університет, м Челябінськ; Ця електронна адреса захищена від спам-ботів. Вам потрібно увімкнути JavaScript, щоб побачити її..
Найман Софія Романівна, студент групи Е-2 елітного освіти Політехнічного інституту, Південно-Уральський державний університет, м Челябінськ; Ця електронна адреса захищена від спам-ботів. Вам потрібно увімкнути JavaScript, щоб побачити її..
Кочетов Микита Євгенович, студент групи Е-2 елітного освіти Політехнічного інституту, Південно-Уральський державний університет, м Челябінськ; шкйакосЬе1; оу98 @ mail.ru.
Воробйов Артем Володимирович, студент групи Е-2 елітного освіти Політехнічного інституту, Південно-Уральський державний університет, м Челябінськ; АГ; етуогоЬуеу1999 @ mail.ru.
Моїсеєв Ігор Олексійович, студент групи Е-2 елітного освіти Політехнічного інституту, Південно-Уральський державний університет, м Челябінськ; moiseevigor.99 @ mail.ru.
Надійшла до редакції 17 січня 2019 р.
ЗРАЗОК ЦИТУВАННЯ
FOR CITATION
A Mathematical Model of Changes in the Composition of Grains during Cooling a Two-Component Melt / A.D. Drozin, S.N. Naiman, N.E. Kochetov et al. // Вісник ЮУрГУ. Серія «Металургія». - 2019. - Т. 19, № 1. - С. 59-66. DOI: 10.14529 / met190107
Drozin A.D., Naiman S.N., Kochetov N.E., Vorobyov A.V., Moiseev I.A. A Mathematical Model of Changes in the Composition of Grains during Cooling a Two-Component Melt. Bulletin of the South Ural State University. Ser. Metallurgy, 2019, vol. 19, no. 1, pp. 59-66. DOI: 10.14529 / met190107
Ключові слова: MATHEMATICAL MODEL /PHYSICAL METALLURGY /STATE DIAGRAM /LIQUATION /МАТЕМАТИЧНА МОДЕЛЬ /МЕТАЛОЗНАВСТВО /ДІАГРАМА СТАНУ /ізоляція
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# A Java Approach: Selection Structures - Use Cases
Hello everyone and welcome back! Up to now we have been concerned to make as complete an overview as possible of the fundamental concepts we need to approach the use of conditional structures. We have therefore introduced the concept of conditional structures, outlining the various situations that could arise. We then went on to outline the main features of Boolean algebra, analysing the use of Boolean variables and Boolean operators.
One concept that is worth repeating is that conditional structures, also called flow control structures, serve to modify the normal flow of program execution.
Since learning to program is not enough to read, let's go and analyse some practical examples so that we can see in action the concepts we have analysed so far. It must be said that, in examining some examples, I will take for granted the theoretical and practical aspects concerning the variables, focusing on the aspects concerning the selection structures.
### Case 1: Nested selection structures
A first question that could be asked is the case of the nested selection structures. One might wonder whether a second if can be grafted within an if.
Let's see how.
```// check that n1 is greater than or equal to n2
if (n1 >= n2) {
// check that n1 is greater than or equal to n3
if (n1 >= n3) {
largest = n1;
}
else {
largest = n3;
}
} else {
// check that n2 is greater than or equal to n3
if (n2 >= n3) {
largest = n2;
}
else {
largest = n3;
}
}
System.out.println("Largest Number: " + largest);
```
This example may seem complex on the surface, but if we read it carefully, we realize that the task it performs is quite simple: it is concerned with finding the variable that has the maximum value within a set of three variables.
To do this, the reasoning is as follows: to be the maximum, one value must be greater than the other two. So what happens is that you check if n1 is greater than n2. If this is true, then you check that n1 is also greater than n3. If two conditions are also true, then you value the variable largest with the value contained in n1.
If n1 is less than n2, we know that n1 cannot be the maximum. The check with n2 and n3 therefore remains to be done. If n2 is greater than n3, then n2 is the maximum of the three variables. Otherwise, the maximum is n3.
We can easily see that it is possible to engage more if one inside the other. This is allowed, though not always encouraged.
### Syntax note: the curly brackets
Technically, Java allows us to omit the curly brackets from both the if blocks and the other blocks, as long as they contain only one instruction and no more.
Often this practice is used, and I admit that I do it myself: it has to be said that, in my opinion, we are losing legibility and clarity. Obviously, we are going to gain in compactness of the code. In the case of nested ifs, I recommend using them all the time.
### Case 2: Selection structures with composite conditions
Another interesting case to analyse is the one where Boolean operators are used within the conditions. The classic example that can be analysed is one where you want to check that a given number falls within a range. Think, for example, of a software to make statistical evaluations of the school performance of a class.
Suppose we want to count how many grades are sufficient on a scale from 1 to 10, i.e. with a value greater or equal to six.
So let's see how we can do this, assuming we have already declared and initialised a variable called grade.
```int counter = 0;
counter = counter + 1;
}
```
The idea is therefore that a grade should be considered sufficient if and only if it is greater than or equal to 6 and at the same time less than or equal to 10. On a first attempt, someone might argue that a simple grade check >= 6 is sufficient. The problem arises, however, if there is an insertion error and a grade greater than ten is inserted. Our application would consider it sufficient, even if it is not a valid grade. One would then find oneself considering legitimate a situation that should be handled as a problem.
### Case 3: condition with negation
Often, it may be convenient to use conditions that contain a denial within them. This may be for different reasons. It may be developmental convenience or mere readability.
Let's see an example. As a case of use we can always keep the case of grade analysis. Let's assume that we want to check that a grade is valid. A solution might be similar to the one produced before. A far more readable and simple solution, in my opinion, is the following.
```boolean voteValid = vote >= 6 && vote <= 10;
if(!voteValid){
//Manage error
}
```
So we notice that the code here is very easy to understand. In natural language, the condition can be read as "if not votoValido", which translated into Italian natural language, can be "if the vote is not valid". So we understand that the if will be executed only if the vote is not valid. This is the case when using a Boolean variable greatly simplifies code development.
### Conclusions
This brief examination leaves us with some salient aspects. In particular, let us see how operators can be combined to form complex expressions.
It is important, in these situations as in the rest of programming, to pay attention to code indentation. The risk is to write a code that is very complicated to read, if you don't insert the necessary spacing at the beginning of the line.
We will see later on that there is a valid alternative to the massive use of if constructs: the switch statement.
It should be pointed out and reiterated that the analysis made is by no means exhaustive. The idea is to first give the fundamental tools to start writing code, and then to deepen some peculiar aspects.
To practice, it may be useful to solve the following exercise:
Write a SmallEven program that asks the user to enter a whole number and displays the message "Even and Small" if the number is even and is between 0 and 100, otherwise it prints the message "Not even and small". (Note: this program can be done using nested if-else or Boolean expressions. Try both versions).
The exercise comes from the following document.
by Alessio Mungelli Date: 20-10-2020 java jdk jre structures conditionals if usecase tutorial source code developement hits : 1538
#### Alessio Mungelli
Computer Science student at UniTo (University of Turin), Network specializtion, blogger and writer. I am a kind of expert in Java desktop developement with interests in AI and web developement. Unix lover (but not Windows hater). I am interested in Linux scripting. I am very inquisitive and I love learning new stuffs.
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# Finding the matrix of $T(p(x)) = p(2x-1)$ with respect to the basis $B =$ {$1+x, 1-x, x^2$}
Finding the matrix of $$T(p(x)) = p(2x-1)$$ with respect to the basis $$B =$$ {$$1+x, 1-x, x^2$$}
To find the matix of a transformation with respect to a given basis, I find the images of the basis vectors under the transformation, then use the columns of those images with respect to $$R^n$$ as the columns of the matrix I'm looking for. Here's what I mean:
$$T(1+x) = 1 + (2x - 1) = 2x$$
$$T(1-x) = 1 - (2x - 1) = 2 - 2x$$
$$T(x^2) = 1 - (2x - 1)^2 = 1 - 4x + 4x^2$$
With respect to the basis {$${e_1, e_2, e_3}$$} of $$R^3$$, those vectors would be:
$$[0,2,0], [2,-2,0],[1,-4,4]$$
respectively.
And so the matrix of $$T$$ with respect to $$B$$ is:
$$\left[ \begin{array}{ccc} 0&2&1\\ 2&-2&-4\\ 0&0&4 \end{array} \right]$$
This is apparently completely wrong. The correct matrix is:
$$\left[ \begin{array}{ccc} 1&0&-3/2\\ -1&2&5/2\\ 0&0&4 \end{array} \right]$$
My approach has been working thus far, but here it fails. Is there something different about this example that makes it a situation where I can not apply my strategy? What exactly is it that I found (if anything)?
Any help at all is greatly appreciated.
You should express the image of each vector of $$B$$ as a linear combination of elements of $$B$$. For instance\begin{align}T(1+x)&=2x\\&=(1+x)-(1-x)\\&=1\times(1+x)+(-1)\times(1-x)+0\times x^2.\end{align}That's why the entries of the first column of the matrix are $$1$$, $$-1$$, and $$0$$.
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https://us.metamath.org/mpeuni/hlhilsbase2.html
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Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > hlhilsbase2 Structured version Visualization version GIF version
Theorem hlhilsbase2 39138
Description: The scalar base set of the final constructed Hilbert space. (Contributed by NM, 22-Jun-2015.) (Revised by Mario Carneiro, 28-Jun-2015.)
Hypotheses
Ref Expression
hlhilsbase.h 𝐻 = (LHyp‘𝐾)
hlhilsbase.l 𝐿 = ((DVecH‘𝐾)‘𝑊)
hlhilsbase.s 𝑆 = (Scalar‘𝐿)
hlhilsbase.u 𝑈 = ((HLHil‘𝐾)‘𝑊)
hlhilsbase.r 𝑅 = (Scalar‘𝑈)
hlhilsbase.k (𝜑 → (𝐾 ∈ HL ∧ 𝑊𝐻))
hlhilsbase2.c 𝐶 = (Base‘𝑆)
Assertion
Ref Expression
hlhilsbase2 (𝜑𝐶 = (Base‘𝑅))
Proof of Theorem hlhilsbase2
StepHypRef Expression
1 hlhilsbase2.c . . 3 𝐶 = (Base‘𝑆)
2 hlhilsbase.k . . . . 5 (𝜑 → (𝐾 ∈ HL ∧ 𝑊𝐻))
3 hlhilsbase.h . . . . . 6 𝐻 = (LHyp‘𝐾)
4 eqid 2824 . . . . . 6 ((EDRing‘𝐾)‘𝑊) = ((EDRing‘𝐾)‘𝑊)
5 hlhilsbase.l . . . . . 6 𝐿 = ((DVecH‘𝐾)‘𝑊)
6 hlhilsbase.s . . . . . 6 𝑆 = (Scalar‘𝐿)
73, 4, 5, 6dvhsca 38278 . . . . 5 ((𝐾 ∈ HL ∧ 𝑊𝐻) → 𝑆 = ((EDRing‘𝐾)‘𝑊))
82, 7syl 17 . . . 4 (𝜑𝑆 = ((EDRing‘𝐾)‘𝑊))
98fveq2d 6655 . . 3 (𝜑 → (Base‘𝑆) = (Base‘((EDRing‘𝐾)‘𝑊)))
101, 9syl5eq 2871 . 2 (𝜑𝐶 = (Base‘((EDRing‘𝐾)‘𝑊)))
11 hlhilsbase.u . . 3 𝑈 = ((HLHil‘𝐾)‘𝑊)
12 hlhilsbase.r . . 3 𝑅 = (Scalar‘𝑈)
13 eqid 2824 . . 3 (Base‘((EDRing‘𝐾)‘𝑊)) = (Base‘((EDRing‘𝐾)‘𝑊))
143, 4, 11, 12, 2, 13hlhilsbase 39135 . 2 (𝜑 → (Base‘((EDRing‘𝐾)‘𝑊)) = (Base‘𝑅))
1510, 14eqtrd 2859 1 (𝜑𝐶 = (Base‘𝑅))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2115 ‘cfv 6336 Basecbs 16472 Scalarcsca 16557 HLchlt 36546 LHypclh 37180 EDRingcedring 37949 DVecHcdvh 38274 HLHilchlh 39128 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-rep 5171 ax-sep 5184 ax-nul 5191 ax-pow 5247 ax-pr 5311 ax-un 7444 ax-cnex 10578 ax-resscn 10579 ax-1cn 10580 ax-icn 10581 ax-addcl 10582 ax-addrcl 10583 ax-mulcl 10584 ax-mulrcl 10585 ax-mulcom 10586 ax-addass 10587 ax-mulass 10588 ax-distr 10589 ax-i2m1 10590 ax-1ne0 10591 ax-1rid 10592 ax-rnegex 10593 ax-rrecex 10594 ax-cnre 10595 ax-pre-lttri 10596 ax-pre-lttrn 10597 ax-pre-ltadd 10598 ax-pre-mulgt0 10599 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ne 3014 df-nel 3118 df-ral 3137 df-rex 3138 df-reu 3139 df-rab 3141 df-v 3481 df-sbc 3758 df-csb 3866 df-dif 3921 df-un 3923 df-in 3925 df-ss 3935 df-pss 3937 df-nul 4275 df-if 4449 df-pw 4522 df-sn 4549 df-pr 4551 df-tp 4553 df-op 4555 df-uni 4820 df-int 4858 df-iun 4902 df-br 5048 df-opab 5110 df-mpt 5128 df-tr 5154 df-id 5441 df-eprel 5446 df-po 5455 df-so 5456 df-fr 5495 df-we 5497 df-xp 5542 df-rel 5543 df-cnv 5544 df-co 5545 df-dm 5546 df-rn 5547 df-res 5548 df-ima 5549 df-pred 6129 df-ord 6175 df-on 6176 df-lim 6177 df-suc 6178 df-iota 6295 df-fun 6338 df-fn 6339 df-f 6340 df-f1 6341 df-fo 6342 df-f1o 6343 df-fv 6344 df-riota 7096 df-ov 7141 df-oprab 7142 df-mpo 7143 df-om 7564 df-1st 7672 df-2nd 7673 df-wrecs 7930 df-recs 7991 df-rdg 8029 df-1o 8085 df-oadd 8089 df-er 8272 df-en 8493 df-dom 8494 df-sdom 8495 df-fin 8496 df-pnf 10662 df-mnf 10663 df-xr 10664 df-ltxr 10665 df-le 10666 df-sub 10857 df-neg 10858 df-nn 11624 df-2 11686 df-3 11687 df-4 11688 df-5 11689 df-6 11690 df-7 11691 df-8 11692 df-n0 11884 df-z 11968 df-uz 12230 df-fz 12884 df-struct 16474 df-ndx 16475 df-slot 16476 df-base 16478 df-sets 16479 df-plusg 16567 df-starv 16569 df-sca 16570 df-vsca 16571 df-ip 16572 df-dvech 38275 df-hlhil 39129 This theorem is referenced by: hlhils0 39141 hlhils1N 39142 hlhillvec 39147 hlhilsrnglem 39149 hlhilphllem 39155
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1. ## Evaluate the limit..
1. The problem statement, all variables and given/known data
lim.......... (1 + 1/n^2)(7/n+1)
n->infinity
3. The attempt at a solution
lim.......... (1 + 1/n^2)(7/n+1)
n->infinity
= (7/n + 1) + (7/n^3 + n^2)
bring out 7 because constant..
7lim.......... (1/n + 1) + (1/n^3 + n^2)
n->infinity
.
.
.
.
.
7lim.......... (n^2 + n + 1) / (n^3 + 2n^2 +n)
n->infinity
limit does not exist.. am i correct?
2. $\displaystyle \displaystyle \lim_{n\rightarrow\infty}\left(1+\frac{1}{n^{2}}\r ight)\left(\frac{7}{n}+1\right)=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)\lim_{n\right arrow\infty}\left(\frac{7}{n}+1\right)$
$\displaystyle =\left(\lim_{n\rightarrow\infty}1+\lim_{n\rightarr ow\infty}\frac{1}{n^{2}}\right)\left(\lim_{n\right arrow\infty}\frac{7}{n}+\lim_{n\rightarrow\infty}1 \right)=(1+0)(0+1)=1$
3. Originally Posted by nameck
1. The problem statement, all variables and given/known data
lim.......... (1 + 1/n^2)(7/n+1)
n->infinity
3. The attempt at a solution
lim.......... (1 + 1/n^2)(7/n+1)
n->infinity
= (7/n + 1) + (7/n^3 + n^2)
Note that as the limit goes to infinity that
$\displaystyle \frac{1}{n^2}$ and $\displaystyle \frac{7}{n} \to\0$
bring out 7 because constant..
7lim.......... (1/n + 1) + (1/n^3 + n^2)
n->infinity
.
.
.
.
.
7lim.......... (n^2 + n + 1) / (n^3 + 2n^2 +n)
n->infinity
limit does not exist.. am i correct?
First let me ask that this does not represent a series? as you can perform a variety of test on different series to determine convergence, as the standard variable used in a series to n.
$\displaystyle \lim_{n\to\infty}(1+\frac{1}{n^2})(\frac{7}{n}+1)$
Note that:$\displaystyle \frac{1}{n^2}$ and $\displaystyle \frac{7}{n} \to 0$
as we goes towards infinity leaving us with 1 * 1 = 1 the limit is 1
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https://www.cfd-online.com/Forums/main/191792-how-compute-weno-reconstructed-flux-local-characteristic-field.html
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# How to compute WENO-reconstructed flux in local characteristic field?
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August 20, 2017, 04:02 How to compute WENO-reconstructed flux in local characteristic field? #1 Member Oleg Sutyrin Join Date: Feb 2016 Location: Russia Posts: 41 Rep Power: 9 I'm trying to apply finite-difference characteristic-wise WENO method to 1D Euler equations: Physical values are set at grid points , so we need to approximate fluxes at mid-points : where are sought-for approximation of physical fluxes . My algorithm is the following: 1) For each , we calculate simple average state and, using it, local eigenvalues , right eigenvector matrix and it's left counterpart . (Subscript will be omitted below) 2) Now we transform , its differences and flux differences to local characteristic field: It is done only for relevant grid points which in my case () are . 3) Then we reconstruct characteristic variable values by WENO method using , and . We get two "candidates": where are a convex combinations of corresponding polynomial functions obtained by WENO method. The next step would be to compute 2-point flux function using these reconstructed values. I'm using Lax-Friedrich's function: where is maximum value of eigenvalue (different for each of equations, since they are decoupled now). But how to compute these and using ? In component-wise approach it is simple: we reconstruct base values - , for example - and the use explicit formulas to compute and , but we don't have such formulas for because is expressed in local characteristic field...
August 22, 2017, 06:03 #2 Member Oleg Sutyrin Join Date: Feb 2016 Location: Russia Posts: 41 Rep Power: 9 It seems that I was misunderstanding how the flux is reconstructed: My idea was to reconstruct base values - (or if we are performing characteristic decomposition) - and then substitute them to formulas of (or ) to obtain reconstructed fluxes. It looks like the correct way is to apply reconstruction procedure to (or which is obtained by transformation ) itself. Some quick tests I performed with this approach show very plausible results.
September 4, 2017, 14:24 #3 Senior Member Join Date: Sep 2015 Location: Singapore Posts: 102 Rep Power: 10 Hi there, Here's what I am doing: after reconstructing the characteristic variables at the face, I use the left eigenvector matrix to convert them back to primitive variables. Then, I use these to compute the flux using the flux function, e.g. Lax-Friedrich method. It has performed well in several 1D test cases. USV
Tags charateristic-wise, finite-difference, weno
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https://practice.geeksforgeeks.org/problems/k-sum-in-a-linked-list/0
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##### Submissions: 86 Accuracy: 1.17% Max. Score: 100
1. Consider nodes in groups of size k. In every group, replace value of first node with group sum.
2. Also, delete the elements of group except the first node.
3. During traversal, if the remaining nodes in linkedlist are less than k then also do the above considering remaining nodes.
Input Format:
First line of input contains number of testcases T. For each testcase, first line contains length of linked list N and K. Second line contains elements of linked list.
Output Format:
For each testcase, print the modified linked list.
Constraints :
1 <= T <= 100
1 <= N <= 104
1 <= Elements of Linked List <= 106
1 <= K <= N
Example :
Input :
1
6 2
1 2 3 4 5 6
Output:
3 7 11
Explanation:
Testcase 1: k = 2.
We have denoted grouping of k elements by (). The elements inside () are summed.
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null
(1 -> 2) -> 3 -> 4 -> 5 -> 6 -> null
(3) -> 3 -> 4 -> 5 -> 6 -> null
3 -> (3 -> 4) -> 5 -> 6 -> null
3 -> (7) -> 5 -> 6 -> null
3 -> 7 -> (5 -> 6) -> null
3 -> 7 -> (11) -> null
3 -> 7 -> 11 -> null
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02 April 2016
RRB - NTPC Exam Questions Held in March - April 2016
1. Pollination by wind is called?
A) Hydrophily B) Pollinophily
C) Anemophily D) Herbophily
2. The oath of office of a President of India is conducted by .......
A) Prime Minister
B) Speaker of Lok Sabha
C) Chief Election Commissioner
D) Chief Justice of India
3. Which of the following natural events cause an earthquake?
A) Grass hopper invasions
B) Hurricane C) Eclipse of the moon
D) Volcanic eruptions
4. If your weight is 38 kg on Earth, how much would you weigh on the planet Mercury?
A) 19 kg B) 760 kg
C) 10 kg D) 14.3 kg
5. There are 3 heaps of rice weighing 120 kg. 144 kg and 204 kg. Find the maximum capacity of a bag so that the rice of each heap can be packed in exact number of bags?
A) 12 B) 10 C) 15 D) 18
6. Who wrote Malavikagnimitram?
A) Bhasa B) Kaviraya
C) Banabhatta D) Kalidasa
7. Which revolutionary took his own life?
A) Khudiram Bose B) Rash Behari Bose
C) Bhagat Singh D) Chandrashekhar Azad
8. Find the area of a circular field with a circumference of 22 cm.
A) 22 sq.cm B) 11 sq.cm
C) 44 sq.cm D) 38.5 sq.cm
9. The HCF of two numbers is 24, which of the following can be its LCM?
A) 118 B) 144 C) 128 D) 136
10. A dealer sell two chairs for Rs.462. On one he gains 12% and on the other loses 12%. Which of the following is true?
A) He gains Rs.110
B) He gains Rs.13.50
C) He neither gains nor loses
D) He loses Rs.13.50
11. For what value of x, if mode for the following data would be 52 ?
52, 45, 49, 54, 56, 52, x − 3, 56
A) 52 B) 55 C) 54 D) 56
12. What is the retirement age of Supreme Court judges?
A) 61 years B) 63 years
C) 65 years D) 68 years
13. Which individual has won the maximum gold medals in an Olympic games?
A) Mark Spitz B) Michael Phelps
C) Matt Biondi D) Michael Smith
14. In an examination the mean of marks scored by a class of 40 students was calculated as 72.5. Later on it was discovered that marks of one student was wrongly taken as 48 instead of 84. Find the corrected mean?
A) 78.25 B) 60.25 C) 72.70 D) 73.4
15. What is Graphene?
A) An allotrope of carbon
B) A popular graphics software
C) A mythical animal
D) An infected wound
16. Which of the following has the maximum number of legs?
A) Spider B) Millipede
C) Centipede D) Dragonfly
17. What is the maximum runs a batsman can score in an over and still retain strike for the next over, excluding no balls, wides or overthrows?
A) 36 B) 34 C) 35 D) 31
18. Below are given statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance with the commonly known facts.
Statements:
1. Apples are red.
2. No red colored fruits are cheap.
Conclusions:
I. All apples are cheap.
II. Red colored apples are not cheap.
Decide which of the given conclusions logically follow(s) from the given statements.
A) Only conclusion I follows
B) Only conclusion II follows
C) Both I and II follow
D) Neither I nor II follow
19. A lady introduces a man and says ''his daughter is the only grandchild of my father''. How is the man related to the lady?
A) Husband B) Father C) Uncle D) Son
20. Which is the only organ in the human body that can regrow, regenerate?
A) Spleen B) Brain C) Liver D) Pancreas
21. What can split sunlight into its constituent colours?
A) Refraction B) Reflection
22. Where can you find the Kunchikal waterfalls?
A) Kerala B) Karnataka
23. From a point 20 m away from the foot of a tower the angle of elevation from the top of the tower is 600. Find the height of the tower.
A) 30.6 m B) 34.6 m C) 36.4 m D) 36 m
24. A train is running at a speed of 160 km/hr, and it is 180 m long. In what time will it pass
a pole?
A) 4.05 sec B) 3 sec C) 3.2 sec D) 8 sec
25. Name the official language of Brazil?
A) Portugese B) German C) French D) Brazilian
26. Which is the brightest star in our night sky?
A) Canopus B) Sirius A C) Vega D) Spica
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Reduced Row Echelon Form Of A 4x3 Matrix
To find the reduced row echelon form of this matrix:
(1)
\begin{align} \left(\begin{array}{cc} 5 & 4 & 5 & 2 \\ 6 & 5 & 3 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}
1. We start by permuting the first two rows.
(2)
\begin{align} \left(\begin{array}{cc} 6 & 5 & 3 & 2\\ 5 & 4 & 5 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}
2. Subtract the 2nd row from the 1st.
(3)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 5 & 4 & 5 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}
3. Multiply the first row by -5 and add it to the 2nd.
(4)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & -1 & 15 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}
4. Multiply the first row by -7 and add it to the 3nd.
(5)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & -1 & 15 & 2 \\ 0 & 1 & 19 & 6 \end{array}\right) \end{align}
Now we are done with the first column with a pivot in the first row and zeros under that pivot.
5. We now permute the 2nd and 3rd row.
(6)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & -1 & 15 & 2 \end{array}\right) \end{align}
6. Add the 2nd row to the 3rd row.
(7)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & 0 & 34 & 8 \end{array}\right) \end{align}
7. Divide the 3rd row by 34.
(8)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}
8. Multiply the 3rd row by -19 and add it to the 2nd row.
(9)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}
9. Multiply the 3rd row by 2 and add it to the first row.
(10)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 0 & 8/17 \\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}
10. Multiply the 2nd row by -1 and add it to the first row.
(11)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & -18/17 \\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}
-Erick Rodriguez
page revision: 4, last edited: 03 Oct 2011 15:55
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Re: Why does this pattern fail to match?
• To: mathgroup at smc.vnet.net
• Subject: [mg114166] Re: Why does this pattern fail to match?
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 26 Nov 2010 05:26:14 -0500 (EST)
```expr = Times[-1, Power[s, -1]];
Attributes[Times]
{Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected}
Note that Times has attribute Orderless and pattern for Power can only have one or two blanks not three.
{MatchQ[expr, Times[___, Power[__]]],
MatchQ[expr, Times[___, Power[_]]],
MatchQ[expr, Times[__, Power[_]]],
MatchQ[expr, Times[Power[__], ___]],
MatchQ[expr, Times[Power[_], ___]],
MatchQ[expr, Times[Power[_], __]],
MatchQ[expr, Times[Power[__], _]]}
{True, True, True, True, True, True, True}
And, as you discovered in part
{MatchQ[expr, Times[___, _Power]],
MatchQ[expr, Times[__, _Power]],
MatchQ[expr, Times[_, _Power]],
MatchQ[expr, Times[_Power, ___]],
MatchQ[expr, Times[_Power, __]],
MatchQ[expr, Times[_Power, _]]}
{True, True, True, True, True, True}
However, given that these all work, it is not clear why the following fail
{MatchQ[expr, Times[_, Power[_]]],
MatchQ[expr, Times[__, Power[__]]],
MatchQ[expr, Times[Power[__], __]]}
{False, False, False}
Bob Hanlon
---- kj <no.email at please.post> wrote:
=============
I'm tearing my hair out over this one. Could someone please explain
to me why all the following MatchQ expressions
MatchQ[Times[ -1, Power[s, -1] ],
Times[___, Power[___], ___]]
MatchQ[Times[ -1, Power[s, -1]],
Times[___, Power[___] ]]
MatchQ[Times[ -1, Power[s, -1] ],
Times[___, _Power , ___]]
return False? (In all cases, s is undefined). And yet, this succeeds:
MatchQ[Times[ -1, Power[s, -1]],
Times[___, _Power ]]
Is there a systematic way to debug/troubleshoot such pattern matching
problems? Trace is useless here.
Also, whatever the reason is for the failure of this pattern to
match, where is this reason documented? I've gone blind poring
over the documentation trying to find an answer, without success.
TIA!
~kj
```
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Zener diode in PIXHawk
On the power connection page, it is suggested that we use zener diode for protection when using back up power supply. The connection suggested is to connect zener across the +5v and GND of the servo rails directly.
http://plane.ardupilot.com/wiki/common-pixhawk-wiring-and-quick-start/
However when I read about zener diode, then the circuit to control voltage is suggested as follows.
Notice the resistor in series with the zener. I understand that this resistor will consume the voltage which is excess to the required voltage. However in the connection suggested for PXHawk, there is no resistor. In this case how is it able to regulate the voltage? My knowledge of electronics is limited, so I would appreciate an explanation of this.
Views: 11625
Replies to This Discussion
The second circuit if for voltage control circuit. Note that here we are not using zener for voltage stabilization. If for some reason the voltage spikes above the safe limit then we want a pass circuit to prevent the PIXHawk from being damaged. This is the purpose of this zener. So you do not need resistor. Hope this helps.
Putting it another way, the zener will act as a very low value resistor if voltage gets too high and clamp the voltage, but it's not meant to do this for long periods of time since it will burn out if you send that amount of current through it for more than short spikes.
I have no idea how long my zener can withstand current before burning out though, whether it's milliseconds, seconds, or longer.
In the zener diode documentation the diagram is intended to drop the voltage from a higher voltage (say 10 volts) to 5 volts using a resisor and the zener, but in the pixhawk quick start guide, the zener is intended to cut off the voltage spikes that are for very short durations, and the voltage on average is alreay 5V...
If you look at the screen shot of the oscilloscope reading they provided:
The time frame is for one millisecond (that's the whole reading), and the spike is aproximatelly 1/10 of that, so this means that the spike duration is about 1/10,000 of a second, and a zenner diode can handle that with no problems.
I also suggest to put a capacitor in parallel with the zenner of any value between 100uF and 1,000uF (and the cap should be at least 6 volts of course), this way the zener will cut off the higher peaks, and the cap will make the voltage more smooth (more like a straight line)
By the way, the zener should not be exactly 5V, I think a 5.2V zener should be used... in the pixhawk quick start guide I see only one reference the zener voltage, and it says 5.6V, and then later on it says "digital servo causes the voltage on the rail to rise above the critical 5.7V level", so that's a little too close... only 0.1 voltage difference until it gets "critical"...
I just want to point out a mistake in the photo of the servo connector: it shows that the diode is connected between positive and signal, instead of ground.
You can probably flip a JR connector around in the pixhawk, so for safety sake, I'd recommend using a Futaba connector and the diode in the correct pins.
In relation to this..
http://plane.ardupilot.com/wiki/common-pixhawk-overview/#Powering
"To complement the diode, it is also useful to add a capacitorin parallel to the diode. The capacitor will smooth out eventual voltage ripples. As advised for the diode, the capacitor should be connected with as short wires as possible. Do not oversize the capacitor."
1) what is the recommended capacitory? Will a 25v 1000uf work or is that 'oversized'?
2) How do we parallel it.. should it be done this way?
--------(+)-----/[capacitor]----/[zener diode with rev polarity]
--------(-)------/------------------/
or this way:
--------(+)-----/[zener diode with rev polarity]----/[capacitor]
--------(-)------/-------------------------------------------/
thanks for the help!
I use a 6.3V 200uF capacitor.
Actually, in my case when using a redundant backup 5V supply to power the ESC output rail, my PixHawk will only boot with the capacitor added.
If I don't use a capacitor, the Pixhawk just makes cricket-like sounds with crazy flashing of the LEDs instead of booting. See it happen after 40secs here:
I suspect that it has something to do with the sequence in which voltage rises at the power module input and the ESC connector during power up, and the power selection mechanism in the Pixhawk.
The capacitor in my case, serves to slow the down the initial voltage rise at the ESC input.
I've reported it a month ago to 3DR support, but haven't heard anything of it since.
thanks.. may I know how you soldered it up? is it this way:
A) capacitor then zener in reverse
--------(+)-----/[capacitor]----/[zener diode with rev polarity]
--------(-)------/------------------/
B) zener in reverse then capacitor
--------(+)-----/[zener diode with rev polarity]----/[capacitor]
--------(-)------/-------------------------------------------/
If I'm interpreting your ASCII art correctly, then B makes no sense, because either your circuit is open, or you want to use the capacitor in series.
The capacitor must be connected in parallel, so it doesn't even matter in which ESC/servo plug you stick it. You can make one plug for the zener diode as shown in the documentation, and another one for the capacitor.
The only reason I'm even using a capacitor is simply to make it boot. Otherwise I wouldn't even bother, because I don't have noisy equipment connected to my Pixhawk. My ESCs are opto isolated too, so it's as clean as can be.
*slap on forehead* =) thanks for taking time to clarify that! I've been searching exactly what it means by 'parallel' and I just never thought of doing it in a separate servo lead!
Actually, A or B will work:
Hi,
I have the same issue on my Tricopter, I've had to power the servo rail via an external BEC. Cant explain why the BEC from the ESC is not liked yet the external one is all good.
I've tried a 200uf capacitor but it didnt work. I tried them 2 in parallel (for 400uf) and it didn't work, did you get a resolution in the end?
Chris
Hi guys,
Is this a suitable capacitor to use in parallel with the zener?
https://www.elfaelektronikk.no/elfa3~no_no/elfa/init.do?init=2&...
Cheers,
Sandy
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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
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let X be RealNormSpace; :: thesis: for Y being Subset of X
for Z being Subset of st Y = Z holds
Cl Y = Cl Z
let Y be Subset of X; :: thesis: for Z being Subset of st Y = Z holds
Cl Y = Cl Z
let Z be Subset of ; :: thesis: ( Y = Z implies Cl Y = Cl Z )
assume A1: Y = Z ; :: thesis: Cl Y = Cl Z
ex Z0 being Subset of st
( Z0 = Y & Cl Y = Cl Z0 ) by defcl;
hence Cl Y = Cl Z by A1; :: thesis: verum
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A254114 a(n) = A010060(A255056(n)). 4
0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 0 LINKS Antti Karttunen, Table of n, a(n) for n = 0..8590 FORMULA a(n) = A010060(A255056(n)). PROG (Scheme) (define (A254114 n) (A010060 (A255056 n))) CROSSREFS Binary complement: A254113. Cf. A010060, A255056, A255064. Sequence in context: A200244 A261185 A093692 * A105384 A288694 A292077 Adjacent sequences: A254111 A254112 A254113 * A254115 A254116 A254117 KEYWORD nonn AUTHOR Antti Karttunen, Feb 14 2015 STATUS approved
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Last modified August 13 17:42 EDT 2024. Contains 375144 sequences. (Running on oeis4.)
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Algorithmic Modelling with Grasshopper by Mohamad Khabazi
II
ALGORITHMIC MODELLING With GRASSHOPPER Introduction Have you ever played with LEGO Mindstorms NXT robotic set? Associative modelling is something like that! While it seems that everything tends to be Algorithmic and Parametric why not architecture? During my Emergent Technologies and Design (EmTech) master course in the Architectural Association (AA), I decided to share my experience in realm of Algorithmic design and Associative Modelling with Grasshopper as I found it a powerful platform for design in this way. I did this because it seems that the written, combined resources in this field are limited (although on‐line resources are quiet exciting). This is my first draft and I hope to improve it and I also hope that it would be helpful for you. Mohamad Khabazi © 2009 Mohamad Khabazi This book produced and published digitally for public use. No part of this book may be reproduced in any manner whatsoever without permission from the author, except in the context of reviews. m.khabazi@gmail.com www.khabazi.com/flux
III
Contents Chapter_1_Algorithmic Modelling ........................................................................................................ 1 Chapter_2_The very Beginning .............................................................................................................. 5 2_1_Method ....................................................................................................................................... 6 2_2_The very basics of Grasshopper .................................................................................................. 7 2_2_1_Interface, workplace ........................................................................................................... 7 2_2_2_Components ........................................................................................................................ 8 2_2_3_Data matching ................................................................................................................... 14 2_2_4_Component’s Help (Context pop‐up menu)...................................................................... 16 2_2_5_Type‐In component searching / adding ............................................................................ 17 2_2_6_Geometry Preview Method .............................................................................................. 17 2_3_Other Resources ........................................................................................................................ 18 Chapter_3_Data sets and Math ........................................................................................................... 19 3_1_Numerical Data sets .................................................................................................................. 20 3_2_On Points and Point Grids ......................................................................................................... 22 3_3_Other Numerical Sets ................................................................................................................ 23 3_4_Functions ................................................................................................................................... 25 3_5_Boolean Data types ................................................................................................................... 28 3_6_Cull Patterns .............................................................................................................................. 30 3_7_2D Geometrical Patterns ........................................................................................................... 35 Chapter_4_Transformation.................................................................................................................. 46 4_1_Vectors and planes .................................................................................................................... 48 4_2_On curves and linear geometries .............................................................................................. 49 4_3_Combined Experiment: Swiss Re ............................................................................................... 57 4_4_On Attractors ............................................................................................................................ 68
IV
Chapter_ 5_Parametric Space .............................................................................................................. 80 5_1_One Dimensional (1D) Parametric Space .................................................................................. 81 5_2_Two Dimensional (2D) Parametric Space .................................................................................. 83 5_3_Transition between spaces ....................................................................................................... 84 5_4_Basic Parametric Components .................................................................................................. 85 5_4_1_Curve Evaluation ............................................................................................................... 85 5_4_2_Surface Evaluation ............................................................................................................ 86 5_5_On Object Proliferation in Parametric Space ............................................................................ 88 Chapter_6_ Deformation and Morphing ............................................................................................. 96 6_1_Deformation and Morphing ...................................................................................................... 97 6_2_On Panelization ......................................................................................................................... 99 6_3_Micro Level Manipulations ..................................................................................................... 102 6_4_On Responsive Modulation ..................................................................................................... 106 Chapter 7_NURBS Surface and Meshes ............................................................................................. 112 7_1_Parametric NURBS Surfaces .................................................................................................... 113 7_2_Mesh vs. NURBS ...................................................................................................................... 124 7_2_1_Geometry and Topology ................................................................................................. 124 7_3_On Particle Systems ................................................................................................................ 126 7_4_On Colour Analysis .................................................................................................................. 135 7_5_Manipulating Mesh objects as a way of Design ...................................................................... 139 Chapter_8_Fabrication ....................................................................................................................... 141 8_1_Datasheets .............................................................................................................................. 143 8_2_Laser Cutting and Cutting based Fabrication .......................................................................... 155 Chapter_9_Design Strategy ............................................................................................................... 170 Bibliography .................................................................................................................................... 174
Chapter_1_Algorithmic Modelling
2
Associative Modelling
Chapter_1_Algorithmic Modelling If we look at architecture as an object represented in the space, we always deal with geometry and a bit of math to understand and design this object. In the History of architecture, different architectural styles have presented multiple types of geometry and logic of articulation and each period have found a way to deal with its geometrical problems and questions. Since computers started to help architects, simulate the space and geometrical articulations, it became an integral tool in the design process. Computational Geometry became an interesting subject to study and combination of programming algorithms with geometry yielded algorithmic geometries known as Generative Algorithm. Although 3D softwares helped to simulate almost any space visualized, it is the Generative Algorithm notion that brings the current possibilities of design, like ‘parametric design’ in the realm of architecture.
Contemporary architecture after the age of “Blob” seems to be even more complex. Architectural design is being affected by the potentials of algorithmic computational geometries with multiple hierarchies and high level of complexity. Designing and modelling free‐form surfaces and curves as building elements which are associated with different components and have multiple patterns is not an easy job to do with traditional methods. This is the time of algorithms and scripts which are forward pushing the limits. It is obvious that even to think about a complex geometry, we need appropriate tools, especially softwares, which are capable of simulating these geometries and controlling their properties. As the result architects feel interested to use Swarms or Cellular Automata or Genetic Algorithms to generate algorithmic designs and go beyond the current pallet of available forms and spaces. The horizon is a full catalogue of complexity and multiplicity that combines creativity and ambition together.
Chapter 1
Architects started to use free form curves and surfaces to design and investigate spaces beyond the limitations of the conventional geometries of the “Euclidian space”. It was the combination of Architecture and Digital that brought ‘Blobs’ on the table and then push it further. Although the progress of the computation is extremely fast, architecture has been tried to keep track with this digital fast pace progress.
3
Associative Modelling
Fig.1.1. Parametric Modelling for Evolutionary Computation and Genetic Algorithm, Mohamad khabazi, Emergence Seminar, AA, conducted by Michael Weinstock, fall 2008. A step even forward, now embedding the properties of material systems in design algorithms seems to be more possible in this parametric notion. Looking forward material effects and their responses to the hosting environment in the design phase, now the inherent potentials of the components and systems should be applied to the parametric models of the design. So not only these generative algorithms does not dealing only with form generation, but also there is a great potential to embed the logic of material systems in them. “The underlying logic of the parametric design can be instrumentalised here as an alternative design method, one in which the geometric rigour of parametric modelling can be deployed first to integrate manufacturing constraints, assembly logics and material characteristics in the definition of simple components, and then to proliferate the components into larger systems and assemblies. This approach employs the exploration of parametric variables to understand the behaviour of such a system and then uses this understanding to strategise the system’s response to environmental conditions and external forces” (Hensel, Menges, 2008).
Generally speaking, Associative modelling relates to a method in which elements of design being built gradually in multiple hierarchies and at each level, some parameters of these elements being extracted to be the generator for other elements in the next level and this goes on, step by step to produce the whole geometry. So basically the end point of one curve could be the center point of another circle and any change in the curve would change the circle accordingly. Basically this method of design deals with the huge amount of data and calculations and runs through the flow of algorithms. Instead of drawing objects, Generative Algorithmic modelling usually starts with numbers, mathematics and calculations as the base data to generate objects. Even starting with objects, it extracts parametric data of that object to move on. Any object of design has infinite positions inside,
Chapter 1
To work with the complex objects, usually a design process starts from a very simple first level and then other layers being added to it; complex forms are comprised of different hierarchies, each associated with its logics and details. These levels are also interconnected and their members affect each other and in that sense this method called ‘Associative’.
4
Associative Modelling
and these positions could be used as the base data for the next step and provide more possibilities to grow the design. The process called ‘Algorithmic’ because of this possibility that each object in the algorithm generated by previously prepared data as input and has output for other steps of the algorithm as well. The point is that all these geometries are easily adjustable after the process. The designer always has access to the elements of the design product from the start point up to details. Actually, since the design product is the result of an algorithm, the inputs of the algorithm could be changed and the result would also be updated accordingly. In conventional methods we used to modify models and designs on paper and model the final product digitally, to avoid changes which was so time‐ consuming. Any change in the design affected the other geometries and it was dreadful to fix the problems occurred to the other elements connected with the changed element and all those items should be re‐adjusted, re‐scaled, and re‐orientated if not happened to re‐draw. It is now possible to digitally sketch the model and generate hundreds of variations of the project by adjusting some very basic geometrical parameters. It is now possible to embed the properties of material systems, Fabrication constraints and assembly logics in parameters. It is now even possible to respond to the environment and be associative in larger sense. “… Parametric design enables the recognition of patterns of geometric behaviour and related performative capacities and tendencies of the system. In continued feedback with the external environment, these behavioural tendencies can then inform the ontogenetic development of one specific system through the parametric differentiation of its sub‐locations” (Hensel, Menges, 2008).
Grasshopper is a platform in Rhino to deal with this Generative Algorithms and Associative modelling. The following chapters are designed in order to combine geometrical subjects with algorithms and to address some design issues in architecture in an ‘Algorithmic’ method.
Chapter 1
Fig.1.2. A. form‐finding in membranes and minimal surfaces, physical model, B. membrane’s movement modelled with Grasshopper, Mohamad Khabazi, EmTech Core‐Studio, AA, Conducted by Michael Hensel and Achim Menges, fall 2008.
Chapter_2_The very Beginning
6
The very Beginning
Chapter 2
http://grasshopper.rhino3d.com/
7
The very Beginning
2_2_The very basics of Grasshopper 2_2_1_Interface, workplace Beside the other usual Windows menus, there are two important parts in the Grasshopper interface: Component panels and Canvas. Component panels provide all elements we need for our design and canvas is the work place. You can click on any object and click again on canvas to bring it to work place or you can drag it on to the work place. Other parts of the interface are easy to explore and we will be familiar with them throw using them later on. (If you like to know more, just go to http://grasshopper.rhino3d.com/2008/05/interface‐explained_26.html for more details)
Chapter 2
Fig.2.1. Grasshopper Component menu and Canvas.
8
The very Beginning
2_2_2_Components There are different types of objects in Grasshopper component menu which we use to design stuff. You can find them under nine different tabs called: Params, Logic, Scalar, Vector, Curve, Surface, Mesh, Intersect and XForm.
Parameters are objects that represent data, like a point or line. We can define them manually from Rhino objects as well. Components are objects that do actions with them like move, orientate, and decompose. We usually need to provide relevant data for them to work. In this manual I used the term component to talk about any objects from the component panel to make life easier! and I always use <> to address them clearly in the text, like <Point>.
<Point> component
Chapter 2
If you right‐click on a component a menu will pop‐up that contains some basic aspects of the component. This menu called “context pop‐up menu”.
Context pop‐up menu of <Pt> component
9
The very Beginning
Defining external geometries Most of the time we start our design by introducing some objects from Rhino workplace to the Grasshopper; A point, a curve, a surface up to multiple complex objects to work on them. Since any object in Grasshopper needs a component in canvas to work with, we can define our external geometries in canvas by components in the Params tab under Geometry. There is a list of different types of geometries that you can use to define your object. After bringing the proper geometry component to the canvas, define a Rhino object by right‐click on the component (context menu) and use “set one ... / set multiple … “ to assign abject to the component. By introducing an object/multiple objects to a component it becomes a Grasshopper object which we can use it for any purpose. It means we can use our manually created objects or even script generated objects from Rhino in Grasshopper.
Let’s have a simple example. We have three points in the Rhino viewport and we want to draw a triangle by these points. First we need to introduce these points in Grasshopper. We need three <point> components from Params > Geometry > Point and for each we should go to their context menu (right click) and select ‘set one point’ and then select the point from Rhino viewport (Fig.2.6).
Chapter 2
Fig.2.2. Different geometry types in the Params > Geometry menu
10
The very Beginning
Fig.2.3. Set point from Rhino in Grasshopper component
Chapter 2
Fig.2.4. The Grasshopper canvas and three points defined in the canvas which turned to (x) in the Rhino workplace. I renamed the components to point A/B/C by the first option of their menu to recognize them easier in Grasshopper canvas.
11
The very Beginning
Components and connections There are so many different actions that we can perform by components. Generally a component takes some data from another source (like parameters) and gives the result back. We need to connect the component which includes the input data to the processing component and connect the result to the other component that needs this result and so on. Going back to the example, now if you go to the Curve tab of components, in the Primitive section you will see a <line> component. Drag it to the canvas. Then connect <point A> to the A port of the <line> and <point B> to the B port (just click on the semi‐circle and drag it up to the other semi‐circle on the target. You can see that Rhino draws a line between these points).
Fig.2.5. Connecting the <point> components to the <line> component by dragging from output of the <point B> to the input of the <line>.
Chapter 2
Now add another <line> component for <point B> and <point C>. Do it again for <point C> and <point A> with the third <line> component. Yes! There is a triangle in Rhino.
Fig.2.6. The <line> components draw lines between <point> components. As you see any component could be used more than once as the source of information for other actions.
12
The very Beginning
Fig.2.7. Now if you change the position of the points manually in Rhino viewport, the position of points in Grasshopper (X ones) and the triangle will change accordingly and you do not need to redraw your lines any more. As you can see in this very first example, the associative modelling technique made it possible to manipulate the points and still have the triangle between these points without further need to adjustment. We will do more by this concept. Input / Output As mentioned before, any component in the grasshopper has input and output which means it processes the given data and gives the processed data back. Inputs are at left part of the component and outputs at right. The data comes from any source attached to the input section of the component and the output of the component is the result of that specific function.
Chapter 2
You have to know that what sort of input you need for any specific function and what you get after that. We will talk more about the different sort of data we need to provide for each component later on. Here I propose you to hold your mouse or “hover” your mouse over any input/output of the components. A tooltip will pop up and you will see the name, sort of data you need to provide for the component, is any predefined data there or not, and even what it for is.
Fig.2.8. Pop‐up tooltip
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The very Beginning
Multiple connections Sometimes you need to feed a component by more than one source of data. Imagine in the above example you want to draw two lines from point A to point B and C. you can use two different <line> components or you can use one <line> component and attach both point B and C as the second point of the <line> component. To do this, you need to hold Shift key when you want to connect the second source of data to a component otherwise Grasshopper would substitute it (Fig.2.12). When holding shift, the arrow of the line appear in a green circle with a tiny (+) icon while normally it is gray. You can also use Ctrl key to disconnect a component from another (or use menu) to disconnect an existing unwanted connection. In this case the circle around the arrow appears in red with a tiny (‐) icon.
Fig.2.9. Multiple connections for one component by holding shift key Colour coding
Chapter 2
There is a colour coding system inside the Grasshopper which shows the components working status.
Fig.2.10. The colour coding. Any gray component means there is no problem and the data defined correctly/the component works correctly. The orange shows warning and it means there is at least one problem that should be solved but the component still works. The red component means error and the component does not work in this situation. The source of the error should be found and solved in order to make
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The very Beginning
component works properly. You can find the first help about the source of error in the component’s context menu (context menu > Runtime warning/error) and then search the input data to find the reason of the error. The green colour means this component selected. The geometry which is associated with this component also turns into green in Rhino viewport (otherwise all Grasshopper geometries are red). Preview The components that produce objects in Rhino have the ‘Preview’ option in their menu. We can use it to hide or unhide it in the scene. Any unchecked preview make the component black part become hatched. We usually use preview option to hide the undesired geometries like base points and lines in complex models to avoid distraction. 2_2_3_Data matching For many Grasshopper components it is always possible to provide a list of data instead of just one input. So in essence you can provide a list of points and feed a <line> component by this list and draw more lines instead of one. It is possible to draw hundreds of objects just by one component if we provide information needed. Look at this example:
Chapter 2
I have two different point sets each with seven points. I used two <point> components and I used ‘set multiple points’ to introduce all upper points in one component and all lower ones in another component as well. As you see, by connecting these two sets of points to a <line> component, seven lines being generated between them. So we can generate more than one object with each component (Fig.2.14)
Fig.2.11. Multiple point sets and generating lines by them.
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The very Beginning
But what would happen if the number of points would not be the same in two point (data) sets? In the example below I have 7 points in top row and 10 points in the bottom. Here we need a concept in data management in Grasshopper called ‘Data matching’. If you have a look at the context menu of the component you see there are three options called: Shortest list Longest list Cross reference Look at the difference in the Figure 0.15
Chapter 2
Fig.2.12. Data matching A: shortest list, B: longest list and C: cross reference
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The very Beginning
It is clear that the shortest list uses the shortest data set to make the lines with, and the longest list uses the longest data set while uses an item of the shortest list more than once. The cross reference option connects any possible two points from the lists together. It is very memory consuming option and sometimes it takes a while for the scene to upgrade the changes. Since the figures are clear, I am not going to describe more. For more information go to the following link: http://grasshopper.rhino3d.com/2008/06/description‐of‐data‐stream‐matching.html 2_2_4_Component’s Help (Context pop‐up menu) As it is not useful to introduce all components and you will better find them and learn how to use them gradually in experiments, I recommend you to play around, pick some components, go to the components context menu (right‐click) and read their Help which is always useful to see how this component works and what sort of data it needs and what sort of output it provides. There are other useful features in this context menu that we will discuss about them later.
Chapter 2
Fig.2.13. Context pop‐up menu and Help part of the component
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The very Beginning
2_2_5_Type‐In component searching / adding If you know the name of the component that you want to use, or if you want to search it faster than shuffling the component tab, you can double‐click on the canvas and type‐in the name of the component to bring it to the canvas. For those who used to work with keyboard entries, this would be a good trick!
Fig.2.14. Searching for line component in the component‐pop‐up menu by double clicking on the canvas and typing the name of it. The component will be brought to the canvas.
Chapter 2
2_2_6_Geometry Preview Method
Fig.2.15. In order to enhance the working speed and get faster updates, whenever your project becomes heavy to calculate, use the Wireframe Preview option. It is always faster.
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The very Beginning
2_3_Other Resources There are so many great on‐line resources and creative ideas that you can check and learn from them. Here are some of them: Main Grasshopper web page: http://grasshopper.rhino3d.com/2008/06/some‐examples‐of‐grasshopper.html Some resources on McNeel Wiki WebPages: http://en.wiki.mcneel.com/default.aspx/McNeel/ArchitectureCommunity.html http://en.wiki.mcneel.com/default.aspx/McNeel/ExplicitHistoryExamples.html (Links to other resources) As mentioned before, the Grasshopper Primer from Lift Architects: http://www.liftarchitects.com/journal/2009/1/22/the‐grasshopper‐primer.html And hundreds of on‐line video tutorials which you can search easily.
Chapter 2
Chapter_3_Data sets and Math
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Data sets and Math
Chapter_3_Data sets and Math Although in 3D softwares we used to select our geometry from menus and draw them explicitly by clicking without thinking of the mathematical aspects of what we design, in order to work with Generative Algorithms, as the name sounds, we need to think a bit about data and math to make inputs of algorithm and generate multiple objects. Since we do not want to draw everything manually, we need some sources of data as the basic ingredients to make this generation possible. The way algorithm works is simple and straightforward. As I said, instead of copying by clicking 100 times in the screen, we can tell the algorithm, copy an item for 100 times in X positive direction. To do that you need to define the 100 as number of copying and X Positive direction for the algorithm, and it performs the job automatically. All we are doing in geometry has some peace of math behind. We can use these simple math functions in our algorithms, in combination of numbers and objects, generate infinite geometrical combinations. Let’s have a look; it is easier than what it sounds!
First of all we should have a quick look at numerical components to see how we can generate different numerical data sets and then the way we can use them. One numerical value The most useful number generator is <Number slider> component (Params > Special > Number slider) that generates one number which is adjustable manually. It could be integer, real, odd, even and with limited lower and upper values. You can set them all by ‘Edit’ part of the context menu. For setting one fixed numeric value you can go to the Params > Primitive > Integer / Number to set one value.
Chapter 3
3_1_Numerical Data sets
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Data sets and Math
Series of numbers We can produce a list of discrete numbers by <series> component (Logic > Sets > Series). This component produces a list of numbers which we can adjust the start point, step size of the numbers, and the number of values. 0, 1, 2, 3, … , 100 0, 2, 4, 6, … , 100 10, 20, 30, 40, … , 1000000
Rang of numbers We can divide a numerical range between a low and high value by evenly spaced numbers and produce a range of numbers. We need to define an interval to set the lower and upper limit and also the number of steps between them (Logic > Sets > Range). 1, 2, 3, … , 10 1, 2.5, 5, … , 10
Intervals Intervals provide a range of all real numbers between a lower and upper limit. There are one dimensional and two dimensional intervals that we talk about them later. We can define a fixed interval by using Params > Primitive > Interval/interval2 component or we can go to the Scalar > Interval which provides a set of components to work with them in more flexible ways. Intervals by themselves do not provide numbers, they are just extremes, upper and lower limits. As you now there are infinite real numbers between any two numbers. We use different functions to divide them and use division factors as the numerical values.
Chapter 3
1, 5, 10
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3_2_On Points and Point Grids Points are among the basic elements for geometries and Generative Algorithms. As points mark a specific position in the space they can be a start point of a curve or multiple curves, centre of a circle, origin of a plane and so many other roles. In Grasshopper we can make points in several ways. ‐ We can simply pick a point/bunch of points from the scene and introduce them to our workplace by <point> component (Params > Geometry > point) and use them for any purposes (These points could be adjusted and moved manually later on in Rhino scene and affect the whole project. Examples on chapter_2). ‐ We can introduce points by <point xyz> component (vector > point > point xyz) and feed the coordinates of the points by different datasets, based on our needs. ‐ We can make point grids by <grid hexagonal> and <grid rectangular> components. ‐ We can extract points from other geometries in many different ways like endpoints, midpoints, etc. ‐ Sometimes we can use planes (origins) and vectors (tips) as points to start other geometries and vice versa. You have seen the very first example of making points in chapter_2 but let’s have a look at how we can produce points and point sets by <series>, <range> and <number slider> components and other numerical data providers.
Fig.3.1. feeding a <point xyz> component by three <number slider> to make a point by manually feeding the X,Y and Z coordinates.
Fig.3.2. Making a grid of points by <series> and <point xyz> components while the first <number sliders> controls the distance between points and the second one controls the number of points by controlling the number of values in <series> component (The data match of the <pt> set into cross reference to make a grid of points but you can try all data matching options).
Chapter 3
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Fig.3.3. Dividing a numerical range from 0 to 1 by 5 and feeding a <pt> component with ‘Longest list’ data match. You can see we have 6 points which divided the range by 5 and all points drawn between the origin point and (1, 1) on the Rhino workplace (you can change the lower and upper limit of the <range> to change the coordinates of the point). Since the first experiments look easy, let’s go further, but you can have your own investigations around these components and provide different point grids with different positions and distances. 3_3_Other Numerical Sets Random data sets
Fig.3.4. Making a random point set. The <random> component produces 10 random numbers which is controlled by <number slider> and then this list is shuffled by <jitter> component (Logic > Sets > Jitter) for Y coordinate of the points once, and again for Z coordinates, otherwise you could see some sort of pattern inside your grid (check it!). The data match set to longest list again to avoid these sorts of patterns in the grid. In the figure 3.4 all points are distributed in the space between 0 and 1 for each direction. To change the distribution area of the points we should change the numerical domain in which random component produces the numbers. This is possible by manually setting the “domain of random numeric range” on Rhino command line or by defining the domain intervals adjustable by sliders. (Fig.3.5)
Chapter 3
I was thinking of making a randomly distributed set of points for further productions. All I need is a set of random numbers instead of a <series> to feed my <pt> component (I use <pt> instead of <point xyz> because it is shown on the component). So I pick a <random> component from Logic > sets. To avoid the same values for X,Y and Z, I need different random numbers for each.
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Fig.3.5. Setting up a domain by an <interval> component (Scalar > Interval > Interval) to increase the distribution area of the points (look at the density of the scene’s grid in comparison with the Fig.3.4). If you connect only one <number slider> to the domain of the <random> component it just adjusts the upper interval of the domain (with lower as 0). Fibonacci series What about making a point grid with non‐evenly spaced increasing values? Let’s have a look at available components. We need series of numbers which grow rapidly and under Logic tab and Sets section we can see a <Fibonacci> component. A Fibonacci is a series of numbers with two first defined numbers (like 0 and 1) and the next number is the sum of two previous numbers. N(0)=0, N(1)=1, N(2)=1, N(3)=2, N(4)=3, N(5)=5, … , N(i)=N(i‐2)+N(i‐1) Here are some of the numbers of the series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
Here I use <Fibonacci> series (Logic > Sets > Fibonacci) to produce incremental numbers and feed the <pt> component with them.
Fig.3.6. Using <Fibonacci> series to produce increasing distances (none‐evenly spaced series of numbers) to make points. The number of points could be controlled with a <number slider>.
Chapter 3
As you see the numbers grow rapidly.
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3_4_Functions Functions are components that are capable of performing math functions. There are functions from one to eight variables (Scalar > Expressions). You need to feed a function with different data (not always numeric but also Boolean, coordinate, etc) and it performs the user defined function on the input data. To define the function you can right‐click on the (F) part of the component and type it or go to the Expression Editor. Expression editor has so many predefined functions and a library of math functions for help. Math functions Using the predefined components is not always what we aimed for, but in order to get the desired result we can use mathematical functions to change the data sets and feed them for making geometries. A simple example is the mathematical function of a circle that is X=Sin(t) and Y=Cos(t) while (t) is a range of numbers from 0 to 2 Pi. I am producing it by a <range> of numbers which is starts from 0 to 1 by N number in between, times 2Pi by <function> that means a range of numbers from 0 to 2pi that makes a complete circle in radian.
Chapter 3
Fig.3.7. Parametric circle by mathematical functions. You have <Sin> and <Cos> functions in the Scalar > Trig. (F(x)=x * 2Pi).
Fig.3.8. Series of points which is defined by <Fibonacci> series and simple mathematical functions (x‐>F(x)=x/100, y‐>F(x)=x/10). The selected green F(x) is a simple function to add 1 to the <number slider> (x+1) in order to make the values of <series> numbers equal to the Fibonacci numbers. The aim is to show you that we can simply manipulate these data sets and generate different geometries accordingly.
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Fig.3.9. A <range> of numbers from 0 to 2 times by 2Pi with <Function> that make it a numerical range from 0 to 4Pi that feeds the <pt> component by the following math function (X=t * Sin(t), Y=t * Cos(t)). You can reduce all components between <range> and <pt> by two functions to feed the <pt> by defining the whole process in Expression Editor. X of pt > F(x)=X * Sin (x*2*Pi) Y of pt > F(x)=X * Cos(x*2*Pi)
Fig.3.10. Inter tangent spirals from two inverted spiral point sets (<range> interval from 0 to 4 multiplied by 2Pi, makes the data set from 0 to 8Pi which is inverted for the second spiral by <Reverse list> component from Logic > Lists as 8pi to 0). First <pt>: X=t * Sin(t), Y=t * Cos(t) in which t=0 to 8Pi Second <pt>: X=t’ * sin(t), Y=t’ * Cos(t) in which t’=8Pi to 0 (<reverse list>)
Chapter 3
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Data sets and Math
Fig.3.11. Moebius by points X= Sin(u)*(‐2+v*sin(u/2)) Y= Cos(u)*(‐2+v*sin(u/2)) Z= v*Cos(u/2) While u=0 to 8Pi and v=‐2 to 2 Playing around the math functions could be endless. You can find so many mathematical resources to match your data sets with them. The important point is that you can manipulate the original data sets and generate different numerical values and feed other components by them. So as you see by some simple set of numerical data we can start to generate different geometries. Since we need to work with these data sets as a source of our geometries lets go further with them.
Fig.3.12.Enneper surface, by Rhino’s Math function plug‐in. Designing surfaces with mathematical equations.
Chapter 3
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Data sets and Math
3_5_Boolean Data types Data is not limited to Numbers. There are other data types that are useful for other purposes in programming and algorithms. Since we are dealing with algorithms, we should know that the progress of an algorithm is not always linear. Sometimes we want to decide whether to do something or not. Programmers call it conditional statements. And we want to see whether a statement meets certain criteria or not to decide what to do next. The response of the conditional ‘question’ is a simple yes or no. in algorithms we use Boolean data to represent these responses. Boolean values are data types which represent True (yes) or False (no) values only. If the statement meets the criteria, the response is True, otherwise False. As you will see later, this data type is very useful in different cases when you want to decide about something, select some objects by certain criteria, sort objects, etc.
Fig.3.13. Here I generated ten <random> values and by a <function> component I want to see if these numbers are less than a certain <Upper_limit> or not. As you see the <function> is simply X>Y and whenever the numbers meet the criteria, the function passes True to the <panel>.
Chapter 3
Fig.3.14.a. For the next step, I used a <Modulus> component (Scalar > Operators > Modulus) to find the remainder of the division of the Random values by <2> and I pass the result to a <function> to see if this remainder =0 or not (f(x)=x=0), simply means whether the number is even or not. As you see the result is another <panel> of True/False values.
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Fig.3.14.b.Here I used a <Gate And> component (Logic > Boolean > Gate And) and I attached both <function>s to perform Boolean conjunction on them. The result is True when both input Boolean values are True, otherwise it would be False. As you see, those numerical values which are both even and bigger than the <Upper_limit> are meeting the criteria and pass True at the end. We will discuss how to use these Boolean values later. There are multiple Boolean operators on Boolean section of the Logic tab that you can use to create your criteria and combine many of them.
Chapter 3
Fig.3.15. we can manually define a set of Boolean data by <Boolean> component from Params tab under Primitive section or use <Boolean Toggle> in Special section to use one manually changeable Boolean value.
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3_6_Cull Patterns There are many reasons that we might want to select some of the items from a given data set and do not apply a function to all elements. To do this we either need to select some of the specific items from a list or omit other items. There are different ways to achieve this but let’s start with omitting or culling lists of data. Up to now there are two <cull> components to cull a list of data in Grasshopper. While <cull Nth> omit every N item of the given list of data, <cull pattern> takes a pattern of Boolean values (True/False) and cull a list of data, based on this pattern, means any item of the list that associates with True value in Boolean list will pass and those that associate with False will omit from the list. If the number of values in the data list and Boolean list are the same, each item of the data list being evaluated by the same item in the Boolean list. But you can define a simple pattern of Boolean values (like False/False/True/True which is predefined in the component) and <cull> component would repeat the same pattern for all items of the data list. Distance logic
As mentioned before, <Cull pattern> component takes a list of generic data and a list of Boolean data and omits the members of the generic list of data who associate with the false value of the Boolean list. So the output of the <Call pattern> component is a set of points that associate with True values which means they are closer than the specified number shown on the <number slider>, to the reference point, because the X>Y function always pass True for the smaller values. To show them better I just connected them to the reference point by a simple line.
Fig.3.16. Selection of points from a point set based on their distance from a reference point with <Cull pattern> component.
Chapter 3
I am thinking of selecting some points from a point set based on their distance to another point (reference point). Both point set and the reference point are defined each by a <point> component. First of all what we need is a <distance> component (Vector > Point > Distance) that measures the distance between points and the reference. I compared these distances by a user defined number (<number slider>) with a <F2> component. This comparison generates Boolean values as output (True/False) to show whether the value is smaller (True) or bigger (False) than the upper limit. I am going to use these Boolean values to feed the <Cull pattern> component (the function of <F2> component defined as f=x>y).
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Topography Having tested the first distance logic, I am thinking of selecting some points which are associated with contour lines on a topography model, based on their height.
Fig.3.17. Topography with points associated with contour lines. I want to select these points based on their height. What I have is a point set which is defined by a <point> component (named topography). I need the height of the point with the same logic as the above example to select the specific points. Here I used a <Decompose> component (Vector > Point > Decompose) to get the Z values of these points. I compared these values with a given number (<number slider>) with a <F2> component to produce a list of associative Boolean values. The <Cull pattern> component passes those who associated with the True values which means selected points are higher than the user defined height value (the function of <F2> component defined as f=x>y).
Chapter 3
Fig.3.18. Selected points which are higher than 4.7550 units! (A user defined value). These points are now ready to plant your Pine trees!!!!
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Connectivity logic: Triangles Let’s have another example of culling this time with <Cull Nth>. Imagine we have a network of points and we want to draw lines to make triangles with a pattern like Figure.3.19.
Fig.3.19. Making triangles by a network of points. The first step is to simply create a grid of points by <series> and <pt> components. The next step is to find the proper points to draw lines in between. Each time we need a line starts from a point and ends at the next point on the next column, then another line goes from there to the back column but at the next row and final line goes back to the start point. To do this, it seems better to make three different lists of points, one for all first points, one for all second points and another for all third points and then draw line between them.
The first second point is the second point on the point set and then the list goes on one by one. So to select the second points I just shifted the original list by <Shift list> component (Logic > List > Shift list) by shift offset=1 to shift the data set by one value and make the second points list. (Go to the <shift list> help to learn more about component). Since the original data is a set of points which make the start points in our example, the shifted data would be the second point of the triangles (all second points in the network). The third point of triangles is in the same column as the start point but in next row, so I shifted the original list of points again by shift offset=the number of columns (the value comes from the <number slider>) to find these points for all point set and make a list of all third points.
Chapter 3
I can use the original points as the list for all start pints.
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Data sets s and Matth
s pointts by the sh hift offset va alue of equa al to the nu umber of Fig.3.20. Selected ittem is the shifted d points of th he triangles. columnss which produces all third To comp plete the task I need to o omit some points in each h set. First off all the poin nts in the last column never co ould be first points of triaangles so I need to omit them from tthe list of staart points. Th he points on the ffirst column also never ccould be thee second poiints, so I neeed to omit tthem from th he list of second points and tthe same forr last column n again as th hird points. SSo basically I attached all points’ Cull Nth> component wh hich omits specific mem mbers of a daata set by a number lists eacch to one <C which iss cull frequen ncy (Fig.3.17 7). In this casse all data se ets culled by the numberr of columns which is clear wh hy. So I just cconnected th he <number sslider> to each <Cull Nth> componen nt as frequen ncy.
Chapter 3
Fig.3.21. Using <Culll Nth> to omit o A. last column, B. first f column and c. last column of the first, second a and third poiints’ lists. The nexxt step is jusst to feed th hree <line> components c ond, then to connect first points to the seco second p points to thee third and finally third points to the first again.
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Fig.3.22. Making lines by connecting culled lists of points to the <Line> component.
Although there are still some problems with our design and we now that we should not start any triangle from the points of the last row, but the concept is clear…… so let’s go further. We will come back to this idea while talking about mesh geometries and then I will try to refine it.
Chapter 3
Now by changing the <number slider> you can have different grids of points which produces these triangles accordingly.
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3_7_2D Geometrical Patterns
Fig.3.23. Extracting the concept of a pattern by simple geometries. I still insist on working on this models by data sets and simple mathematical functions instead of other useful components just to see how these simple operations and numerical data have the great potential to generate shapes, even classical geometries.
Chapter 3
Geometrical Patterns are among the exciting experiments with the Generative Algorithms and in Grasshopper. We have the potential to design a motif and then proliferate it as a pattern which could be used as a base of other design products and decorations. In case of designing patterns we should have a conceptual look at our design/model and extract the simple geometry that produces the whole shape while being repeated. So by producing the basic geometry we can copy it to produce the pattern as large as we need (Fig.3.23).
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g o Sheikh Lottfolah Mosq of que’s tile wo ork comprisees of simple patterns Fig.3.24. Complex geometries athematical‐‐geometricall calculationss. Sheikh Lotffolah Mosqu ue, Isfahan, Irran. which crreated by ma Simple linear pattern n Here I decided d to make m a simp ple pattern by b some inte ersecting linees and my aaim is to draaw some patternss similar to Figure.3.25.
Chapter 3
of simple con ncepts to makke patterns. Fig.3.25. Examples o I started d my definitiion by a <series> which I am able to o control thee number of f values (here points) and the step size (h here distancee between p points). By th his <series> I generated a set of points and I also gen nerated anotther three different d setss of points with w differen nt Y values w which I am adjusting a them reelatively by aa <number sllider> and <F1> compon nents (y=‐x/3 3, y=x, y=x/3, y=x+(x/3) if x is the number coming from m the <numb ber slider>).
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Fig.3.26. Generating four set of points, all associated with a one <series> component which controls the number and distance of points and another <number slider> which controls the Y value of the point sets (I am using longest list for points data matching). To get the “zig‐zag” form of the connections I need to cull the point sets with <cull pattern> one with True/False and another one with False/True pattern and then connect them together (Fig.3.20).
Fig.3.27. Cull pattern and selected points as a base for “zig‐zag” pattern. Since I want to draw poly line on this culled point sets, I cull all point sets with the same logic and merge these points to make a one merged stream by <merge 02> component (Logic > Streams > Merge 02).
Chapter 3
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Data sets and Math
Fig.3.28. Cull pattern for the middle points inverted to make the mirrored pattern of the first set. To make the second rows of points, I used the same points of the first row and merge them again with the second row to have four merged data streams at the end. If you connect these merged data sets to a <poly line> component you will get a z‐shaped poly line and that is because the points are not in a desired order and the <merge> component just add the second list of points at the end of first one. So I need to sort the points in the desired way. A <Sort> component sorts some generic data based on a sortable key. What do we have as a sortable key?
Chapter 3
If we look at the order of the points we can see that the X dimension of the points increases incrementally, which seems suitable as an item to sort our points with. To do this, we need to extract the X coordinate of the points. The <decompose> component make it possible. So we connect the X coordinate of the points as a sortable key to the <sort> component and then sort the points with that. Finally we can use these sorted points to feed <polyline> components, make our pattern with them (Fig.3.29).
Fig.3.29. Sorting points by their X component as key and then making polyline with them. I sorted mirrored geometry by another <sort> component because they are from different cull pattern.
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Data sets and Math
Chapter 3
Fig.3.30. Later on we will discuss how we could create repetitive patterns by simple components and the way we can array the simple motif to create complex geometries.
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Circular patterns There are endless possibilities to create motifs and patterns in this associative modelling method. Figure.3.31 shows another motif which is based on circular patterns rather than the linear one. Since there are multiple curves which all have the same logic I will just describe one part of the algorithm and keep the rest for you.
Fig.3.31. Circular geometrical patterns which is repeated in the second picture.
Chapter 3
The start point of this pattern is a data set which produces a bunch of points along a circle, like the example we have done before. This data set could be rescaled from the centre to provide more and more circles around the same centre. I will cull these sets of points with the same way as the last example. Then I will generate a repetitive ‘zig‐zag’ pattern out of these rescaled‐circular points to connect them to each other, make a star shape curve. Overlap of these stars could make motifs and using different cull patterns make it more interesting.
Fig.3.32. Providing a range of 0 to 2Pi and by using Sin/Cos functions, making a first set of points. The second <number slider> changes the radius of the circle (power of the X and Y).
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Fig.3.33. Increasing the numbers of Sin/Cos functions by a <number slider> making the second set of points with bigger radius.
Fig.3.34. First and second circles made by points. In order to cull the points, we can simply use the <Cull pattern> for the points and use True/False like the last example. But how we can sort the list of points after all? If you connect the culled points to a <poly line> component you will not get a star shape poly line but two offset polygon connected to each other. Here I think it is better to sort the points based on their index number in the set. Because I produced the points by a <range> component, here we need a <series> component to provide the indices of the points in the list. The N parameter of the <range> factor defines the number of steps of the range so the <range> produces N+1 number. I need a <series> with N+ 1 value to be the index of the points (Fig.3.34) and cull and sort these points based on their indices.
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Fig.3.35. Generating index number of the points.
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Fig.3.36. We need to cull indices and points the same and merge them together. Although the result of the merging for <series> could be again the numbers of the whole data set, the order of them is like the points, and so by sorting the indices as sortable keys we can sort the points as well. The only thing remain is to feed a <polyline> component by sorted points.
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Fig.3.37. Generating Polyline by sorted points.
Fig.3.38. Star‐shaped polyline. The same logic could be used to create a more complex geometry by simply generating other point sets, culling them and connecting them together to finally produce patterns. We can use these patterns as inputs for other processes and design other decorative shapes.
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Fig.3.39. You can think about different possibilities of the patterns and linear geometries in applications. Although I insisted to generate all previous models by data sets and simple mathematical functions, we will see other simple components that made it possible to decrease the whole process or change the way we need to provide data.
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Fig.3.40. Final model.
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Chapter_4_Transformation
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Chapter_4_Transformation Transformations are essential operations in modelling and generating geometries. They can enable us to get variations from the initial simple geometries. Transformations help us to scale and orientate our objects, move, copy and mirror them, or may result in accumulation of objects, that could be the desired model we. There are different types of transformations but to classify it, we can divide it to main branches, the first division is linear and the second is spatial transformations. Linear transformations perform on 2D space while spatial transformations deal with the 3D space and all possible object positioning.
In order to transform objects, conceptually we need to move and orientate objects (or part of objects like vertices or cage corners) in the space and to do this we need to use vectors and planes as basic constructs of these mathematical/geometrical operations. We are not going to discuss about basics of geometry and their mathematical logic here but first let’s have a look at vectors and planes because we need them to work with.
Fig.4.1. Transformations provide great potential to generate forms from individuals. Nature has some great examples of transformations in its creatures.
Chapter 4
In other sense we can classify transformations by status of the initial object; transformations like translation, rotation, and reflection keep the original shape but scale and shear change the original state of the object. There are also non‐linear transformations. In addition to translation, rotation and reflection we have different types of shear and non‐uniform scale transformations in 3D space, also spiral and helical transformations and projections which make more variations in 3D space.
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4_1_Vectors and planes Vector is a mathematical/geometrical object that has magnitude (or length) and direction and sense. It starts from a point, go toward another points with certain length and specific direction. Vectors have wide usage in different fields of science and in geometry and transformations as well.
Fig.4.2. A: Basic elements of a Vector, B: point displacement with a vector.
Simply if we have a point and a vector, this vector could displace the point with the distance of vector’s magnitude and toward its direction to create a new position for the point. We use this simple concept to generate, move, scale and orientate geometries in our associative modelling method. Planes are another useful set of geometries that we can describe them as infinite flat surfaces which has an origin point. Construction planes in Rhino are these types of planes. We can use these planes to put our geometries on them and do some transformations based on their orientation and origin. For example in the 3D space we cannot orientate an abject on a vector! but we need two vector to make a plane to be able to put geometry on it. Vectors have direction and magnitude while planes have orientation and origin. So they are two different types of constructs that can help us to create, modify, transform and articulate our models. Grasshopper has some of the basic vectors and planes as predefined components. These are including X, Y and Z vectors and XY, XZ, and YZ planes. There are couple of other components which we can produce and modify them which we talk about them in our experiments. So let’s jump into design experiments and start with some of the simple usage of vectors and go step by step forward.
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4_2_On curves and linear geometries As we have experimented with points that are 0‐Dimension geometries now we can start to think about curves as 1‐Dimensional objects. Like points, curves could be the base for constructing so many different objects. We can extrude a simple curve along another one and make a surface, we can connect different curves together and make surfaces and solids, we can distribute any object along a curve with specific intervals and so many other ways to use a curve as a base geometry to generate other objects. Displacements We generated so many point grids in chapter 3. There is a component called <Grid rectangular> (Vector > Point > Grid rectangular) which produces a grid of points which are connected together make some cells also. We can control the number of points in X and Y direction and the distance between points.
Fig.4.3. a simple <Grid Rectangular> component with its predefined values. You can change the size of grid by a <number slider>. I want to change the Z coordinates of the points as well. So I need to change the base plane of the grid. To do this, I introduced a <XY plane> component (Vector > Constants > XY plane) which is a predefined plane in the orientation of the X and Y axis and I displaced it in Z direction by a <Z unit> component (Vector > Constants > Z unit) which is a vector along Z axis with the length (magnitude) of one. I can change the height of this displacement by the size of the vector through a <number slider> that I connected to the input of the <Z unit> component. So by changing the position of the <XY plane> along the Z axis the height of the grid also changes.
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Fig.4.4. Manipulated Grid (selected in green) with one <number slider> for scale of the grid and another with a <Z unit> and <XY plane> to change the Z coordinate of the grid’s points. (Further you can just connect the <Z> vector component to the <Grid rectangular> component and get the same result).
Chapter 4
Now if you look at the output of the <grid rectangular> you can see that we have access to the whole points as well as grid cells and cell centres. I am looking for a bunch of lines that start from the grid cells’ centre points and spread out of it to the space. Although we can simply connect these points from the two <grid> component M part to a <line> component, the length of lines in this case would be different. But as I want to draw lines with the same length, I need another strategy. Here, I am going to use a <line SDL> component. This component draws a line by Start point(S), Direction (D), and Length (L). So exactly what I need; I have the start points (cell’s midpoint), and length of my lines. What about the direction? Since the direction of my lines are in the direction of the lines that connect the mid points of the cells of the two grids, I am going to make a set of vectors by these to point sets.
Fig.4.5. Making vectors from the cells midpoints of the first grid toward the cells midpoints of the second grid by <vector 2pt> component (Vector > Vector > vector 2pt). This component makes vectors by the start and end point of vectors.
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Fig.4.6. The <line SDL> component generates bunch of lines from the grid cell midpoints that spread out into space. I can change the length of lines by <number slider>.
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Fig.4.7. By simply using an <end points> component (Curve > Analysis > End points) and using these ‘end points’ as the ‘base points’ for a set of <circle> components (Curve > Primitive > Circle) and extruding these circles by <extrude point> component we can generate a set of cones, pointing toward same direction and we can finish our first experiment.
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m displacemeents Random I decideed to make a a set of rand domly distributed pipes (lines) with random len ngth, woven n to each other, m make a funnyy element. I ssketched it like Figure.4.8 8.
There are different ways to dessign this mo odel. I am th hinking of maaking a circle as the basse curve, divide itt into desire parts and th hen generatee some random curves frrom these points and then make pipes byy these curvees. But let’s ssee how we ccan do it in G Grasshopper. As I said d, first I wan nt to do it w with a circle tto divide it aand create the base poin nts. I used aa <circle> compon nent (Curve >> Primitive >> Circle) and I attached a <number sslider> for fu urther changges of its radius. TThen we attaach our circlee to a <divid de curve> com mponent (Cu urve > Divisio on > Divide ccurve) to divide th he circle. Aggain we can control the number of d divisions by an integer <<number slid der>. The <Divide curve> component gives me the po oints on the curve (as the division po oints). These e are the first set of points forr generating our base lines.
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Fig.4.8. First sketchees of model
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Fig.4.9. Dividing the base circle. In this step, for some reasons! I want to generate these base lines in different segments. In order to draw our first step, I need the second set of points, above the first set (to make the lines semi‐ horizontal !!!!) and in a randomly distributed field. To do this, I want to make a set of random vectors in Z direction and displace the first set of points with these random vectors. By connecting the first and second sets of points we would have our first part of our lines. Because I need the same amount of vectors as the base points, I used the value of the <number slider> that controls the amount of base points (circle divisions), to generate the same amount of random vectors. So I connect the <number slider> of <divide curve> component to a <random> component N part to generate N random values. Then I used a <unit Z> component but I feed this <Unit Z> vector component with the <random> component so it produces N random length vectors in Z direction accordingly.
Chapter 4
Fig.4.10. Generating random length vectors in Z direction. The <number slider> which is connected to the S part of the <random> component differs the random numbers as ‘seed’ of random engine. So now we are ready to displace the points by these vectors. Just bring a <move> component (XForm > Euclidian > Move) to the canvas. Basically a <move> component moves geometries by given vectors. You can move one object/a group of objects with one vector/a group of vectors. Since you still have the component of the source geometry, the <move> component works like a ‘copy’ command in Rhino. To see how these vectors displaced the points, we can use a <line> component to see the result.
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Fig.4.11. The first set of lines that made by the randomly displaced points.
Fig.4.12. Shuffling the end points and making totally random lines. The rest of the process is simple. I think three segments of lines would be fine. So I just repeat the move and shuffling concept to make second and third set of displaced points.
Chapter 4
As you see in the Figure.4.11 the lines are vertical and I am looking for more random distribution of points to make the whole lines penetrating each other. I will do it by a <jitter> component to simply shuffle the end points.
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Fig.4.13. Generating 3 line segments for each path to be converted to pipes. To clean up the scene you can uncheck the other components and just preview the lines.
Chapter 4
Ok. We have the base geometry. Now just add a <pipe> component (Surface > Freeform > pipe) and attach all <line> components (by holding shift key!) to the ‘base curve’ part of the component and use a <number slider> to control the radius of pipes.
Fig.4.14. Final pipes. Uncheck the preview of lines. It speeds up your processing time. That’s it. Now you can change the radius of the base circle to distribute the pipes in larger/smaller areas, you can change the number of pipes (curves), you can change the random seed and pipe’s radius. To do all and check the result you can go to the ‘View’ menu of the Grasshopper and select ‘Remote Control Panel’ to have the control panel for your adjustments which is much more easier than the sliders inside the canvas when you want to observe the changes in Rhino scene. To hide/Unhide the canvas, just double‐click on its window title bar.
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Fig.4.15. Remote Control Panel (from view menu) and observing the changes of the model.
Chapter 4
Fig.4.16. Although the connections of pipes need a bit more elaboration, for an experiment it is enough.
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4_3_Combined Experiment: Swiss Re Today it is very common to design the concept of towers with this associative modelling method. It allows the designer to generate differentiated models simple and fast. There are so many potentials to vary the design product and find the best concepts quiet quickly. Here I decided to model a tower and I think the “Swiss Re” tower from ‘Foster and partners’ seems sophisticated enough to start, for some modelling experiments. Let me tell you the concept. I am going to draw a simple plan of the tower and copy it to make the floors. Then I will rescale these floors to match the shape, and then I will make the skin of the surface and finally the façade’s structural elements. I will do the process with very simple geometries for this step and also to save time. Let’s start with floors. I know that the Swiss Re’s floors are circles that have some V‐shaped cuts around it, but I just use a simple circle to make the section of the tower. Since I know that it has 41 floors I will copy this section for 41 times In Z direction with 4 meters space in between and I will play around the proportions because I don’t know the real dimensions, but I will try to manage it visually.
Chapter 4
Fig.4.17. The <circle> component (plan_section) with radius of 20 which is copied by <move> component along Z direction by a <Z unit> vector component for 41 times above. To get this I used a <series> component (floors) starts from 0 and has the step size=4 with 41 values. (I renamed the components to recognize them easily).
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Now the first thing we need is to rescale these circles to make the proper floors. I need a <scale> component (XForm > Affine > Scale) to rescale them. The <scale> component needs the geometry to scale, centre for scaling and the factor of scaling. So I need to feed the geometry part of it by our floors or circles which is <move> component. The predefined centre of scaling is the origin point, but if we scale all our floors by the origin as centre, the height of the tower would rescale and the plane of each floor would change. So we need the centre of rescaling at the same level at the floor level and exactly at the centre of it. So I used a <Centre> component (Curve > Analysis > Centre) which gives me the centre of the circles. By connecting it to the <scale> you can see that all circles would rescale in their level without movement.
Fig.4.18. Rescaling floors from their centre point as the centre of scaling, using a <centre> component. The scaled circles are selected in green. Although I rescaled the whole circles the same, but we know that all floors are not in the same size, so we need to rescale our floors different from each other; and we know that from the circle which is grounded on earth they first become bigger up to certain height, look constant on the middle parts and then become smaller and smaller up to the top point of the tower. So I need to provide a list of scale factors for all floors which are 41 and again I now that this list has three different parts that if we say starts from 1 then increases up to certain factor, then remain constant in some floors and then decreases. If you look at Figure 4.19 you cans see the pictures that give you a sense of these scaling factors. So basically I need to provide a list of data (41 values) which is not strait forward this time.
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Fig.4.19. Swiss Re HQ, 30 St Mary Axe, London, UK, 1997‐2004, (Photos from Foster and Partners website, http://www.fosterandpartners.com). Scale Intervals
As I also mentioned that we have three different parts for the scaling factors of the tower, we need three different set of numbers, the first and the last ones are intervals and the middle part is just a real number which is constant. Let’s have a look at Figure 4.20. Here I used two <interval> component (Scalar > Interval > Interval) to define two numerical range, one increasing and one decreasing. The increasing one starts from 1 which I assumed that the ground floor is constant and then increases up the number of the <number slider>. The second <interval> starts from <number slider> and ends at another <number slider> which is the lower limit. By using the same <number slider> for the middle part, I am sure that middle part of the data set is the same from both sides.
Chapter 4
As I mentioned before, intervals are numeric ranges. They are real numbers from lower limit to upper limit. Since I said real numbers, it means we have infinite numbers in between. They are different types of usage for these mathematical domains. As we experimented before, we can divide a numerical interval by a certain number and get divisions as evenly distributed numbers between two numbers.
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Fig.4.20. Two <interval> components, top one increasing and the bottom one decreasing. I do this because I don’t know the exact proportions, but this help us to discuss about more interesting stuff! Know I have the increasing and decreasing numeric intervals, but to produce the scaling factors I need numbers not ranges. So I am going to use <range> components to divide these numeric intervals up to certain numbers.
Because I don’t know the projects data, still I do not know in which floor it starts to remain constant and where it decreases. So in order to assign these factors correctly, I am going to split the floors in two parts. I am using a <Split list> component (Logic > List > Split list) and I attach the <series> component which I named it floors to it to split the floors in two different parts. Then I need to know how many floors are there in each part that later on I can change it manually and correct it visually. <List length> component (Logic > List > List length) do this for me and passes the number of items in the list, so I know that how many floors are in the increasing scale part and how many in the decreasing scale part. The only remaining part is the constant floors. I just assumed that there are 8 floors which their scale does not change and I need to omit them from the floors in the increasing and decreasing scaling part. Since two of these floors are included in increasing and decreasing lists as the maximum scale I need to omit another 6 floors. So I just need to get the number of the floors from <list length> and apply (‐3) function to each, to omit these 6 floors and distribute them between both two lists. And the final trick! Since the <range> component divides the domain to N parts, it produces N+1 number at the end which means 1 more number than we need. So all together I need to add a <function> component by (X‐4) expression to reduce the number of steps that each <range> component wants to divide its numeric range.
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Fig.4.21. Generating the scale factors. And finally as you can see in the Figure.4.22, I merged all my data by a <merge 8> component to make a unified list of data which is the numeric factors for the increasing parts, 6 constant number just coming from the < number slider> (the constant factor between the two range) and the decreasing factor. The <merge 8> component includes 41 scaling factor that now I can attach to the <scale> component and rescale all floors. Chapter 4
Fig.4.22. The <merge 8> component includes <range> of increasing numbers, 6 constant numbers (which is the maximum number of the increasing and decreasing parts), and the <range> of decreasing numbers as one unified list of data.
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Chapter 4
Fig.4.23. Scaling factors
Fig.4.24. Rescaled floors. Now I can change the position of the constant floors and the scaling factors to visually match the model by the original building. Ok! Let’s go for façade elements.
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The steel elements around the façade are helical shapes that have the cross section like two connected triangles but again to make it simple, I just make the visible part of it which is almost like a triangle (in section). I want to generate these sections and then ‘loft’ them to make a surface. I started with a <polygon> component (Curve > Primitive > Polygon). I used an <end points> component to get the start/end points of my floors. By attaching these points as the base for <polygon> I can generate couple of polygons on the start points of my floors. I attached a <number slider> to the <polygon> to control its radius and I set the number of segments to 3 manually. I renamed the <scale> component to the <rescaled_floors>.
Now I want to make the helical transformation for the polygons. For some reasons I think that every floors rotate for 5 degree! So I need to rotate all polygons for 5 degree around the centre of the floors. So I brought a <rotate> component to the canvas (XForm > Euclidian > Rotate). Geometry is my <polygon> and the base of rotation is the centre of the floors / <centre> of circles. To produce the rotation angle I need a list of incremental factors that each time adds 5 degree. So I used a <series> starts from 0 with the step size of 5 and with 41 values which come from the floors and number of <polygon> also. The only thing remain is because <rotate> component works with Radian I need to convert Degree to Radian by a <function> which is Radian=Degree * Pi / 180 (There is a predefined function called RAD(x) that converts degree to radian also. Check the functions library).
Chapter 4
Fig.4.25. Polygons positioned on the facade.
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Fig.4.26. Rotating the polygons around the centre of floors, each for 5 degree from the previous one. Now just use a <loft> component to make a surface by connecting all these sections together. To make the scene clean, uncheck the preview of any unnecessary geometry.
Chapter 4
As you can see, we don’t have the top point of the tower because we don’t have any floor there. We know that the height of tower is 180m. So I added a simple <point> component (0, 0, 180) to the canvas and I attached it to the <polygon> component (by holding shift). So the <polygon> component produces another polygon at the top point of the tower and the number of polygons becomes 42. So I changed the <series> component to produce 42 numbers as well. I know that the top part of the “Swiss Re” is more elegant than this model but for our purpose this is fine.
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To generate these elements all around the building, I am using a <rotate> component which I attached the <loft> object as source geometry and I used a <range> from 0 to 2Pi divided by a <number slider> as the number of elements, to rotate it all around the circle. Since the centre of rotation is the Z axis at the centre of the tower, and it is pre‐defined on the component, I do not need to change it or introduce any plane for rotation.
Fig.4.28. first set of spiral elements around the tower.
Chapter 4
Fig.4.29. I need to <mirror> (XForm > Euclidian > Mirror) the rotated geometry by <YZ plane> (Vector > Constants > YZ plane) to have the lofted elements in a mirrored helical shape. So at the end I have a lattice shape geometry around the tower.
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Glass cover To cover the whole tower simply with glass, I should go back to the floors component and <loft> them. Again since we don’t have any floor at the top point of the tower, I used another <circle> component and I feed it by the top point as the position and I attached it to the <loft> object to make the loft surface complete up to top point.
Fig.4.30.a. lofting floor curves to make the façade’s glass cover.
Chapter 4
Fig.4.30.b. The lofted surface covers the whole façade. In the Swees Re project, there is two colours of glass. If once we decided to make this effect, we can use façade structure to produce different surfaces and render them differently.
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Chapter 4
4.31. Final model. Although it is not exactly the same, but for a sketch model in a short time, it would work.
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4_4_On A Attractors m toward wh hich that system tends to o evolve, “Attractor is a set off states of a dynamic physical system point attractor is an attrractor consissting of a regardleess of the staarting condittions of the system. A p single sttate. For example, a marble rolling in n a smooth, rrounded bow wl will alwayys come to re est at the lowest point, p in thee bottom cen nter of the bowl; b the fin nal state of position p and d motionlessness is a point atttractor.”
Fig.4.32. Strange Atttractor. In the caase of design and geom metry, attracttors are elem ments (usuallly points bu ut could be ccurves or any otheer geometry) that affect the other geeometries in the space, cchange their behaviour aand make them diisplace, re‐o orientate, reescale, etc. They T can arrticulate thee space arou und themselves and introducce fields of actions with h specific rad dius of power. Attractors have diffeerent applications in paramettric design since they haave the poteential to change the who ole objects o of design co onstantly. Definingg a field, attrractors could d also affect the multiple e agent systeems in multiiple actions. The way they cou uld affect th he product and a the pow wer of attracctors are all adjustable. We go thro ough the concept of attractorrs in differentt occasions sso let’s have some very ssimple experiments first.
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Point Attractors I have a grid of points that I want to generate a set of polygons on them. I also have a point that I named it <attractor_1> and I draw a <circle> around it just to realize it better. I want this <attractor_1> affects all my <polygon>s on its field of action. It means that based on the distance between each <polygon> and the <atractor_1>, and in the domain of the <attractor_1>, each <polygon> respond to the attractor by change in its size.
Fig.4.33. Base <point_grid> and the <polygon>s and the <attractor_1>.
The algorithm is so simple. Based on the <distance> between <attractor_1> and the <Pt‐grid>, I want to affect the radius of the <polygon>, so the ‘relation’ between attractor and the polygons define by their distance. I need a <distance> component to measure the distance between <attractor_1> and the polygon’s center or <pt_grid>. Because this number might become too big, I need to <divide> (Scalar > Operators > Division) this distance by a given number from <number slider> to reduce the power of the <attractor_1> as much as I want.
Fig.4.34. <Distance> divided by a number to control the ‘power’ of the <attractor_1>. I also made a Cluster by <attractor_1> and its <circle> to have one component as attractor in the canvas. You can convert any group of related geometries to clusters by selecting them and using ‘make cluster from selection’ from the canvas toolbar (or Arrange menu or Ctrl+G).
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Now if you connect this <div> component to the Radius (R) part of the <polygon> you can see that the radius of polygons increases when they go farther from the <attractor_1>. Although this could be good for the first time, we need to control the maximum radius of the polygons, otherwise if they go farther and farther, they become too big, intersecting each other densely (it also happens if the power of the attractor is too high). So I control the maximum radius value of the polygons manually.
Fig.4.35. By using a <minimum> component (Scalar > Util > Minimum) and a user defined number, I am telling the algorithm to choose the value from the <div> component, if it is smaller than the number that I defined as a maximum radius by <number slider>. As you can see in the pictures, those polygons that are in the power field of attractor being affected and others are constant. Now if you change the position of the <attractor_1> in the Rhino workplace manually, you can see that all polygons get their radius according to the <attractor_1> position.
Chapter 4
Fig.4.36. The effect of the <attractor_1> on all polygons. Displacement of the attractor, affects all polygons accordingly.
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Fig.4.37. Wiyh the same concept, I can displace polygons in Z direction based on the numbers coming from the <Min> component or changing it by mathematical functions, if necessary. Simple. I can do any other function on these polygons like rotate, change colour, etc. But let’s think what would happen if I would have two attractors in the field. I made another cluster which means another point in Rhino associated with a <point> and <circle> in Grasshopper. It seems that the first part of the algorithm is the same. Again I need to measure the distance between this <attractor_2> and the polygons’ center or <pt_grid> and then find the <min> of these distances and the previously defined maximum number for the radius.
Chapter 4
Fig.4.38. Introducing second <attractor_2> to and applying the same algorithm to it. Now we have two different data lists that include the distance from the polygon to each attractor. Since the closer attractor would affect the polygon more, I should find one which is closer, and use that one as the source of action. So I will use a <min> component to find which distance is minimum or which point is closer.
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4.39. Finding the closer attractor. After finding the closer one by <min> component, the rest of the process would be the same. Now all <polygon>s are being affected by to attractors.
Chapter 4
Fig.4.40. Again you can change the position of the attractors and see how all polygons reacting accordingly. We can have more and more attractors. The concept is to find the attractor which is closer for each polygon and apply the effect by selecting that one. Selection in terms of distance happens with <min> functions, but we will talk about other types of selection later. There are other ways of dealing with attractors like using <cull> component. In this method you need to provide different lists of data from the distance between points and attractors and then culls those far, select the closer one by simple Boolean function of a>b. since there are multiple examples on this topic on‐line, I hope you will do them by yourself.
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Curve Attractors: Wall project Let’s complete this discussion with another example but this time by Curve attractors because in so many cases you need to articulate your field of objects with linear attractors instead of points. My aim here is to design a porous wall for an interior space to let me have a multiple framed view to the other side. This piece of work could be cut from sheet material. In my design space, I have a plane sheet (wall), two curves and bunch of randomly distributed points as base points of cutting shapes. I decided to generate some rectangles by these points, cutting them out of the sheet, to make this porous wall. I also want to organize my rectangles by this two given curve so at the end, my rectangles are not just some scattered rectangles, but randomly distributed in accordance to these curves which have a level of organisation in macro scale and controlled randomness in micro scale. What I need is to generate this bunch of random points and displace them toward the curves based on the amount of power that they receive from these lines. I also decided to displace the points toward both curves so I do not need to select closer one, but I displace the points based on their distance to the curve. Then I want to generate my rectangles over these points and finally I will define the size of these rectangles in relation to their distance to the attractors.
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Fig.4.41. Generating a list of randomly distributed <point>s and introducing the attractors by two <curve> component (Params > Geometry > Curve) over a sheet. I used an <interval> component to define the numeric interval between 0 and <number slider> for the range of random points. I will make a cluster by <interval>, <random>, <jitter> and <point> to make the canvas more manageable.
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Fig.4.42. When the attractor is a point, you can simply displace your geometry towards it. But when the attractor is a curve, you need to find a relative point on curve and displace your geometry towards that specific point. And this point must be unique for each geometry, because there should be a one to one relation between attractor and any geometry in the field. If we imagine an attractor like a magnet, it should pull the geometry from its closest point to the object. So basically what I first need is to find the closest point of <Rnd_pt_grid> on both attractors. These points are the closest points on the attractors for each member of the <Rnd_Pt_Grid> separately. I used <Curve CP> component (Curve > Analysis > Curve CP) which gives me the closest point of the curve to my <Rnd_Pt_Grid>. Chapter 4
Fig.4.43. In order to displace the points towards the attractors, I need to define a vector for each point in <Rnd_Pt_Grid>, from the point to its closest point on the attractors. Since I have the start and end point of the vector I am using a <vector 2Pt> component to do that. The second point of the vector (B port of the component) is the closest point on the curve.
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Fig.4.44. Now I connected all my <Rnd_Pt_Grid> to two <move> components to displace them towards the attractors. But if I use the vector which I created in the last step, it displaces all points onto the curve and that’s not what I want. I want to displace the points in relation to their distance to the attractor curves. If you look at the <Curve CP> component it has an output which gives us the distance between the point and the relevant closest point on the curve. Good. We do not need to measure the distance by another component. I just used a <Function 2> component and I attached the distance as X and a <number slider> to Y to divide the X/Log(Y) to control the factor of displacement (Log function change the linear relation between distance and the resulting factor for displacement).
I just used a <multiply> component (Vector > Vector > Multiply), I attached the <vector 2P>as base vector and I changed its size by the factor I created by distance, and I attached the resulting vector to the <move> components which displaces the <Rnd_Pt_Grid> in relation to their distance to the attractors, and towards them.
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Fig.4.45. The <number slider> changes the power with which attractors displace objects towards themselves.
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Fig.4.46. I used a <rectangle> component and I attached the <move>d or displaced points to it as the base point (planes) for my rectangle components. But as I told you, I want to change the size of the <rectangle>s based on their distances to each <attractor>. So I used the same numerical values which I used for vector magnitude and I changed them by two functions. I divided this value by 5 for the X value of the rectangles and I divided by 25 for the Y value. As you can see, rectangles have different dimensions based on their original distance from the attractor.
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Fig.4.47. Manipulating the variables would result in differentiated models that I can choose the best one for my design purpose.
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Fig.4.48. Different sh hadow effectts of the fina al design prod duct as a porrous wall sysstem.
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Fig.4.49. Final design product.
Chapter_ 5_Parametric Space
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Chapter_ 5_Parametric Space Our survey in Geometry looks for objects in the space; Digital representation of forms and tectonics; different articulation of elements and multiple processes of generations; from classical ideas of symmetry and pattern up to NURBS and curvature continuity.
We formulate the space by coordinate systems to identify some basic properties like position, direction and measurement. The Cartesian coordinate system is a 3 dimensional space which has an Origin point O=(0,0,0) and three axis intersecting at this point which make the X, Y and Z directions. But we should consider that this 3D coordinate system also includes two ‐ dimensional system ‐ flat space (x, y) ‐ and one dimension‐linear space (x) ‐ as well. While parametric design shifts between these spaces, we need to understand them as parametric space a bit. 5_1_One Dimensional (1D) Parametric Space The X axis is an infinite line which has some numbers associated with different positions on it. Simply x=0 means the origin and x=2.35 a point on the positive direction of the X axis which is 2.35 unit away from the origin. This simple, one dimensional coordinate system could be parameterised in any curve in the space. So basically not only the World X axis has some real numbers associated with different positions on it, but also any curve in the space has the potential to be parameterized by a series of real numbers that show different positions on the curve. So in our 1D parameter space when we talk about a point, it could be described by a real number which is associated with a specific point on the curve we are dealing with. It is important to know that since we are not working on the world X axis any more, any curve has its own parameter space and these parameters does not exactly match the universal measurement systems. Any curve in the Grasshopper has a parameter space starts from zero and ends in a positive real number (Fig.5.1).
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We are dealing with objects. These objects could be boxes, spheres, cones, curves, surfaces or any articulation of them. In terms of their presence in the space they generally divided into points as 0‐dimensional, curves as 1‐dimensional, surfaces as 2‐dimensional and solids as 3‐dimensional objects.
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Fig.5.1. 1D‐parameter space of a curve. Any ‘t’ value is a real number associated with a position on the curve.
So talking about a curve and working and referencing some specific points on it, we do not need to always deal with points in 3D space with p=(X,Y,Z) but we can recall a point on a curve by p=t as a specific parameter on it. And it is obvious that we can always convert this parameter space to a point in the world coordinate system. (Fig.5.2)
Fig.5.2. 1D‐parmeter space and conversion in 3D coordinate system.
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5_2_Two Dimensional (2D) Parametric Space Two axis, X and Y of the World coordinate system deals with the points on an infinite flat surface that each point on this space is associated with a pair of numbers p=(X,Y). Quite the same as 1D space, here we can imagine that all values of 2D space could be traced on any surface in the space. So basically we can parameterize a coordinate system on a curved surface in the space, and call different points of it by a pair of numbers here known as UV space, in which P=(U,V) on the surface. Again we do not need to work with 3 values of (X,Y,Z) as 3D space to find the point and instead of that we can work with the UV “parameters” of the surface. (Fig.5.3)
Fig.5.3. UV (2D) parameter space of surface. These “Parameters” are specific for each surface by itself and they are not generic data like the World coordinate system, and that’s why we call it parametric! Again we have access to the 3D equivalent coordinate of any point on the surface (Fig.5.4).
Fig.5.4. Equivalent of the point P=(U,V) on the world coordinate system p=(X,Y,Z).
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5_3_Transition between spaces It is a crucial part in parametric thinking of design to know exactly which coordinate system or parameter space we need to work with, in order to design our geometry. Working with free form curves and surfaces, we need to provide data for parameter space but we always need to go back and forth for the world coordinate system to provide data for other geometry creations or transformations etc. It is almost more complicated in scripting, but since Grasshopper has a visual interface rather than code, you simply identify which sort of data you need to provide for your design purpose. Consider that it is not always a parameter or a value in a coordinate system that we need in order to call geometries in Generative Algorithms and Grasshopper, sometimes we need just an index number to do it. If we are working with a bunch of points, lines or whatever, and they have been generated as a group of objects, like point clouds, since each object associated with a natural number that shows its position in a list of all objects, we just need to call the number of the object as index instead of any coordinate system. The index numbering like array variables in programming is a 0‐based counting system which starts from 0 (Fig.5.5).
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Fig.5.5. Index number of a group of object is a simple way to call an on object. This is 0‐based counting system which means numbers start from 0. So as mentioned before, in Associative modelling we generate our geometries step by step as some related objects and for this reason we go into the parameter space of each object and extract specific information of it and use it as the base data for the next steps. This could be started from a simple field of points as basic generators and ends up at the tiny details of the model, in different hierarchies.
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5_4_Basic Parametric Components 5_4_1_Curve Evaluation The <evaluate> component is the function that can find the point on a curve or surface, based on the parameter you feed. The <evaluate curve> component (Curve > Analysis > Evaluate curve) takes a curve and a parameter (a number) and gives back a point on curve on that parameter.
Fig.5.6. The evaluated point on <curve> on the specific parameter which comes from the <number slider>. Chapter 5
Fig.5.7. We can use <series> of numbers as parameters to <evaluate> instead of one parameter. In the above example, because some numbers of the <series> component are bigger than the domain of the curve, you see that <Evaluate> component gives us warning (becomes orange) and that points are located on the imaginary continuation of the curve.
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Fig.5.8. Although the ‘D’ output of the <curve> component gives us the domain of the curve (minimum and maximum parameters of the curve), alternatively we can feed an external <curve> component from Param > Geometry and in its context menu, check the Reparameterize section. It changes the domain of the curve to 0 to 1. So basically I can track all <curve> long by a <number slider> or any numerical set between 0 and 1 and not be worry that parameter might go beyond the numerical domain of the curve.
There are other useful components for parameter space on curves on Curves > Analysis and Division that we talk about them later. 5_4_2_Surface Evaluation While for evaluating a curve we need a number as parameter (because curve is a 1D‐space) for surfaces we need a pair of numbers as parameters (U, V), with them, we can evaluate a specific point on a surface. We use <evaluate surface> component (Surface > Analysis > Analysis) to evaluate a point on a surface on specific parameters. We can simply use <point> components to evaluate a surface, by using it as UV input of the <Evaluate surface> (it ignores Z dimension) and you can track your points on the surface just by X and Y parts of the <point> as U and V parameters.
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Fig.5.9. A point <Evaluate>d on the <surface> base on the U,V parameters coming from the <number slider> with a <point> component that make them a pair of Numbers. Again like curves you can check the ‘Reparameterize’ on the context menu of the <surface> and set the domain of the surface 0 to 1 in both U and V direction. Change the U and V by <number slider> and see how this <evaluated> point moves on the surface (I renamed the X,Y,Z inputs of the component to U,V,‐ manually).
Fig.5.10. Since we can use <point> to <evaluate> a <surface> as you see we can use any method that we used to generate points to evaluate on the <surface> and our options are not limited just to a pair of parameters coming from <number slider>, and we can track a surface with so many different ways.
Fig.5.11. To divide a surface (like the above example) in certain rows and columns we can use <Divide surface> or if we need some planes across certain rows and columns of a surface we can use <surface frame> both from Surface tab under Util section.
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5_5_On Object Proliferation in Parametric Space For so many design reasons, designers now use surfaces to proliferate some other geometries on them. Surfaces are flexible, continues two dimensional objects that prepare a good base for this purpose. There are multiple methods to deal with surfaces like Penalisation, but here I am going to start with one of the simplest one and we will discuss about some other methods later. We have a free‐form surface and a simple geometry like a box. The question is, how we can proliferate this box over the surface, in order to have a differentiated surface i.e. as an envelope, in that we have control of the macro scale (surface) and micro scale (box) of the design separately, but in an associative way. In order to do this, we should deal with this surface issue by dividing it to desired parts and generate our boxes on these specific locations on the surface and readjust them if we want to have local manipulation of these objects. Generating the desired locations on the surface is easy. We can divide surface or we can generate some points based on any numerical data set that we want. About the local manipulation of proliferated geometries, again we need some numerical data sets which could be used for transformations like rotation, local displacement, resize, adjustment, etc.
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Fig.5.12. A free‐form, reparameterized, <surface> being <evaluate>d by a numeric <range> from 0 to 1, divided by 30 steps by <number slider> in both U and V direction. (Here you can use <divide surface> but I still used the <point> component to show you the possibilities of using points in any desired way).
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Fig.5.13. As you see the <evaluate> component gives ‘Normal’ and ‘plane’ of any evaluated points on the surface. I used these frames to generate series of <box>es on them while their sizes are being controlled by <number slider>s.
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In order to manipulate the boxes locally, I just decided to rotate them, and I want to set the rotation axis the Y direction of the coordinate system so I should use the XZ plane as the base plane for their rotation (Fig.5.13).
Fig.5.14. Local rotation of the box.
Fig.5.15. The <rotate> component needs ‘geometry’ which I attached <box>es and ‘rotation angle’ that I used random values (you can rotate them gradually or any other way) and I set the Number of random values as much as boxes. Finally to define the plane of axis, I generated <XZ plane>s on any point that I <evaluate>d on the <surface> and I attached it to the <rotate> component.
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Fig.5.16. Final geometry.
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Non‐uniform use of evaluation During a project this idea came to my mind that why should I always use the uniform distribution of the points over a surface and add components to it? Can I set some criteria and evaluate my surface based on that and select specific positions on the surface? Or since we use the U,V parameter space and incremental data sets (or incremental loops in scripting) are we always limited to a rectangular division on surfaces? There are couple of questions regarding the parametric tracking a surface but here I am going to deal with a simple example to show how in specific situations we can use some of the U,V parameters of a surface and not a uniform rectangular grid over it. Social Space I have two Free‐form surfaces as covers for a space and I think to make a social open space in between. I want to add some columns between these surfaces but because they are free‐form surfaces and I don’t want to make a grid of columns, I decided to limit the column’s length and add as many places as possible. I want to add two inverted and intersected cone as columns in this space just to make the shape of them simple.
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Fig.5.17. Primary surfaces as covers of the space.
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Fig.5.18. I introduced surfaces to Grasshopper by <srf_top> and <srf_bottom> and I Reparameterized them. I also generated a numerical <range> between 0 and 1, divided by <number slider>, and by using a <point> component I <evaluate> these surfaces at that <points>. Again just to say that still it is the same as surface division.
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Fig.5.19. I generated bunch of <line>s between all these points, but I also measured the distance between any pair of points (we can use line length also), as I said I want to limit these lines by their length.
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Fig.5.20. Here I used a <dispatch> component (Logic > Streams > Dispatch) to select my lines from the list. A <dispatch> component needs Boolean data which is associated with the data from the list to sent those who associated with True to the A output and False one to the B output. The Boolean data comes from a simple comparison function. In this <function> I compared the line length with a given number as maximum length of the line (x>y, x=<number slider>, y=<distance>). Any line length less than the <number slider> creates a True value by the function and passes it through the <dispatch> component to the A output. So if I use the lines coming out the output of the <dispatch> I am sure that they are all less than the certain length, so they are my columns.
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Fig.5.21. The geometry of columns is just two inverted cones which are intersecting at their tips. Here because I have the axis of the column, I want to draw to circles at the end points of the axis and then extrude them to the points on the curve which make this intersection possible.
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Fig.5.22. By using an <end points> component I can get the both ends of the column. So I attached these points as base points to make <circle>s with given radius. But you already know that these circles are flat but our surfaces are not flat. So I need to <project> my circles on surfaces to find their adjusted shape. So I used a <project> component (Curve > Util > Project) for this reason.
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Fig.5.23. The final step is to extrude these projected circles towards the specified points on column’s axis (Fig.5.20). So I used <extrude point> component (Surface > Freeform > Extrude point) and I connected the <project>ed circles as base curves. For the extrusion point, I attached all columns’ axis to a simple <curve> component and I ‘Reparameterized’ them, then I <evaluate>d them in two specific parameter of 0.6 for top cones and 0.4 for bottom cones.
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Fig.5.24. Although in this example, again I used the grid based tracking of the surface, I used additional criteria to choose some of the points and not all of them uniformly.
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Fig.5.25. Final model.
Chapter_6_ Deformation and Morphing
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Chapter_6_ Deformation and Morphing 6_1_Deformation and Morphing Deformation and Morphing are among the powerful functions in the realm of free‐form design. By deformations we can twist, shear, blend, … geometries and by Morphing we can deform geometries from one boundary condition to another. Let’s have a look at a simple deformation. If we have an object like a sphere, we know that there is a bounding‐box (cage) around it and manipulation of this bounding‐box could deform the whole geometry.
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Fig.6.1. Deformation by Bounding‐box (cage). Based on different manipulations, we might call it shear or blend or free deformation. For any deformation function, we might need the whole bounding‐box, or just one of its sides as a plane or even one of the points to deform. If you check different deformation components in Grasshopper you can easily find the base geometrical constructs to perform the deformations. Morphing in animation means transition from one picture to another smoothly or seamlessly. Here in 3D space it means deformation from one state or boundary condition to another. The Morphing components in Grasshopper work in the same fashion. There are two <morph> components, one deform an object from a reference box (Bounding Box) to a target box, the other component works with a surface as a base on that you can deform your geometry, on the specified domains of the surface and height of the object.
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The first one is <Box Morph> and the next one is <Surface Morph> both from XForm tab under the Morph section. Since we have couple of commands that deform a box, if we use these deformed boxes as target boxes then we can deform any geometry in Grasshopper by combination with Box Morph component. As you see in Figure.6.2 we have an object which is introduced to Grasshopper by a <Geometry> component. This object has a bounding‐box around it which I draw here just to visualize the situation. I also draw another box by manually feeding values.
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Fig.6.2. Object and manually drawn box.
Fig.6.3. The <Box morph> component (XForm > Morph > Box morph) deforms an object from a reference box to a target box. Because I have only one geometry I attached it as a bounding box or reference box to the component but in other cases, you can use <Bounding box> component (Surface > Primitive > Bounding box) to use as the source box. I unchecked the preview of the <Box> component to see the morphed geometry better.
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Fig.6.4. Now if you simply change the size of the target box you can see that the morphed geometry would change accordingly.
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Fig.6.5. here you see that instead of one box, if I produce bunch of boxes, we can start to morph our object more and more. As you see the differentiated boxes by the <series> component in their Y dimension, show the differentiation in the morphed object as well. 6_2_On Panelization One of the most common applications of the morphing functions is Panelization. The idea of panelization comes from the division of a free‐form surface geometry into small parts especially for fabrication issues. Although free‐form surfaces are widely being used in car industry, it is not an easy job for architecture to deal with them in large scales. The idea of panelization is to divide a surface into small parts which are easier to fabricate and transport and also more controllable in the final product. Sometimes the reason is to divide a curve surface into small flat parts and then get the curvature by the accumulation of the flat geometries which could be then fabricated from sheet materials. There are multiple issues regarding the size, curvature, adjustment, etc. that we try to discuss some of them.
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Let’s staart with a sim mple surface and a compo onent as the e module to m make this surface.
Fig.6.6. A A simple dou uble‐curve su urface for pa anelization.
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ant to prolifeerate on the surface……. Not special, jjust for exam mple !!! Fig.6.7. The component that I wa
we need to in ntroduce ourr surface and d module as Grasshopperr componentts. Based Fig.6.8. First of all, w p com mponents in the Grasshopper, the idea i is to generate cou uple of boxes on the on the possible surface and use theese boxes ass target boxees and morp ph or modulle into them. So I used the <box morph> and I used tthe geometrry itself as bo ounding‐boxx. Now we neeed to generrate our targ get boxes to morph the compo onent into them.
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Fig.6.9. The component that we need to make our target boxes is <surface box> (XForm > Morph > Surface box). This component generates multiple boxes over a surface based on the intervals on the surface domain and height of the box. So I just attached the surface to it and the result would be the target boxes for the <box morph> component. Here I need to define the domain interval of the boxes, or actually number of boxes in each U and V direction of the surface.
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Fig.6.10. Now I connected <divide interval2> which tells the <surface box> that how many divisions in U and V directions we need. Another <number slider> defines the height of the target boxes which means height of the morphed components. So basically the whole idea is simple. We produce a module (a component) and we design our surface. Then we make certain amount of boxes over this surface (as target boxes) and then we morph the module into these boxes. After all we can change the number of elements in both U and V direction and also change the module which updates automatically on the surface.
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Fig.6.11. Final surface made up of our base module 6_3_Micro Level Manipulations Although it is great to proliferate a module over a surface, it still seems a very generic way of design. It is just a ‘bumpy’ surface. We know that we can change the number of modules, or change the module by itself, but still the result is a generic surface and we don’t have local control of our system. Now I am thinking of making a component based system that we could apply more local control over the system and avoid designing generic surfaces which are not responding to any local, micro scale criteria. In order to introduce the concept, let’s start with a simple example and proceed towards a more practical one. We used the idea of attractors to apply local manipulations to a group of objects. Now I am thinking to apply the same method to design a component based system with local manipulations. The idea is to change the components size (in this case, their height) based on the effect of a (point) attractor.
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As you have seen, the <surface box> component has the height input which asks for the height of the boxes in the given intervals. The idea is to use relative heights instead of constant one. So instead of a constant number as height, we can make a relation between the position of each box in relation to the attractor’s position. What I need is to measure the distance between each box and the attractor. Since there is no box yet, I need a point on surface at the center of each box to measure the distance.
Fig.6.13. Here I used the same <divide interval2> for an <Isotrim> component (Surface > Util > Isotrim). This component divides the surface into sub‐surfaces. By these sub‐surfaces I can use another component which is <BRep Area> (Surface > Analysis > BRep area) to actually use the by‐ product of this component that is ‘Area Centroid’ for each sub‐surface. I measured the distance of these points (area centroids) from the <attractor> to use it as the reference for the height of the target boxes in <surface box> component.
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Fig.6.12. A double‐curve surface introduced as <Base_Srf> and a cone which is introduced as <component> to the Grasshopper, a <divide interval2> for surface divisions, and a <bounding box> as the reference box of the <component>. Here I used a <scale> component for my bounding box. Now if I change the size of the bounding box, I can change the size of all components on the <base_srf> because the reference box has changed.
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Fig.6.14. I just divided the distances by a given number to control the effect of the attractor and I used the result as ‘height’ to generate target boxes with <surface box> component. The surface comes from the <base_srf>, the <divide interval2> used as surface domain and heights coming from the relation of box position and the attractor. As you see, the height of boxes now differ, based on the position of the point attractor.
Fig.6.15. The only remaining part, connecting the <component>, <scale>d bounding box and <surface box> to a <morph box> component and generate the components over the surface. By changing the scale factor, you can change the size of the all components and like always, the position of the attractor is also manually controllable.
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Fig.6.16. Final model. As you see, the size of components started to accept local manipulations, based on an external property which is here a point attractor. Although the idea is a simple attractor, the result could be interesting and you could differentiate your reference boxes in so many other ways as well. Now we know that the morphing concept and panelization is not always generic. Having tested the concept, let’s go for another practical experiment.
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6_4_On R Responsive M Modulation The ideaa for the nexxt step is to m modulate a ggiven surface e with contrrol over each h module, means any module of this systeem, has to b be responsible for some certain criteeria. So even n more than regional here I want tto have a more specific control overr my system by given differentiation of the modules, h criteria. These could d be environm mental, funcctional, visuaal or any other associativve behaviourr that we ur module bee responsiblee for. want ou In the next examp ple, in ordeer to make a building’’s envelope more resp ponsive to the t host environm ment, I just wanted the system be rresponsive to the sun ligght. In your experimentss it could be wind, rain or inteernal function ns or any oth her criteria that you are llooking for.
The Grasshopper deefinition doess not have anything new w, but it is thee concept th hat allows us to make variation n over the eenvelope insttead of making a genericc surface. Baasically when n the surface e is free‐ form an nd it turns arround and has h differentt orientations, it also has different aangles with the t main sun lightt at each part. So based on the anglle differentiaation betweeen the surface and the ssun light, this variation in the componentss happens to o the system..
Fig.6.17 7.First sketchees of responssive moduless on façade ssystem.
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Here I have a surfacce, simply as the envelop pe of a building which I w want to coveer with two different types off componentts. One whicch is closed aand does no ot allow the penetration of the sun light and over my envvelope based on the the otheer has opening. These components c should be proliferated p main dirrection of the sun light aat the site. I sset a user de efined angle to say the algorithm thaat for the certain d degrees of su un light we sshould have cclosed comp ponents and for the others, open one es.
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Fig.6.18. Surface of the building as envelope.
Fig.6.19. Two different types of components for building envelope.
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Fig.6.20. The first step is similar to the previous experiments. I introduces <surface> and I used <divide interval2> to divide it in U and V directions and I generated target boxes by <surface box>. I also used the <isotrim> with the same intervals and I used <BRep area> to find the centroid of each area (which is selected in green). At the same time I used a <curve> component to introduce the main sun‐light angle of the site and whit its <end points> I made a <vector 2pt> which specify the direction of the sun light. You can manipulate and change this curve to see the effect of sun light in different directions on components later on.
Fig.6.21. in order to evaluate the angle between the sun‐light and the surface, I want to measure this angle between sun light and normals of the surface at the position of each component. So I can decide for each range of angles what sort of component would be useful. So after generating the center points, I need normals of the surface at those points. that’s why I used a <surface CP> to find the closest point of each center point on the surface to get its UV parameters and use these parameters to <evaluate> the surface at that points to actually get the normals of surface at that points.
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Fig.6.22. I used an <angle> component (Vector > Vector > Angle) to evaluate the angle between the sun direction and the façade. Then I converted this angle to degree and I used a <function> to see whether the angle is bigger than the <max_angle> or not. This function (x>y) gives me Boolean data, True for smaller angles and False for bigger angles. Chapter 6
Fig.6.23. Based on the Boolean data comes from the angle comparison, I <dispatch> the data which are target boxes (I have the same amount of target box as the center points and normals). So basically I divided my target boxes in two different groups whose difference is the angle they receive the sun light.
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The rest of the algorithm is simple and like what we have done before. I just need to morph my components into the target boxes, here for two different ones.
Fig.6.24. Here I introduced two different components as single geometries and I used two <morph box> components each one associated with one part of the <dispatched> data to generate <C_close> or <C_open> components over the façade.
6.25. Now if you look closer, you can see that in different parts of the façade, based on its curvature and direction, different types of components are generated.
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Fig.6.26. Final model. The bifurcation of the target boxes (and the components) could be more than two in the algorithm. It depends on the design and the criteria that we use. We can think about a component based façade, in which some components are closed, and some of them are open, in which open ones have adjustable parts that orientate towards external forces, and have apertures that adjust themselves to the internal functions of the building and so on and so forth. You see that the idea is to have module based control over the system. We still have the global control (form) and regional control (affecting components height or scale regionally) over the system as well.
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Surface geometries seems to be the final products in our design, like facades, walls etc. and that’s why we need lots of effort to generate the data like points and curves that we need as the base geometries and finally surfaces. Here I decided to design a very simple schematic building just to indicate that the multiple surface components in the Grasshopper have the potential to generate different design products by very simple basic constructs. 7_1_Parametric NURBS Surfaces In the areas of Docklands of Thames in London, close to the Canary Wharf, where the London’s high rises have the chance to live, there are some potential to build some towers. I assumed that we can propose one together, and this design could be very simple and schematic, here just to test some of the basic ideas of working with free‐form surfaces. Let’s have a look at the area.
Fig.7.1. Arial view, Canary Wharf, London (image: www.maps.live.com, Microsoft Virtual Earth).
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We have had some experiments with surfaces in the previous chapters. We used Loft and Pipe to generate some surfaces. We also used free form surface and some surface analysis accordingly. Usually by surfaces, we mostly mean Free‐Form surfaces which we generate them by curves or sometimes bunch of points. So usually generating surfaces depends on the curves that we provide for our surface geometries. There are multiple surface components in the Grasshopper and if you have a little bit of experience in working with Rhino you should know how to generate your surface geometries by them.
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The site that I have chosen to design my project is in the bank of Thames, with a very good view to the river and close to the entrance square to the centeral part of Canary Wharf (Westferry Road). I don’t want to go through the site specifics so let’s just have a look at where I am going to propose my tower and continue with the geometrical issues.
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Fig.7.2. Site of the proposed tower. Manual drawings There are multiple ways to start this sketch. I can draw the ground floor and copy it and start to add detail like what we have done for Swiss Re. Another way is to draw some boundary curves for the tower and use this curves as base geometry for the building volume and then add more details to complete the sketch. I would prefer the second way since we already know about the first one. I am going to draw the boundary curves in Rhino. I can draw these lines in Grasshopper but because I want them to match the geometry of the site so I need some data from the site which I can get in this way. I have a vague idea in mind. My tower has some linear elements in façade which are generic, but I also want to have some hollow spaces between the outside and inside, positioned on the façade. I also want to design a public space close to the river and connected to the tower with the same elements as façade, continuous from tower towards the river bank.
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Fig.7.3. The basic lines of the tower associated with the site. As you see in the Figure.7.3 I have drawn my base curves manually in Rhino. These curves correspond to the site specifics, height limitations, site’s shape and borders, etc, etc. four curves drawn for the main building and another two curves as the borders of the public space, which started from the earth level and then went up to be parallel to the tower edges. These curves are very general. You could draw whatever you like and go for the rest of the process. Basic façade elements
Fig.7.4. For the first step I imported these four corner curves into Grasshopper by <curve> components and then I used <divide curve> to divide these curves into <N_div> parts.
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Fig.7.5. Now I have the same amount of division points on each curve and I draw <line> between each curve’s points and the next one, so basically I generated <N_div> number of quads like floor edges.
Fig.7.6. By connecting these <line>s to <extrude> component (Surface > Freeform > Extrude) and extrusion in Z direction, I generated surfaces across façade which reminds me Mies high‐rises but in a differentiated way!
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Fig.7.7. There are four straight lines that connect the four corners of the tower in plan. I introduced these lines by another four <curve> components and I generated four surfaces as general façade surfaces that could be glass. Here because I have the planner section curve and two curves that define the boundary of the façade on each face, I used a <Sweep 2> component to use Rails for making a surface by a section curve. So basically after I attached each <Crv_p_n> to the <Sweep2> component as the section curve, I attached the corner curves as the rails to make each face’s surface. For each section curve in the plan, I used two edge curves that start from its end points. So the order is <crv_1> and <crv_2> for the <crv_p_1> and so on.
Fig.7.8. Using tower’s edge curves as Rails to make façade surfaces behind the ribs shaped façade elements that we made before.
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Fig.7.9. With the same method, I just divided another two curves that I have drawn for the public space and I attached them to a line component to generate some lines as the base geometry for extrusion. I will attach it to the same <extrude> component as the façade (don’t forget to hold the Shift key when you want to add a new component to the input of an already existing component).
Fig.7.10. Now you can alter the corner curves if you like, the number of divisions and the height of extrusion to get the best schematic design.
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Local changes Up to this point you already know that how to design this tower with differentiation and gradual changes across the façade elements. Here I designed very generic surface elements and the only parameters that make the global differentiations are edge curves of the tower. Now I want to add some elements to make some local changes in these generic elements to show how you can change the behaviour of these elements you designed. As you see, the main element in our case is some lines that I generated between division points of the edge curves and then I extruded them to make the façade elements. There are many ways that you can change and manipulate these lines and then extrude them, so you can get different forms outside, and different spaces inside.
To add these hollow spaces, I want to add them manually as solids to the tower and then subtract them from the already existing forms. As an example I want to add some very simple ellipsoids to the façade, and randomly distribute them alongside the its face. These ellipsoids could be added by a certain spatial criteria but here the aim is just to explore modelling properties and techniques and I don’t want to go further in design issues.
Fig.7.11. As you can see, I randomly added some ellipsoids o the tower that could be special hollow spaces in the building. I will do it for all surfaces of the tower.
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Here I decided to add some hollow spaces between these elements and make the façade with these general linear elements with locally omitted parts as the combination of general and local design effects.
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Fig.7.12. after distributing all ellipsoids alongside the façade surfaces now I import them to the Grasshopper by <Geometry> components and for each side of the façade I use one <geometry> component and introduce them as ‘Set multiple Geometries’. So each side of the façade has one group of ellipsoid geometries.
Fig.7.13. By using a <Trim with BReps> component (Curve > Util > Trim with BReps) I can trim these lines across the façade by these randomly distributed ellipsoids. The output of the component are curves inside the BRep or outside it and I use the outside curves to extrude with the same <extrude> component that I generated before. I hide all geometries in Rhino and uncheck the preview in Grasshopper to see the result better. I will do it for all faces and I will connect the output curves to the extrude component.
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Fig.7.14. Now the facade elements are all have their trimmed ellipsoids inside. The next step is to add these hollow spaces to the façade surface as well. The process is simple. I just need to find the difference between these ellipsoid geometries and façade surfaces. The only point is that the direction of the normal of the surface affects the ‘difference’ command so you should be aware of that and if needed, change this direction.
Fig.7.15. The <Solid difference> component (Intersect > Boolean > Solid difference) takes two different solids and performs solid difference on them. So I attached the façade surface and hollow geometries to find the difference of them. As I told you here I also used <Flip> component (Surface > Util > Flip) to flip the Normal of the surface and then attach it to the <Solid difference> component otherwise the hollow space would face the inside of the tower. I will do it for all surfaces of the tower.
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Fig.7.16. All trimmed façade surfaces. Now you only need to burn your geometries, cap the tower and render it to complete your sketch. Again I should mention here that this was just an example to show that how you can generate series of elements and differentiate them in global and local scale with surface components.
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Fig.7.17.a/b. Final model.
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7_2_Mesh h vs. NURBSS Up to now n we havve used diffferent comp ponents thatt worked wiith the NUR RBS surfacess. But as mention ned before there t are otther types of o surfaces which w are usseful in otheer contexts. It is not always tthe smooth b beauty of thee NURBS thaat we aimed ffor, but we m might need m more precise e control, easier processing p or simpler eq quations. Beeside the claassical surfacce types of revolution, ruled or pipes, w we have different free forrm surfaces llike Besier orr B‐Splines. B But here I am m going to taalk a little bit abou ut meshes wh hich are quieet different ttypes of surfaaces.
If we loo ok at a mesh h, first we seee its faces. FFaces could be triangle, quadrant or hexagon. Byy looking closer w we can see a grid of poin nts which make these faaces. These p points are b basic elements of the mesh su urface. Any tiny group off these points (for examp ple any threee in triangulaar mesh) make a face with wh hich the who ole geometryy become su urface. These points aree connected together byy straight lines. There arre two important issues about meshes: position of these points and the connectivity of these points. TThe position n of the poin nts related to o the geome etry of the m mesh and thee connectivity of the points reelated to thee topology. 7_2_1_G Geometry an nd Topology While th he geometryy deals with the position n of the stuff !! in the space, s topollogy deals with w their relationss. Mathematically speakking, topologgy is a prop perty of the object that transformation and deformaation cannott change it. SSo for instan nce circle and d ellipse are topologically the same and they have only geometriccal differencee. Have a loo ok at Figure 7.20. As you u see there aare four poin nts which nected to eaach other. In the first imaage, both A and B have the same to opology becaause they are conn have th he same rellation between their points p (same e connection). But theey are geom metrically different, because of the displaccement of on ne point. Butt in the secon nd image, the geometry of points is the saame but the cconnectivity is different aand they are e not topologgically equivaalent.
and Geometry ry. Fig.7.20. Topology a
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Meshes are anotherr type of freee‐form surfaces but mad de up of small parts (facees) and accumulation whole surfacce. So there is no internaal, hidden mathematical function of thesee small parts makes the w that gen nerates the shape of the surface, but these faces define the shape of the surface all to ogether.
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The idea of topology is very important in meshes. Any face in the mesh object has some corner points and these corner points are connected to each other with an order in a same way for all faces of the mesh object. So we can apply any transformation to a mesh object and displace the vertices of the mesh in the space even non‐uniformly, but the connectivity of the mesh vertices should be preserved to remain the faces otherwise it collapses.
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Fig.7.21. A: Both red and grey surfaces are meshes with the same amount of faces and vertices, but in the gray one, the vertices are just displaced, make another geometrical configuration of the mesh, but the connectivity of the mesh object is not changed and both surfaces are topologically the same. B: Two Mesh surfaces are geometrically the same and have the same amount of vertices but connectivity of vertices are different, faces are triangular in the gray one but quad in the red, make different topology for them. Knowing the importance of topological aspects of mesh objects, they are powerful geometries while we have bunch of points and we need a surface type to represent them as a continuous space. Different types of algorithms that working with points could be applied to the mesh geometry since we save the topology of the mesh. For instance, using finite element analysis or specific applications like dynamic relaxation, and particle systems, it is easier to work with meshes than other types of surfaces since the function can work with mesh vertices. Mesh objects are simple to progress and faster to process; they are capable of having holes inside and discontinuity in the whole geometry. There are also multiple algorithms to refine meshes and make smoother surfaces by mesh objects. Since different faces could have different colours initially, mesh objects are good representations of analysis purposes (by colour) as well. There are multiple components that dealing with the mesh objects in ‘mesh’ tab in Grasshopper. Let’s start a mesh from scratch and then try to grow our knowledge.
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7_3_On Particle Systems I have a group of points and I want to make a surface by these points. In this example the group of points is simplified in a grid structure. To discuss the basic concept of the ‘particle systems’ here I will try to project this idea on a mesh geometry in a simple example. I am thinking of a vertical grid of points that represent the basic parameters of a surface which is being affected by an imaginary wind pressure. I want to displace the points by this wind factor (or any force that has a vector) and represent the resultant deformed surface. Basically by changing the wind factor, we can see how the surface changes.
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(Actually I want to deal with a surface geometry and an extra force that applies to this surface, with each point as a particle separately, and observe the result with changes of the main force)
Fig.7.22. The first step is simple. By using a <series> component with controlled number of points <N_pt>, and distance between them <distance_pt> I generated a grid of cross referenced <point>s. The pressure of the imaginary wind force, affects all points in the grid but I assumed that the force of the wind increases when goes up, so the wind pressure become higher in the higher Z values of the surfaces. And at the same time, the force affects the inner points more than the points close to the edges. The points on the edges in the plan section do not move at all (fix points). In particle system there is a relation between particles as well, and we should define another set of rules for them, but here I just set the external forces just to show how we can work with points as particles.
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Fig.7.23. Diagram of the wind force affected the surface. A: section; the vertical effect of the force, B: plan; the horizontal effect. Basically I need two different mechanisms to model these effects, one for the section diagram and another for the plan. I simplified the mechanism to an equation just to show the way we want the force affects points. For the first mechanism, to produce points displacements by using (X^2) while X is the Z value of the point being affected by the force. So for each point I need to extract the Z coordinate of the point. To make everything simple, I just assumed that the force direction is in the Y direction of the world coordinate system. So for each point on the grid, I need to generate a vector in Y direction and I set its force by the number that I receive from the first equation. For the second diagram we need a bit more of an equation to do. Let’s have a look at part one first.
Fig.7.24. The Z coordinate of the point extracted by a <decompose> component and then powered by (x^2) and then divided by a given <number slider> just to control the movement generaly. The result is a factor to <multiply> the vector (Vector > Vector > Multiply) which is simply a world <unit Y> vector as force. So basically I generated force vectors for each point in the grid.
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Fig.7.25. If I displace the points by these vectors you can see the resultant grid of points that satisfy the first step of this modelling.
Now if we look at the second part of the force modelling, as I said, I assumed that in the planner section, the points on the edge are fixed and the points on the middle displaced the most. Figure 7.26 show this displacement for each row of the point grid.
Fig.7.26. Displacement of points in rows (planner view). Since I have the force vectors for each point, I need to control them and set a value again, to make sure that their displacement in the planner section is also met the second criteria. So for each row of the points in the grid, I want to generate a factor to control the force vector’s magnitude. Here I assumed that for the points in the middle, the force vector’s power are maximum that means what they are, and for the points on the edges, it become zero means no displacement and for the other points a range in between.
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Fig.7.27. For the second mechanism, I need a <range> of numbers between 0 and 1 to apply to each point; 0 for the edge, 1 for the middle. I need a range from 0 to 1 from one edge to the middle and then from 1 to 0 to go from the middle to the other edge. I need this <range> component generates values as half of the numbers of points in each row. I set the <N_pt> to the even numbers, and I divided it by 2, then minus 1 (because the <range> component takes the number of divisions and not number of values). You see the first <panel> shows four numbers from 0 to 1 for the first half of the points. then I <reverse>d the list and I merge these two list together and as you see in the second <panel> I generated a list from 0 to 1 to 0 and the number of values in the list is the same as number of points in each row. The final step is to generate these factors for all points in the grid. So I <duplicate>d the points as much as <N_pt> (number of rows and columns are the same). Now I have a factor for all points in the grid based on their positions in their rows.
Fig.7.28. Now I need to <multiply> the force vector again by the new factors. If I displace the points by these new vectors, we can see how two different mechanisms affected the whole point grid.
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Actually this part of the example needed a little bit of analytical thinking. In reality, methods like Particle spring systems or Finite Element Analysis use this concept that multiple vectors affecting the whole points in the set and points affecting each other also. So when you apply a force, it affects the whole points and points affecting each other simultaneously. These processes should be calculated in iterative loops to find the resting position of the whole system. Here I just make a simple example without these effects of particle systems and I just want to show a very simple representation of such a system dealing with multiple forces which in real subjects are a bit more complicated!
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Fig.7.29. Mesh generation. Now if you simply add a <mesh> component (Mesh > Primitive > Mesh) to the canvas and connect the displaced points to it as vertices, you will see that nothing happening in the scene. We need to define the faces of the mesh geometry to generate it. Faces of the mesh are actually a series of numbers who just defines the way these points are connected together to make the faces of the surface. So here vertices are the geometrical part of the mesh but we need the topological definition of the mesh to generate it. Every four corner point of the grid, define a quadrant face for the mesh object. If we look at the point grid, we see that there is an index number for each point in the grid. We know each point by its index number instead of coordinates in order to deal with its topology.
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Fig.7.30. Index number of points in the grid. To define the mesh faces, we need to call every four corners that we assumed to be a face and put them together and give them to the <mesh> component to be able to make the mesh surface. The order of this points in each face is the same for all faces.
Fig.7.31. In a given point grid, a simple quadrant face defined by the order of points that if you connect them by a polyline, you can make the face. This polyline starts from a point in the grid, goes to the next point, then goes to the same point of the next row and then goes to the back column point of that row, and by closing this polyline, you will see the first face of the mesh finds its shape. Here the first face has points [0,1,6,5] in its face definition. The second face has [1,2,7,6] and so on.
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To define all mesh faces, we should find the relation between these points and then make an algorithm that generates these face matrices for us. If we look at the face matrix, we see that for any first point, the second point is the next in the grid. So basically for each point (n) in the grid, the next point of the face is (n+1). Simple! For the next point of the grid, we know that it is always shifts one row, so if we add the number of columns (c) to the point (n) we should get the point at the next row (n+c). So for instance in the above example we have 5 columns so c=5 and for the point (1) the next point of the mesh face is point (n+c) means point (6). So for each point (n) as the first point, while the second points is (n+1), the third point would be (n+1+c). That’s it.
All together for any point (n) in the grid, the face that starts from that single point has this points as the ordered list of vertices: [n, n+1, n+1+c, n+c] while (c) is the number of columns in the grid.
Fig.7.32. After defining all mesh faces, the mesh can be generated. Looking at the mesh vertices, there is a bit more to deal with. If you remember the ‘Triangle’ example of chapter_3, there was an issue to select the points that could be the first points in the grid. If you look at the grid of points in the above example, you see that the points on the last column and last row could not be the start points of any face. So beside the fact that we need an algorithm to generate the faces of the mesh object, we need a bit of data management to generate the first points of the whole grid and pass these first points to the algorithm and generate mesh faces.
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For the last point, it is always stated in one column back of the third point. So basically for each point (n+1+c) as the third point, the next point is (n+1+c‐1) which means (n+c). So for instance for the point (6) as the third point, the next point becomes point (5).
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So basically in the list of points, we need to omit the points of the last row and last column and then start to generate face matrices. To generate the list of faces, we need to generate a list of numbers as the index of points.
Fig.7.33. Generating the index number of the first points in the grid with a <series> component. The number of values in the series comes from the <N_pt> as the number of columns (same as rows) and by using a function of < x * (x‐1)> I want to generate a series of numbers as <columns*(rows‐1)> to generate the index for all points in the grid and omit the last row. The next step is to <cull> the index list by the number of columns (<N_pt>) to omit the index of the last columns as well.
Fig.7.34. Final index number of the possible first points of the mesh faces in the grid (with 8 points in each column).
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Fig.7.35. A <Mesh quad> component (Mesh > Primitive > Mesh quad) is in charge of generating quad faces in Grasshopper. I just attached the list of first numbers to the first point of the <quad>. Now this is the time to generate the list of indices for the faces:
Fig.7.36. While (x) is the index of the first point (<cull> list) and (y) is the number of the columns(from <N_pt> number slider), the second point is (x+1), the third point is ((x+1)+y) (the index of second point + number of columns), and the last point is ((x+1+y)‐1) (the index of the third point ‐1).
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Fig.7.37. The resultant mesh. Chapter 7
Fig.7.38. Changing the parameters, related to the force and manipulating the mesh. 7_4_On Colour Analysis To finish this example, let’s have a look at how we can represent our final mesh with colours as a medium for analysis purposes. There are different components in the Grasshopper that provide us colour representations and these colours are suitable for our analysis purpose. Here in this example, again to bring a concept, I simply assumed that at the end, we want to see the amount of deviation of our final surface from the initial position (vertical surface). I want to apply a gradient of colours start from the points which remained fix with bottom colour up to the points which has the maximum amount of deviation from the vertical position with the higher colour of the gradient.
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Simply, to find the amount of deviation, I need to measure the final state of each point to its original state. Then I can use these values to assign colour to the mesh faces base on these distances.
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Fig.7.39. If I simply go back, I have the initial point grid that we generated in the first step and I have also the final displaced point grid that I used to generate the mesh vertices. I can use a <distance> component to measure the distance between the initial position of the point and its final position to see the deviation of the points.
Fig.7.40. For our analysis purpose I want to use a <Gradient> component (Params > Special > Gradient) to assign gradient of colours to the mesh. I attached my <distance> values to the parameter part (t) of the <Gradient> and I attached it to the Colour input of the <mesh> component. But to complete the process I need to define the lower limit and upper limit of gradient range (L0 and L1). Lower limit is the minimum value in the list and upper limit is maximum value in the list and other values are being divided in the gradient in between. To get the lower and upper limit of the list of deviations I need to simply sort the data and get the first and last values in that numerical range.
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Fig.7.41. By using a <sort> component to sort the distances, I get the first item of the data list (index= 0) as lower limit and the last one (index= <list length> ‐ 1) as the upper limit of the data set (deviation values) to connect to the <gradient> component to assign colours based on this range.
Fig.7.42. By clicking on the small colour icon on the corner of the <gradient> component we can change the colours of the gradient.
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Fig.7.43. Right‐click on the component and on the context pop‐up menu you have more options to manipulate your resultant colours. Different types of colour gradient to suit the graphical representation of your analysis purpose.
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7_5_Manipulating Mesh objects as a way of Design Depends on the object and purpose of the modelling, I personally prefer to get my mesh object by manipulating a simple mesh geometry instead of generating a mesh from scratch since defining the point set and face matrices are not always simple and easy to construct. By manipulating, I mean that we can use a simple mesh object, extract its components and change them and then again make a mesh with varied vertices and faces. So I do not need to generate points as vertices and matrices of faces. Let’s have a look at a simple example.
Modifying Topology Fig.7.45. In this example, I simply used a <mesh plane> component and I extracted its data by using a <mesh components> to have access to its vertices and faces. Then I displaced vertices along Z direction by random values powered by a <number slider> and again attached them to a <mesh> component to generate another mesh. Here I also used a <cull pattern> component and I omitted some of the faces of original mesh and then I used them as new faces for making another mesh. The resultant mesh has both geometrical and topological difference wih its initial mesh and can be used for any design purpose. This idea of geometrically manipulating the vertices and topologically changing the faces has so many different possibilities that you can use in your design experiments. Since mesh objects have the potential to omit some of their faces and still remain as surface, different design ideas like porous surfaces could be pursued by them.
Fig.7.46. Resultant manipulated mesh (just a random case).
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Fig.7.47.a/b. Final model.
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Fabrication
Today there is a vast growing interest on material practice and fabrication in combination with Computer Aided Manufacturing. Due to the changes have happened in design processes, it seems a crucial move and one of the ‘Musts’ in the field of design. Any design decision in digital area, should be tested in different scales to show the ability of fabrication and assembly. Since it is obvious that the new design processes and algorithms do not fit into the traditional building processes, designers now try to use the modern technologies in fabrication to match their design products. From the moment that CNC machines started to serve the building industry up to now, a great relation between digital design and physical fabrication have been made and many different technologies and machineries being invented or adjusted to do these types of tasks. In order to design building elements and fabricate them, we need to have a brief understanding of the fabrication processes for different types of materials and know how to prepare our design outputs for them. This is the main purpose of the fabrication issues in our design process. Based on the object we designed and material we used, assembly logic, transportation, scale, etc. we need to provide the suitable data from our design and get the desired output of that to feed machineries. If traditional way in realization of a project made by Plans, Sections, Details, etc. today, we need more details or data to transfer them to CNC machines, to use them as source codes and datasheets for industries and so on. The point here is that the designer should provide some of the required data, because it is highly interconnected with design object. Designer sometimes should use the feedback of the fabrication‐ data‐preparation for the design readjustment. Sometimes the design object should be changed in order to fit the limitations of the machinery or assembly. Up to this point we already know different potentials of the Grasshopper to alter the design, and these design variations could be in the favour of fabrication as well as other criteria. I just want to open the subject and touch some of the points related to the data‐preparation phase, to have a look at different possibilities that we can extract data from design project in order to fabricate it or sometime readjust it to fit the fabrication limitations.
Chapter 8
Chapter_8_Fabrication
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8_1_Data asheets In orderr to make ob bjects, somettimes we sim mply need a series of meeasurementss, angels, coo ordinates and gen nerally numeerical data. There are multiple components in n Grasshopp per to comp pute the measureements, distaances, angells, etc. the im mportant point is the corrrect and preecise selectio on of the objects that we neeed to addresss for any sp pecific purpo ose. We shou uld be awaree of any geo ometrical oose the dessired points for measureement purpo oses. The complexxity that exitts in the dessign and cho next poiint is to find the position ns that give u us the prope er data for our fabricatio on purpose and avoid to generate lots of tables of numerical daata which co ould be timee consumingg in big projjects but n to expo ort the data from 3D so oftware to th he spreadsheets and useless at the end. Finally we need datasheets for further use. Chapter 8
Paper_sstrip_projectt The ideaa of using paper p strips attracted me m for some investigations, although h it had bee en tested before (like in Morpho‐Ecologiess by Hensel aand Menges,, 2008). To understand th he simple assemblies I started d with very ssimple comb binations forr first level and I tried to o add these simple comb binations together as the seco ond level of assembly. Itt was interessting in the ffirst tries butt soon it beccame out was not what I assumed. SSo I tried to be more preecise to get tthe more of orderr and the ressult object w delicate geometries at the end.
Fig.8.1. Paper strips,, first try.
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In the neext step I tried to make a very simple set up and d understand d the geometrical logic and use it as the base b for digital modelling. I assumed d that by jumping into digital d modeelling I would not be able to m make physicaal model and d I was sure tthat I need to test the eaarly steps witth paper. My aim was to use tthree paper strips and connect them m, one in thee middle and d another tw wo in two ngth, restrictted at their eends to the m middle strip. This could be the basic m module. sides witth longer len
Chapter 8
Fig.8.2. simple papeer strip com mbination to understand d the connecctions and m move toward ds digital modellin ng. Digital m modelling Here I wanted to w m model the paper strip diggitally after my basic m understanding of the physsical one. From th he start poin nt I need a very v simple curve c in the middle as the t base of my design and a I can divide it and by culling these division points (true, false) and moving (false ones) perpendicullar to the middle curve and using all these points (true ones and moved ones) as tthe vertices for two interpolated curves I can model this paper sttrips almost the same as what I described.
Fig.8.3.a a/b. First mo odelling meth hod with inteerpolated currves as side sstrips.
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But it seemed so simple and straightforward. So I wanted to add a gradual size‐differentiation in connection points so it would result in a bit more complex geometry. Now let’s jump into Grasshopper and continue the discussion with modelling there. I will try to describe the definition briefly and go to the data parts.
Chapter 8
Fig.8.4. The <curve> component is the middle strip which is a simple curve in Rhino. I reparameterized it and I want to evaluate it in the decreasing intervals. I used a <range> component and I attached it to a <Graph Mapper> component (Params > Special > Graph Mapper). A <Graph mapper> remaps a set of numbers in many different ways and domains by choosing a particular graph type. As you see, I evaluated the curve with this <Graph mapper> with parabola graph type and the resultant points on the curve are clear. You can change the type of graph to change the mapping of numeric range (for further information go to the component help menu).
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Fig.8.5. After remapping the numerical data I evaluated the middle curve with two different <evaluate> components. First by simply attach it to the data from <graph mapper> as basic points. Then I need to find the midpoints. Here I find the parameters of the curve between each basic point and the next one. I <shift>ed the data to find the next point and I used <dispatch > to exclude the last item of the list (exclude 1) otherwise I would have one extra point in relation to the <shift>ed points. The <function> component simply find the parameter in between ( f(x)=(x+y)/2 ) and you see the resultant parameters being evaluated.
Fig.8.6. Now I want to move the midpoints and make the other vertices of the side strips. Displacement of these points must be always perpendicular to the middle curve. So in order to move the points I need vectors, perpendicular to the middle curve at each point. I already have the Tangent vector at each point, by <evaluate> component. But I need the perpendicular vector. We now that the Cross product of two vectors is always a vector perpendicular to both of them (Fig.8.7). For example unit Z vector could be the cross product of the unit X and Y vectors. Our middle curve is a planer curve so we now that the Z vector at each point of the curve would be always perpendicular to the curve plane. So if I find the cross product of the Tangent of the vector and Z vector at each point, the result is a vector perpendicular to the middle curve which is always lay down in the curve’s plane. So I used Tangent of the point from <evaluate> Component and a <unit Z> vector to find the <XProd> of them which I know that is perpendicular to the curve always. Another trick! I used the numbers of the <Graph Mapper> as the power of these Z vectors to have the increasing factors for the movements of points, in their displacements as well, so the longer the distance between points, the bigger their displacements.
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Fig.8.7. Vector crosss product. Veector A and B B are in basee plane. Vecttor C is the cross productt of the A nd it is perpeendicular to tthe base plan ne so it is alsso perpendicular to both vectors A an nd B. and B an
Fig.8.8. Now I have b both basic po oints and mo oved points. I <merge>d them togeth her and I sortted them based on n their (Y) va alues to generate an <interpolate>d curve which h is one of m my side paperr strip. (If you man nipulate you ur main curvve extremelyy or rotate itt, you should d sort your p points by th he proper factor).
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Fig.8.9. Using a <Mirror Curve> component (XForm > Morph > Mirror Curve) I can mirror the <interpolate>d curve by middle <curve> so I have both side paper strips.
Fig.8.10. Now if I connect middle curve and side curves to an <extrude> component I can see my first paper strip combination with decreasing spaces between connection points.
Fig.8.11. I can simply start to manipulate the middle strip and see how Grasshopper updates the three paper strips which are connecting to each other in six points.
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After I found the configuration that I wanted to make a paper strip model, I needed to extract the dimensions and measurements to build my model with that data. Although it is quiet easy to model all these strips on paper sheets and cut them with laser cutter but here I like to make the process more general and get the initial data needed to build the model, so I am not limited myself to one specific machine and one specific method of manufacturing. You can use this data for any way of doing the model even by hand !!!! as I want to do in this case to make sure that I am not overwhelmed by digital!
Since strips are curve, the <distance > component does not help me to find the measurements. I need the length of curve between any two points on each strip. When I evaluate a parameter on a curve, it gives me its distance from the start point as well. So I need to find the parameter of the connection points of the strips (curves) and evaluate the position of them for each curve and the <evaluate> component would give me the distance of the points from the start point of curve means positions of connection points.
Fig.8.12. Although my file became a bit messy I replaced some components position on canvas to bring them together. As you see I used the first set of points that I called them ‘main curve points’ on the middle strip (initial curve). These are actually connection points of strips. The (L) output of the component gives me the distances of connection points from the start points of the middle strip. I used these points to find their parameter on the side curves as well (<curve cp> component). So I used these parameters to evaluate the side curve on those specific parameters (connection points) and find their distances from the start point. I can do the same to find the distance of the connection points on the other side strip (<mirror>ed one) also. At the end, I have the position of all connection points in each paper strip.
Chapter 8
By doing a simple paper model I know that I need the position of the connection points on the strips and it is obvious that these connection points are in different length in left_side_strip, right_side_strip and middle_strip. So if I get the division lengths from Grasshopper I can mark them on the strips and assemble them.
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Make sure that the direction of all curves should be the same otherwise you need to change the direction of the curve or if it affects your project, you can simply add a minus component to minus this distances from the curve length which mathematically inverses the distance and gives you the distances of points from the start point instead of end point (or vice versa).
Chapter 8
Exporting Data
Fig.8.13. Right‐click on the <panel> component and click on the ‘stream contents’. By this command you would be able to save your data in different formats and use it as a general numeric data. Here I would save it with simple .txt format and I want to use it in Microsoft Excel.
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Fig.8.14. On the Excel sheet, simply click on an empty cell and go to the ‘Data’ tab and under the ‘Get External Data’ select ‘From Text’. Then select the saved txt file from the address you saved your stream content and follow the simple instructions of excel. These steps allow you to manage your different types of data, how to divide your data in different cells and columns etc.
Fig.8.15. Now you see that your data placed on the Excel data sheet. You can do the same for the rest of your strips.
Chapter 8
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Fig.8.16. Table of the connection points alongside the strip. If you have a list of 3D coordinates of points and you want to export them to the Excel, there are different options than the above example. If you export 3D coordinates with the above method you will see there are lots of unnecessary brackets and commas that you should delete. You can also add columns by clicking in the excel import text dialogue box and separate these brackets and commas from the text in different columns and delete them but again because the size of the numbers are not the same, you will find the characters in different columns that you could not align separation lines for columns easily. In such case I simply recommend you to decompose your points to its components and export them separately. It is not a big deal to export three lists of data instead of one.
Fig.8.17. Using <decompose> component to get the X, Y and Z coordinates of the points separately to export to a data sheet.
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You can also use the ‘Format()’ function to format the output text, directly from a point list in desired string format. You need to define your text in way that you would be able to separate different parts of the text by commas in separate columns in datasheet.
Chapter 8
Enough for modelling! I used the data to mark my paper strips and connect them together. To prove it even to myself, I did all the process with hand !!!! to show that fabrication does not necessarily mean laser cutting (HAM, as Achim Menges once used for Hand Aided Manufacturing!!!!). I just spent an hour to cut and mark all strips but the assembly process took a bit longer which should be by hand anyway.
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Fig.8.18.a/b/c. Finall paper‐strip project.
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8_2_Laser Cutting and Cutting based Fabrication
Since the laser cutter is a generic tool, there are other methods also, but all together the important point is to find a way, to reduce the geometry to flat pieces to cut them from a sheet material, no matter paper or metal, cardboard or wood and finally assemble them together (if you have Robotic arm and you can cut 3D geometries it is something different!). Among the different ways discussed here I want to test one of them in Grasshopper and I am sure that you can do the other methods based on this experiment easily. Free‐form surface fabrication I decided to fabricate a free‐form surface to have some experiments with preparing the nested parts of a free‐form object to cut and all other issues we need to deal with.
Fig.8.19. Here I have a surface and I introduced this surface to Grasshopper as a <Geometry> component, so you can introduce any geometry that you have designed or use any Grasshopper object that you have generated. Sections as ribs In order to fabricate this generic free‐form surface I want to create sections of this surface, nest them on sheets and prepare the files to be cut by laser cutter. If the object that you are working on has a certain thickness then you can cut it but if like this surface you do not have any thickness you need to add a thickness to the cutting parts.
Chapter 8
The idea of laser cutting on sheet materials is very common these days to fabricate complex geometries. There are different ways that we can use this possibility to fabricate objects. Laser cutter method suits the objects that built with developable surfaces or folded ones. One can unfold the digital geometry on a plane and simply cut it out of a sheet and fold the material to build it. It is also suitable to make complex geometries that could be reduced to separate pieces of flat surfaces and one can disassemble the whole model digitally in separate parts, nest it on flat sheets, add the overlapping parts for connection purposes (like gluing) and cut it and assemble it physically. It is also possible to fabricate double‐curve objects by this method. It is well being experimented to find different sections of any ‘Blob’ shaped object, cut it at least in two directions and assemble these sections together usually with Bridle joints and make rib‐cage shaped models.
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Fig.8.20. In the first step I used a <Bounding Box> component to find the area that I want to work on. Then by using an <Explode> component (Surface > Analysis > BRep components) I have access to its edges. I selected the first and second one (index 0 and 1) which are perpendicular to each other.
Fig.8.21. In this step I generated multiple perpendicular frames alongside each of selected edges. The number of frames is actually the number of ribs that I want to cut.
Fig.8.22. Closer view of frames generated alongside the length and width of the object’s bounding box. As you see I can start to cut my surface with this frames.
Chapter 8
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Fig.8.23. Now if I find the intersections of these frames and the surface (main geometry), I actually generated the ribs base structure. Here I used a <BRep | Plane> section component (Intersect > Mathematical > BRep | Plane) to solve this problem. I used the <Geometry> (my initial surface) as BRep and generated frames, as planes to feed the section component.
Fig.8.24. Intersections of frames and surface, resulted in series of curves on the surface.
Chapter 8
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Nesting The next step is to nest these curve sections on a flat sheet to prepare them for the cutting process. Here I drew a rectangle in Rhino with my sheet size. I copied this rectangle to generate multiple sheets overlapping each other and I drew one surface that covers all these rectangles to represent them into Grasshopper.
Chapter 8
Fig.8.25. Paper sheets and an underlying surface to represent them in Grasshopper. I am going to use <Orient> component (XForm > Euclidian > Orient) to nest my curves into the surface which represents the sheets for cutting purpose. If you look at the <orient> component you see that we need the objects plane as reference plane and target plane which should be on the surface. Since I used the planes to intersect the initial surface and generate the section curves, I can use them again as reference planes, so I need to generate target planes.
Fig.8.26. I introduced the cutting surface to Grasshopper and I used a <surface Frame> component (Surface > Util > Surface frames) to generate series of frames across the surface. It actually works like <divide surface> but it generates planes as the output, so exactly what I need.
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Frames
Fig.8.27. Orientation. I connected the section curves as base geometries, and the planes that I used to generate these sections as reference geometry to the <orient> component. But still a bit of manipulations is needed for the target planes. If you look at the <surface frame> component results you see that if you divide U direction even by 1 you will see it would generate 2 columns to divide the surface. So I have more planes than I need. So I <split> the list of target planes by the number that comes from the number of reference curves. So I only use planes as much as curve that I have. Then I moved these planes 1 unit in X direction to avoid overlapping with the sheet’s edge. Now I can connect these planes to the <orient> component and you can see that all curves now nested on the cutting sheet.
Fig.8.28. nested curves on the cutting sheet.
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Fig.8.29. After nesting the curves on the cutting sheet, as I told you, because my object does not have any thickness, in order to cut it, we need to add thickness to it. That’s why I <offset> oriented curves with desired height and I also add <line>s to both ends of these curves and their offset ones to close the whole drawing so I would have complete ribs to cut. Joints (Bridle joints) The next issue is to generate ribs in other direction and make joints to assemble them after being cut. Although I used the same method of division of the bounding box length to generate planes and then sections, but I can generate planes manually in any desired position as well. So in essence if you do not want to divide both directions and generate sections evenly, you can use other methods of generating planes and even make them manually.
Fig.8.30. As you see here, instead of previously generated planes, I used manually defined planes for the sections in the other direction of the surface. One plane generated by X value directly from <number slider> and another plane comes from the mirrored plane on the other side of the surface (surface length – number slider). The section of these two planes and surface is being calculated for the next steps.
Chapter 8
Making ribs
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Fig.8.31. since we have the curves in two directions we can find the points of intersections. That’s why I used <CCX> components (Intersect > Physical > Curve | Curve) to find the intersect position of these curves which means the joint positions (The <CCX> component is in cross reference mode). After finding joint’s positions, I need a bit of drawing to prepare these joints to be cut. I am thinking of preparing bridle joints so I need to cut half of each rib on the joint position to be able to join them at the end. First I need to find these intersect position on the nested ribs and then draw the lines for cutting.
Fig.8.32. If you look at the outputs of the <CCX> component you can see that it gives us the parameter in wish each curve intersect with the other one. So I can <evaluate> the nested or <orient>ed curves with these parameters to find the joint positions on the cutting sheet as well.
Chapter 8
Now I can orient these new curves on another sheet to cut which is the same as the other one. So let’s generate joints for the assembly which is the important point of this part.
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Fig.8.33. Now we have the joint positions, we need to draw them. First I drew lines with <line SDL> component with the joint positions as start points, <unit Y> as direction and I used half of the rib’s height as the height of the line. So as you see each point on the nested curves now has a tiny line associated with it.
Fig.8.34. Next step, draw a line in X direction from the previous line’s end point with the length of the <sheet_thickness> (depends on the material).
Chapter 8
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Fig.8.35. This part of the definition is a bit tricky but I don’t have any better solution yet. Actually if you offset the first joint line you will get the third line but as the base curve line is not straight it would cross the curve (or not meet it) so the end point of the third line does not positioned on the curve. Here I drew a line from the end point of the second line, but longer than what it should be, and I am going to trim it with the curve. But because the <trim with BRep> component needs BRep objects not curves, I extruded the base curve to make a surface and again I extruded this surface to make a closed BRep. So if I trim the third line of the join with this BRep, I would get the exact joint shape that I want.
Fig.8.36. Using a <join curves> component (Curve > Util > Join curves) now as you can see I have a slot shaped <join curve> that I can use for cutting as bridle join in the ribs.
Chapter 8
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Right joint slot
Fig.8.37. I am applying the same method for the other end of the curve (second joints on the other side of the curve).
Fig.8.38. Ribs with the joints drawn on their both ends. With the same trick I can trim the tiny part of the base curve inside joint but because it does not affect the result I can leave it. Labelling While working in fabrication phase, it might be a great disaster to cut hundreds of small parts without any clue or address that how we are going to assemble them together, what is the order, and which one goes first. It could be simply a number or a combination of text and number to address the part. If the object comprises of different parts we can name them, so we can use the names or initials with numbers to address the parts also. We can use different hierarchies of project assembly logic in order to name the parts as well. Here I am going to number the parts because my assembly is not so complicated.
Chapter 8
Left joint slot
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Fig.8.39. As you remember I had a series of planes which I used as the target planes for orientating my section curves on the sheet. I am going to use the same plane to make the position of the text. Since this plane is exactly on the corner of the rib I want to displace it first.
Initial planes
Moved planes
Fig.8.40. I moved the corner planes 1 unit in X direction and 0.5 unit in Y direction (as <sum> of the vectors) and I used these planes as the position of the text tags. Here I used <text tag 3D> and I generated a series of numbers as much as ribs I have to use them as texts. The <integer> component that I used here simply converts 12.0 to 12. As the result, you can see all parts have a unique number in their left corner.
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Fig.8.41. I can chang ge the divisio on factors off the cutting surface to co ompress ribss as much ass possible d wasting ma aterial. As yo ou see in thee above exam mple, from th he start poin nt of the sheet_3 ribs to avoid started tto be more fflat and I havve more spacce in between n. Here I can n split ribs in two differen nt cutting surface and change the division points of each to compreess them bassed on their shape. But b because I h lots of parrts I can do this type off stuff manu ually in Rhino o, all parts does not am not dealing with necessarrily to be Asssociative! No ow I have thee ribs in one direction, an nd I am goin ng to do the same for the otheer direction o of ribs as welll. The only th hing that you should con nsider here iss that the dirrection of the joints flip aroun nd here, so basically while w I was working w with h the <orien nt> geometrry in the previouss part here I sshould workk with the <offfset> one. Cutting When all geometries become reeady to cut, II need to burn them and d manage theem a bit morre on my sheets. A As you see in Figure 8.42 they all neested in thre ee sheets. I ggenerated th hree differen nt shapes for the rribs in the wiidth direction of the objeect to check tthem out. Th he file is now w ready to be e cut.
Fig.8.42. Nested ribss, ready to bee cut.
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Fig.8.43. Cut ribs, rea ady to assem mble. Assembly In our case assembly is quiet sim mple. Somettimes you ne eed to checkk your file aggain or even n provide different fabrrication metthods. All some heelp files or eexcel sheets in order to aassemble your parts in d together, here is thee surface that I made.
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8.44.a/b b. Final modeel.
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Fabrication is a wide topic to discuss. It highly depends on what you want to fabricate, what is the material, what is the machine and how fabricated parts going to be assemble and so on. As I told you before, depend on the project you are working on, you need to provide your data for the next stages. Sometimes it is more important to get the assembly logic, for example when you are working with simple components but complex geometry as the result of assembly. Chapter 8
Fig.8.45. Assembly logic; material and joints are simple; I can work on the assembly logic and use the data to make my model.
Chapter_9_Design Strategy
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Chapter_9_Design Strategy Associative modelling is an algorithmic way of dealing with geometry. More than the conventional geometrical objects, with this algorithmic method, now we have all possibilities of computational geometries as well as dealing with the huge amount of data, numbers and calculations. It seems that there is a great potential in this realm. Here the argument is to not limit the design in any predefined experiment, and explore these infinite potentials; there are always alternative ways to do algorithmic modelling. Although it seems that the in‐built commands of these parametric modelling softwares could limit some actions or dictate some methods, but alternative solutions could always be brought to the table, let our creativity fly away of limitations. In order to design something, having a Design Strategy always helps to set up a good algorithm and to find the design solution. Thinking about the general properties of the design object, drawing some parts, even making some physical models, would help to a better understanding of the algorithm so better choice of <components> in digital modelling. Thinking about fix parameters, parameters that might change during the design, numerical data and geometrical objects needed, always helps to improve the algorithm. It would be helpful to analytically understand the design problem, sketch it and then start an algorithm that can solve the problem. We should think in an Algorithmic way to design Algorithmic.
Chapter 9
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Fig.9.1. Weaving project; From Analytical understanding to Associative modelling.
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Fig.9.2. Porous wall project; From m Analytical understandiing to Associiative modellling.
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Bibliography Pottman, Helmut and Asperl, Andreas and Hofer, Michael and Kilian, Axel, 2007: ‘Architectural Geometry’, Bently Institute Press. Hensel, Michael and Menges, Achim, 2008: ‘Morpho‐Ecologies’, Architectural Association. Rutten, David, 2007: ‘Rhino Script 101’, digital version by David Rutten and Robert McNeel and Association. Flake, Gary William, 1998: ‘The computational beauty of nature, computer explorations of fractals, chaos, complex systems, and adaptation’, The MIT Press. Main Grasshopper web page: http://grasshopper.rhino3d.com/ Grasshopper tutorials on Robert McNeel and Associates wiki: http://en.wiki.mcneel.com/default.aspx/McNeel/ExplicitHistoryExamples.html Kilian, Axel and Dritsas, Stylianos: ‘Design Tooling ‐ Sketching by Computation’, http://www.designexplorer.net/designtooling/inetpub/wwwroot/components/sketching/index.html Wolfram Mathworld: http://mathworld.wolfram.com/ Stylianos Dritsas, http://jeneratiff.com/
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1. ## Welcome to the Play; Vectors applied
I recall standing near one of two doors to the auditorium on a Sunday Afternoon, waiting on people to arrive. it was not until 1:45 until people started to swarm in. Ripping up tickets and providing programs, I was busy at work making sure that people would feel welcome inside the dark chamber. The performance itself was based off of a child's book, reimagined, and transformed into the musical known as Frog and Toad. yet all that mattered to me at the time was dealing with ticketwork and getting the viewers in. Yet at the same time, i wasn't responsible for every single viewer. No, instead some funneled in through the alternative door, while others seated themselves onto the balcony on the second floor. So if i didn't have to manage every ticket, how could i predict where people would enter? That's where physics comes in. As it is known, Vectors are quantities of both magnitude and direction. Near my door, i could figure which people would check through me through the direction in which they took to reach the auditorium. If they entered through the main entrance and through the commons (presumably eastward), it would have been unlikely that they would have entered through my door, since it was more situated towards an entranceon the left of the building. Should they have entered through that entrance ( which was southward), they would have more likely checked in with me. Another thing i noticed was the magnitude of their movement. I noticed that generally people were more rushed the closer it was to show time; a sign that they moved faster. In all, by observing this, I was able to take something as absurd as being a volunteer, and use physics to make the job easier. with the assistance of vectors, i could keep myself reminded that not every person would be checking in with me.
2. ## Mini lab news article
Matthew Fiordeliso Grace Saxby Xiao meng jiang Danny Waheibi Breaking News 4 high school students had discovered the acceleration due to gravity what they had used was a dodgeball a ruler to measure the height from the dodgeball to the floor and a stopwatch to calculate the time. What they did was dropped the dodgeball and measured how long it had taken for it to hit the ground. They did that 3 times and found the average time 0.64 seconds. what they discovered was the acceleration due to gravity is 11.9 meters a second however, it seems that they had some error which turned out to be a 21.3% error because some scientists just moments ago found out that the true number is 9.81 meters a second. this seems rather truthful because the students were using a stopwatch which can lead to inaccurate measurements while the scientist had used state of the art equipment so it seems accurate that they would have a large percent error and that’s the end of the breaking news report.
3. ## Intro
Seems nice thatyou have a good interest in physics.
4. ## Introduction Paragraph
I enjoyed chem, even if i found the course a bit too easy.
5. ## Introduction paragraph
Hi, My name is Danny and I have a passion for music and technology. Right now i am in the IHS wind ensemble and taking AP Music theory. I also am currently taking computer programming and AP calculus AB to help with my understanding of computers. I intend to major in music with programming as a back-up plan. I chose to take on Physics as I felt that the course would supplement me with information that would help with my music and technology careers. I'm hoping that we'd delve into understanding sound, since it would help me to analyze the sound played by music. As the course also covers a lot of mathy stuff and such, I feel that it could help me with understanding computers and how they work. Ultimately, I believe that physics will give me a feel as to what i should expect with my careers and such.
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You are Here: Home >< Physics
# AQA (New spec) mechanics question help Watch
1. (a) Show that the change in momentum of the golf ball during the collision isabout 0.5 N s.
(b) The ball strikes the plate with a speed of 7.1 m s–1 and has a mass of 45 g. It leaves the plate with a speed of 3.9 m s–1.Show that this is consistent with a change in momentum of about 0.5 N s.
Can't show this :\ I keep getting 0.144
(c) The ball continues to bounce, each time losing the same fraction of its energy when it strikes the plate. Air resistance is negligible.Determine the percentage of the original gravitational potential energy of the ball that remains when it reaches its maximum height after bouncing three times.
Wtf???
Basically can you help me with parts (b) and (c) please, they're making me want to rip my hair out.
Here's the link to the paper if you want to see the actual question, it is question 3 in section B of the paper.
http://physicsloreto.weebly.com/uplo..._2_qp_v1.1.pdf
2. (b) Change in momentum = (mass * final velocity) - (mass * initial velocity) = mass ( final velocity - initial velocity) = 0.045 * (3.9 - ( -7.1)) = 0.045 * 11 = 0.495 Ns
The initial velocity is in the opposite direction from the final velocity so the sign is negative.
3. (Original post by jessyjellytot14)
(a) Show that the change in momentum of the golf ball during the collision isabout 0.5 N s.
(b) The ball strikes the plate with a speed of 7.1 m s–1 and has a mass of 45 g. It leaves the plate with a speed of 3.9 m s–1.Show that this is consistent with a change in momentum of about 0.5 N s.
Can't show this :\ I keep getting 0.144
(c) The ball continues to bounce, each time losing the same fraction of its energy when it strikes the plate. Air resistance is negligible.Determine the percentage of the original gravitational potential energy of the ball that remains when it reaches its maximum height after bouncing three times.
Wtf???
Basically can you help me with parts (b) and (c) please, they're making me want to rip my hair out.
Here's the link to the paper if you want to see the actual question, it is question 3 in section B of the paper.
http://physicsloreto.weebly.com/uplo..._2_qp_v1.1.pdf
Lol rest assured it was a tricky question.
For part a, its important you use the graph
For part b I = m (v - u). Convert the mass into kg, decide on a positive direction and substitute apt values.
For part c (i actually wrote wtf on this question ) xD but basically, the clue is "conservation of momentum". Momentum must be conserved. Thus as the ball approaches earth, the eart approaches the ball. Make sense? It sounds ridiculous I know but it approaches the earth very very slowly due to its large mass. If you think about this along with conservation and give 3 points, you should get the marks
4. The PE at the start would be the same as the KE just before impact = 0.5 mv^2 = 0.5 * 0.045 * 7.1 * 7.1 = 1.134 J
After the first bounce, KE = 0.5 * 0.045 * 3.9 * 3.9 = 0.342 J. This will convert back into PE at the top of the bounce meaning that the PE has changed from 1.134 to 0.342 i.e. 30% of the original value. It bounces two more times - each time losing 30% of its value so you can work it out from there.
Lol rest assured it was a tricky question.
For part a, its important you use the graph
For part b I = m (v - u). Convert the mass into kg, decide on a positive direction and substitute apt values.
For part c (i actually wrote wtf on this question ) xD but basically, the clue is "conservation of momentum". Momentum must be conserved. Thus as the ball approaches earth, the eart approaches the ball. Make sense? It sounds ridiculous I know but it approaches the earth very very slowly due to its large mass. If you think about this along with conservation and give 3 points, you should get the marks
Thank you, this makes a bit more sense. I would never get that in an exam though S
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Java Program To Calculate Modulus | Mod Java
Java Program To Calculate Modulus – In this article, we will explain the multitude of methods to calculate the modulus of a number in Java Programming. Suitable examples and sample programs have been included in order to make you understand simply. The compiler has also been added so that you can execute the programs yourself.
The methods used in this article are as follows to calculate Java Modulus
• Using Standard Method
• Using Scanner Class
• Using Command Line Arguments
• Using Static Method
• Using Separate Class
The mod of a number can be regarded as the absolute value of any given number. It plays an important role in pretty much all sectors of mathematics.
For example,
The modulus of -22 will be represented as
| -22 | = 22.
As you can see in the image above, it is the graphical representation of the mod function. The absolute value is always positive no matter what.
Thus, the methods used to calculate the mod of a number in Java Programming are as follows:
Java Mod Code
In here, the entire program is written within the main method itself.
Our input is a predetermined integer number and our output is an integer as well which is the mod of the input number.
There are two ways this can be achieved in Java. They are:-
1. We can directly make use of the absolute method or the abs method from the Math package which consists of a predefined function to do this.
2. All one has to do is, use Math.abs(int number) and you get the output as the mod of the integer given as argument.
3. Another way is to first check whether the number is positive or negative.
4. If the number is negative i.e., if(number<0) then we multiply number with “-1”.
5. This is because, a negative number multiplied with -1 gives a positive number of the same digits which is nothing but the mod. Else the number remains the same because, the mod of positive number is the number itself.
Output:
Java Modulus/Mod using Scanner class
In the above method, we have seen that the integer to find mod is predetermined and given at compile time in the code itself and if we need the mod of another number then the change must be made in the code. This kind of method is not efficient at all.
Hence we can make use of the scanner class in java.
The scanner class of Java accepts inputs of any primitive data type at run time itself. This makes things a lot more convenient. So, after taking the input using scanner class the same logic as above can be applied here.
Output1:
Using Command Line Arguments
Command line arguments in Java are the arguments that are passed while running the run command just after the name of the Java program separated by space between them.
Similarly here, while running Java code a single argument whose mod is to be found can be given with the run command separated with the space. This argument (arg[0]) is converted into integer using parseInt() as follows:
After this, the same logic can be made use of to find out the mod of this integer.
Output:
Using Static Method
Static method is a method that makes use of static variables and performs the actions as written in it. We can make use of static method just to differentiate the inputs and logic from each other for better understanding.
Therefore, in the main method, inputs can be taken in with the use of Scanner Class.
These inputs can then be sent as parameters of the static method.
A separate static method (modCal) can be written which consists of any of the two logic written above and return the mod of the integer which can then be displayed in the console screen.
x
X Star Pattern Java Program – Patterns
Java program to print X star pattern program – We have written the below print/draw ...
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# MPG to KM/L and L/100KM Conversion [Charts]
You can convert MPG figures into km/L and L/100km measurements by using these formulas:
• L/100km = 235.21 / MPG (US)
• L/100km = 282.48 / MPG (IMP)
• KM/L = MPG (US) x 0.425144
• KM/L = MPG (IMP) x 0.354006
But if you don’t feel like calculating and would rather just get the values from some charts, this post is for you.
We at PowerSportsGuide have compiled the most common conversion numbers (US and Imperial MPG) into this post!
Contents
## Some Definitions
### What is MPG?
Miles per gallon (a.k.a. MPG) is a fuel consumption indicator; it shows the distance (in miles) that a vehicle can travel per gallon of fuel.
If you want to compare MPG figures, the general rule is that the better fuel economy a car has, the higher MPG number you can expect.
However, it’s good to know that gallons can be US and Imperial as well. As the name suggests, US gallons are used in the United States, while Imperial gallons are used in the United Kingdom, Ireland, Canada, and Australia.
When it comes to conversion numbers, one Imperial gallon is equal to 4.55 liters, while one US gallon is equal to 3.79 liters.
Since these measurements are not the same, the MPG-L/100km conversion formula will differ.
### What does L/100km Mean?
Unlike MPG, L/100 km (liters per 100 kilometers) is frequently used in Europe to describe the fuel consumption data of vehicles. This measurement tells how many liters of fuel a vehicle consumes while traveling a distance of 100 km.
This figure is often referred to as the vehicle’s fuel economy. The general rule is that the higher L/100km a vehicle has, the worse its fuel economy.
### What is Kilometers Per Liter?
The kilometer per liter (KPL) is a close relative of MPG; it measures the distance (in kilometers) that a vehicle can travel on one liter of fuel.
How do you calculate kilometers per liter?
To convert MPG to kilometers per liter, divide 100 by the L/100km figure.
For example, if a vehicle has 8 L/100km, that’s 100 / 8 = 12.5 km/L (kilometers per liter).
## How do You Convert MPG to Kilometers Per Liter (KM/L)?
One US MPG (miles per gallon) equals 0.425144 km/L (kilometers per liter). This means that you can convert US MPG to km/L by multiplying the MPG figure by the conversion ratio of 0.425144.
As an example:
3 MPG (US) x 0.425144 = 1.275 km/L
In case of imperial gallons, the conversion ratio is 0.354006.
3 MPG (IMP) x 0.354006 = 1.062 km/L
### What is the MPG to km/L formula?
In a nutshell, the MPG to KM/L formulas are as follows:
• KM/L = MPG (US) x 0.425144
• KM/L = MPG (IMP) x 0.354006
### MPG to KM/L Conversion Charts
For your convenience, we’ve done the math and listed the numbers in this MPG to KM/L chart:
MPG (US) to KM/L Chart
MPG (IMP) to KM/L Chart
## How to Convert Kilometers Per Liter to Miles Per Gallon
Let’s look at the inverse of the calculations above.
You can turn a km/L (kilometer per liter) figure into an MPG (US) measurement by multiplying the km/L figure by a conversion ratio of 2.35214.
Examples:
8 km/L x 2.35214 = 18.8 MPG (US)
The conversion ratio for Imperial MPG is 2.82481:
8 km/L x 2.35214 = 22.6 MPG (IMP)
### What is the MPG to km/L formula?
Based on these, the km/L to MPG formulas are as follows:
• MPG (US) = km/L x 2.35214
• MPG (IMP) = km/L x 2.82481
### MPG to KM/L Conversion Charts
Let’s see some common KM/L to MPG conversions in one chart:
KM/L to MPG (US) Chart
KM/L to MPG (IMP) Chart
## How do You Convert MPG to L/100KM?
To convert MPG (US) into L/100km, divide 235.21 by the MPG figure. In other words, one MPG (US) equals 235.2 L/100 km.
For example, converting 22 MPG (US) to liters per 100 km:
235.21/22 MPG (US) = 10.7 L/100km
The number to be divided for Imperial MPG is 282.48:
282.48/22 MPG (IMP) = 18.8 L/100km
### What is the MPG to L/100km Formula?
Therefore, MPG to L/100km formulas are as follows:
• 235.21 / MPG (US) = [L/100km]
• 282.48 / MPG (IMP) = [L/100km]
### MPG to 100L/km Conversion Charts
Let’s see some common MPG to 100L/km conversions in one chart:
US MPG to L/100km Chart
Imp. MPG to L/100km Chart
## How do You Convert L/100KM to MPG?
How do you convert liters per 100km to MPG?
To convert L/100km into MPG (US), divide 235.21 by the L/100km figure. In other words, 235.2 L/100 km is equal to 1 MPG (US).
For example, converting 9 L/100km into MPG (US):
235.21 / 9 L/100km = 26.13 MPG (US)
The number to be divided for Imperial MPG is 282.48:
282.48 / 9 L/100km = 31.39 MPG (IMP)
### What is the MPG to L/100km Formula?
Therefore, the L/100km to MPG formulas are as follows:
• 235.21 / [L/100km] = MPG (US)
• 282.48 / [L/100km] = MPG (IMP)
### 100L/km to MPG Conversion Charts
Let’s see some common 100L/km to MPG conversions under one roof:
100L/km to US MPG Chart
100L/km to Imp. MPG Chart
## Takeaway – FAQs
As a takeaway, we’ve answered the most common questions on the topic.
### How do You Convert MPG to L/100km?
To convert US MPG into L/100km, divide 235.21 by the MPG figure. In other words, one MPG (US) equals 235.2 L/100 km. The number to be divided for Imperial MPG is 282.48.
### How do You Convert MPG to KM/L?
One US MPG is equal to 0.425144 km/l. To convert US MPG to km/l, multiply the MPG figure with the conversion ratio of 0.425144. For Imperial MPG, the conversion ratio is 0.354006.
### How to Convert Miles Per Gallon (MPG) to Miles Per Liter (MPL)?
As a rule of thumb, 1 MPG (US) is equal to 0.2642 miles per liter. To convert MPG (US) to miles per liter, multiply the MPG figure with the conversion ratio of 0.2642.
How Many km/L is 10 MPG?
10 MPG (US) is equal to 4.3 km/L or 23.5 L/100km.
How Many km/L is 15 MPG?
15 MPG (US) is equal to 6.4 km/L or 15.7 L/100km.
How Many km/L is 18 MPG?
18 MPG (US) is equal to 7.7 km/L or 13.1 L/100km.
How Many km/L is 20 MPG?
20 MPG (US) is equal to 8.5 km/L or 11.8 L/100km.
How Many km/L is 25 MPG?
25 MPG (US) is equal to 10.6 km/L or 9.4 L/100km.
How Many km/L is 27 MPG?
27 MPG (US) is equal to 11.5 km/L or 8.7 L/100km.
How Many km/L is 28 MPG?
28 MPG (US) is equal to 11.9 km/L or 8.4 L/100km.
How Many km/L is 30 MPG?
30 MPG (US) is equal to 12.8 km/L or 7.8 L/100km.
How Many km/L is 33 MPG?
33 MPG (US) is equal to 14 km/L or 7.1 L/100km.
How Many km/L is 35 MPG?
35 MPG (US) is equal to 14.9 km/L or 6.7 L/100km.
How Many km/L is 38 MPG?
38 MPG (US) is equal to 16.2 km/L or 6.2 L/100km.
How Many km/L is 40 MPG?
40 MPG (US) is equal to 17 km/L or 5.9 L/100km.
How Many km/L is 45 MPG?
45 MPG (US) is equal to 19.1 km/L or 5.2 L/100km.
How Many km/L is 50 MPG?
45 MPG (US) is equal to 21.3 km/L or 4.7 L/100km.
What is 3.2 L/100km in MPG?
3.2 L/100km is equal to 73.5 MPG (US) and 88.5 MPG (IMP).
What is 3.4 L/100km in MPG?
3.4 L/100km is equal to 69.2 MPG (US) and 83.1 MPG (IMP).
What is 4 L/100km in MPG?
5 L/100km is equal to 58.8 MPG (US) and 70.6 MPG (IMP).
What is 4.3 L/100km in MPG?
4.3 L/100km is equal to 54.7 MPG (US) and 65.7 MPG (IMP).
What is 4.6 L/100km in MPG?
4.6 L/100km is equal to 51.1 MPG (US) and 61.4 MPG (IMP).
What is 5 L/100km in MPG?
5 L/100km is equal to 47 MPG (US) and 56.5 MPG (IMP).
What is 5.3 L/100km in MPG?
5.3 L/100km is equal to 44.4 MPG (US) and 53.3 MPG (IMP).
What is 5.4 L/100km in MPG?
5.4 L/100km is equal to 43.6 MPG (US) and 52.3 MPG (IMP).
What is 5.5 L/100km in MPG?
5.5 L/100km is equal to 42.8 MPG (US) and 51.4 MPG (IMP).
What is 5.9 L/100km in MPG?
5.9 L/100km is equal to 39.9 MPG (US) and 47.9 MPG (IMP).
What is 6 L/100km in MPG?
6 L/100km is equal to 39.2 MPG (US) and 47.1 MPG (IMP).
What is 6.1 L/100km in MPG?
6.1 L/100km is equal to 38.6 MPG (US) and 46.3 MPG (IMP).
What is 6.2 L/100km in MPG?
6.2 L/100km is equal to 37.9 MPG (US) and 45.6 MPG (IMP).
What is 6.5 L/100km in MPG?
6.5 L/100km is equal to 36.2 MPG (US) and 43.5 MPG (IMP).
What is 6.6 L/100km in MPG?
6.6 L/100km is equal to 35.6 MPG (US) and 42.8 MPG (IMP).
What is 7 L/100km in MPG?
7 L/100km is equal to 33.6 MPG (US) and 40.4 MPG (IMP).
What is 7.5 L/100km in MPG?
7.5 L/100km is equal to 31.4 MPG (US) and 37.7 MPG (IMP).
What is 7.6 L/100km in MPG?
7.6 L/100km is equal to 30.9 MPG (US) and 37.2 MPG (IMP).
What is 7.7 L/100km in MPG?
7.7 L/100km is equal to 30.5 MPG (US) and 36.7 MPG (IMP).
What is 8 L/100km in MPG?
8 L/100km is equal to 29.4 MPG (US) and 35.3 MPG (IMP).
What is 8.2 L/100km in MPG?
8.2 L/100km is equal to 28.7 MPG (US) and 34.5 MPG (IMP).
What is 8.7 L/100km in MPG?
8.7 L/100km is equal to 27 MPG (US) and 32.5 MPG (IMP).
What is 9 L/100km in MPG?
9 L/100km is equal to 26.1 MPG (US) and 31.4 MPG (IMP).
What is 9.1 L/100km in MPG?
9.1 L/100km is equal to 25.8 MPG (US) and 31.0 MPG (IMP).
What is 9.4 L/100km in MPG?
9.4 L/100km is equal to 25 MPG (US) and 30.1 MPG (IMP).
What is 10 L/100km in MPG?
10 L/100km is equal to 23.5 MPG (US) and 28.3 MPG (IMP).
What is 10.3 L/100km in MPG?
10.3 L/100km is equal to 22.8 MPG (US) and 27.4 MPG (IMP).
What is 10.7 L/100km in MPG?
10.7 L/100km is equal to 22.0 MPG (US) and 26.4 MPG (IMP).
What is 11 L/100km in MPG?
11 L/100km is equal to 21.4 MPG (US) and 25.7 MPG (IMP).
What is 11.6 L/100km in MPG?
11.6 L/100km is equal to 20.3 MPG (US) and 24.4 MPG (IMP).
What is 11.8 L/100km in MPG?
11.8 L/100km is equal to 19.9 MPG (US) and 23.9 MPG (IMP).
What is 11.9 L/100km in MPG?
11.9 L/100km is equal to 19.8 MPG (US) and 23.7 MPG (IMP).
What is 12 L/100km in MPG?
12 L/100km is equal to 19.6 MPG (US) and 23.5 MPG (IMP).
What is 12.3 L/100km in MPG?
12.3 L/100km is equal to 19.1 MPG (US) and 23.0 MPG (IMP).
What is 12.5 L/100km in MPG?
12.5 L/100km is equal to 18.8 MPG (US) and 22.6 MPG (IMP).
What is 13 L/100km in MPG?
13 L/100km is equal to 18.1 MPG (US) and 21.7 MPG (IMP).
What is 13.8 L/100km in MPG?
13.8 L/100km is equal to 17 MPG (US) and 20.5 MPG (IMP).
What is 14 L/100km in MPG?
14 L/100km is equal to 16.8 MPG (US) and 20.2 (IMP).
What is 14.2 L/100km in MPG?
14.2 L/100km is equal to 16.6 MPG (US) and 19.9 MPG (IMP).
What is 14.3 L/100km in MPG?
14.3 L/100km is equal to 16.4 MPG (US) and 19.8 MPG (IMP).
What is 14.5 L/100km in MPG?
14.5 L/100km is equal to 16.2 MPG (US) and 19.5 MPG (IMP).
What is 15 L/100km in MPG?
15 L/100km is equal to 15.7 MPG (US) and 18.8 MPG (IMP).
What is 16 L/100km in MPG?
16 L/100km is equal to 14.7 MPG (US) and 17.7 MPH (IMP).
https://www.inchcalculator.com/convert/mile-per-hour-to-kilometer-per-hour
https://www.wikihow.com/Convert-MPG-to-Liters-per-100km
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[NumPy-Tickets] [NumPy] #1431: accessing multiple fields in a recarray gives fields in wrong order
NumPy Trac numpy-tickets@scipy....
Tue Mar 16 11:44:12 CDT 2010
```#1431: accessing multiple fields in a recarray gives fields in wrong order
------------------------+---------------------------------------------------
Reporter: samtygier | Owner: somebody
Type: defect | Status: new
Priority: normal | Milestone:
Component: numpy.core | Version:
Keywords: |
------------------------+---------------------------------------------------
when using the list in list notation to access multiple columns of a
record array the columns are returned in the order that they are stored in
the array, not in the order asked for.
{{{
>>> x = np.array([(1.5,2.5,(1.0,2.0)),(3.,4.,(4.,5.)),(1.,3.,(2.,6.))],
dtype=[('x','f4'),('y',np.float32),('value','f4',(2,2))])
>>> x[['x','y']]
array([(1.5, 2.5), (3.0, 4.0), (1.0, 3.0)],
dtype=[('x', '<f4'), ('y', '<f4')])
>>> x[['y','x']]
array([(1.5, 2.5), (3.0, 4.0), (1.0, 3.0)],
dtype=[('x', '<f4'), ('y', '<f4')])
}}}
this different to what happens with a 2d array, where they are returned in
the order asked for:
{{{
>>> a = array([0,0.1,0.2,0.3,0.4])
then
>>> a[[0,1,4]]
array([ 0. , 0.1, 0.4])
>>> a[[4,1,0]]
array([ 0.4, 0.1, 0. ])
}}}
I suggest that the it would be more consistent and predictable for the
order to be the one asked for in both cases.
The current behaviour for the recarray was undocumented until today, so
there can't be many people using it. therefore it might be worth break
backwards compatibility to make the interface better.
Please see discussion at
http://thread.gmane.org/gmane.comp.python.numeric.general/36933
a simple patch is attached.
--
Ticket URL: <http://projects.scipy.org/numpy/ticket/1431>
NumPy <http://projects.scipy.org/numpy>
My example project
```
More information about the NumPy-Tickets mailing list
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https://xgboost.readthedocs.io/en/release_1.0.0/tutorials/custom_metric_obj.html
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# Custom Objective and Evaluation Metric¶
XGBoost is designed to be an extensible library. One way to extend it is by providing our own objective function for training and corresponding metric for performance monitoring. This document introduces implementing a customized elementwise evaluation metric and objective for XGBoost. Although the introduction uses Python for demonstration, the concepts should be readily applicable to other language bindings.
Note
• The ranking task does not support customized functions.
• The customized functions defined here are only applicable to single node training. Distributed environment requires syncing with xgboost.rabit, the interface is subject to change hence beyond the scope of this tutorial.
• We also plan to re-design the interface for multi-classes objective in the future.
In the following sections, we will provide a step by step walk through of implementing Squared Log Error(SLE) objective function:
$\frac{1}{2}[log(pred + 1) - log(label + 1)]^2$
and its default metric Root Mean Squared Log Error(RMSLE):
$\sqrt{\frac{1}{N}[log(pred + 1) - log(label + 1)]^2}$
Although XGBoost has native support for said functions, using it for demonstration provides us the opportunity of comparing the result from our own implementation and the one from XGBoost internal for learning purposes. After finishing this tutorial, we should be able to provide our own functions for rapid experiments.
## Customized Objective Function¶
During model training, the objective function plays an important role: provide gradient information, both first and second order gradient, based on model predictions and observed data labels (or targets). Therefore, a valid objective function should accept two inputs, namely prediction and labels. For implementing SLE, we define:
import numpy as np
import xgboost as xgb
def gradient(predt: np.ndarray, dtrain: xgb.DMatrix) -> np.ndarray:
'''Compute the gradient squared log error.'''
y = dtrain.get_label()
return (np.log1p(predt) - np.log1p(y)) / (predt + 1)
def hessian(predt: np.ndarray, dtrain: xgb.DMatrix) -> np.ndarray:
'''Compute the hessian for squared log error.'''
y = dtrain.get_label()
return ((-np.log1p(predt) + np.log1p(y) + 1) /
np.power(predt + 1, 2))
def squared_log(predt: np.ndarray,
dtrain: xgb.DMatrix) -> Tuple[np.ndarray, np.ndarray]:
'''Squared Log Error objective. A simplified version for RMSLE used as
objective function.
'''
predt[predt < -1] = -1 + 1e-6
hess = hessian(predt, dtrain)
In the above code snippet, squared_log is the objective function we want. It accepts a numpy array predt as model prediction, and the training DMatrix for obtaining required information, including labels and weights (not used here). This objective is then used as a callback function for XGBoost during training by passing it as an argument to xgb.train:
xgb.train({'tree_method': 'hist', 'seed': 1994}, # any other tree method is fine.
dtrain=dtrain,
num_boost_round=10,
obj=squared_log)
Notice that in our definition of the objective, whether we subtract the labels from the prediction or the other way around is important. If you find the training error goes up instead of down, this might be the reason.
## Customized Metric Function¶
So after having a customized objective, we might also need a corresponding metric to monitor our model’s performance. As mentioned above, the default metric for SLE is RMSLE. Similarly we define another callback like function as the new metric:
def rmsle(predt: np.ndarray, dtrain: xgb.DMatrix) -> Tuple[str, float]:
''' Root mean squared log error metric.'''
y = dtrain.get_label()
predt[predt < -1] = -1 + 1e-6
elements = np.power(np.log1p(y) - np.log1p(predt), 2)
return 'PyRMSLE', float(np.sqrt(np.sum(elements) / len(y)))
Since we are demonstrating in Python, the metric or objective needs not be a function, any callable object should suffice. Similarly to the objective function, our metric also accepts predt and dtrain as inputs, but returns the name of metric itself and a floating point value as result. After passing it into XGBoost as argument of feval parameter:
xgb.train({'tree_method': 'hist', 'seed': 1994,
'disable_default_eval_metric': 1},
dtrain=dtrain,
num_boost_round=10,
obj=squared_log,
feval=rmsle,
evals=[(dtrain, 'dtrain'), (dtest, 'dtest')],
evals_result=results)
We will be able to see XGBoost printing something like:
[0] dtrain-PyRMSLE:1.37153 dtest-PyRMSLE:1.31487
[1] dtrain-PyRMSLE:1.26619 dtest-PyRMSLE:1.20899
[2] dtrain-PyRMSLE:1.17508 dtest-PyRMSLE:1.11629
[3] dtrain-PyRMSLE:1.09836 dtest-PyRMSLE:1.03871
[4] dtrain-PyRMSLE:1.03557 dtest-PyRMSLE:0.977186
[5] dtrain-PyRMSLE:0.985783 dtest-PyRMSLE:0.93057
...
Notice that the parameter disable_default_eval_metric is used to suppress the default metric in XGBoost.
For fully reproducible source code and comparison plots, see custom_rmsle.py.
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https://de.maplesoft.com/support/help/maple/view.aspx?path=Statistics%2FHeatMap
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HeatMap - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Statistics
HeatMap
generate heat maps
Calling Sequence HeatMap(A, options, plotoptions)
Parameters
A - data options - (optional) equation(s) of the form option=value where option is one of color, range, rowlabels, or columnlabels; specify options for generating the heat map plotoptions - options to be passed to the plots[display] command
Options
The options argument can contain one or more of the options shown below. All unrecognized options will be passed to the plots[display] command. See plot/options for details.
• color : list or range
Specifies colors for the individual data sets. When a range of colors is given, the color associated with a datum is generated by interpolating between the two named colors based on the range of values in the data. For example if color=red..blue then the minimum values in the data will be red and the maximum values blue. If a list of colors is provided, the color associated with a datum is generated by interpolating uniformly through all named colors.
• range : range, list of ranges, all, or bycolumn;
The option range indicates that the data is to interpreted as being in a specified range and also controls whether the mapping from value to color is applied to each column independently or to the data as a whole.
Option Coloring applies Meaning Example range=all To all data Same color range applied to all data, range inferred from data range=all range=a..b To all data Same explicit color range a..b applied to all data range=1..2 range=bycolumn Columnwise Each column has its own color range inferred from data range=bycolumn range=[a..b,c..d,...] Columnwise Each column has its own explicit color range range=[1..2,3..4]
When ranges are inferred, the minimum data value and the maximum data value from the data are used. If a specified range is used and data values are outside of this range, they are truncated to nearest endpoint of the specified range.
• columnlabels : list
Specifies the labels corresponding to each column. The list can contain either strings or names.
• rowlabels : list
Specifies the labels corresponding to each row. The list can contain either strings or names.
Description
• The HeatMap command generates a heat map for the specified data. A heat map is a method of data visualization representing the magnitude of included data as a discrete density plot.
• The first parameter A is a numeric Matrix, a list of lists, or a DataFrame.
Examples
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
Generate a heat map for a random matrix.
> $\mathrm{RM}≔\mathrm{LinearAlgebra}:-\mathrm{RandomMatrix}\left(10\right):$
The command to create the plot from the Plotting Guide using the data above is
> $\mathrm{HeatMap}\left(\mathrm{RM}\right)$
Generate the same heat map, sampling colors from blue to white to red.
> $\mathrm{HeatMap}\left(\mathrm{RM},\mathrm{color}=\left[\mathrm{blue},\mathrm{white},\mathrm{red}\right]\right)$
Generate visualization to compare the correlation between three data sets.
> $U≔⟨\mathrm{seq}\left(0..10\right)⟩:$$V≔⟨\mathrm{seq}\left(\mathrm{sin}\left(i\right),i=0..10\right)⟩:$$W≔⟨\mathrm{seq}\left(\mathrm{cos}\left(i\right),i=0..10\right)⟩:$
> $M≔\mathrm{Matrix}\left(\left[U,V,W\right]\right):$
> $\mathrm{CM}≔\mathrm{CorrelationMatrix}\left(M,\mathrm{ignore}\right)$
${\mathrm{CM}}{≔}\left[\right]$ (1)
> $\mathrm{HeatMap}\left(\mathrm{CM},\mathrm{columnlabels}=\left["n","sin\left(n\right)","cos\left(n\right)"\right],\mathrm{rowlabels}=\left["n","sin\left(n\right)","cos\left(n\right)"\right]\right)$
Compare intensities of absolute crime numbers for Canadian provinces.
> $\mathrm{CanadaCrime}≔\mathrm{Import}\left("datasets/canada_crimes.csv",\mathrm{base}=\mathrm{datadir}\right):$
> $\mathrm{Statistics}:-\mathrm{HeatMap}\left(\mathrm{CanadaCrime}\left[1..10,..\right],\mathrm{color}=\mathrm{white}..\mathrm{red},\mathrm{size}=\left[900,300\right]\right)$
Visualize the same crime statistics using the same color shading for all types of crime.
> $\mathrm{Statistics}:-\mathrm{HeatMap}\left(\mathrm{CanadaCrime}\left[1..10,..\right],\mathrm{color}=\mathrm{white}..\mathrm{red},\mathrm{range}=\mathrm{all},\mathrm{size}=\left[900,300\right]\right)$
Compare the similarity of words for "water" in several languages using the Levenshtein distance.
> $\mathrm{Words}≔\left["acqua","agua","eau","su","Wasser","water","woda","vatten","voda"\right]$
${\mathrm{Words}}{≔}\left[{"acqua"}{,}{"agua"}{,}{"eau"}{,}{"su"}{,}{"Wasser"}{,}{"water"}{,}{"woda"}{,}{"vatten"}{,}{"voda"}\right]$ (2)
> $\mathrm{WordDistances}≔\mathrm{DataFrame}\left(\mathrm{Matrix}\left(\mathrm{numelems}\left(\mathrm{Words}\right),\left(i,j\right)↦\mathrm{StringTools}:-\mathrm{Levenshtein}\left(\mathrm{Words}\left[i\right],\mathrm{Words}\left[j\right]\right)\right),\mathrm{rows}=\mathrm{Words},\mathrm{columns}=\mathrm{Words}\right):$
> $\mathrm{Statistics}:-\mathrm{HeatMap}\left(\mathrm{WordDistances},\mathrm{size}=\left[800,400\right],\mathrm{color}=\mathrm{blue}..\mathrm{white},\mathrm{range}=\mathrm{all}\right)$
Compatibility
• The Statistics[HeatMap] command was introduced in Maple 2016.
• For more information on Maple 2016 changes, see Updates in Maple 2016.
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Comp 110 random(_)
The 'random(_)' Function
Like all of our introcs functions, random needs to be imported!
`import { random } from "introcs"; `
This is what the function definition looks like:
``````let random = (floor: number, ceiling: number): number => {
//...
}``````
random takes in two number arguments that act as the 'floor' and the 'ceiling'.
The number that is returned by the function will return a randomly generated number between the floor and the ceiling. So, calling:
`print(random(1, 10));`
will print a random number between 1 and 10.
The floor and the ceiling are inclusive, meaning the generated number will be >= floor and <= ceiling.
Example of use:
``````let whereShouldIStudy = () : string => {
let randomNumber = random(1,4);
if(randomNumber === 1) {
return "Old Well";
} else if(randomNumber === 2) {
return "Bell Tower";
} else if(randomNumber === 3) {
return "Dean Dome";
} else {
return "Franklin Street";
}
}
//whatever number is generated will determine what string is printed!``````
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https://math.stackexchange.com/questions/2129541/number-of-32-character-alphanumeric-strings-with-certain-conditions/2129559
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# Number of $32$-character alphanumeric strings with certain conditions
I'm seeking a solution of one of the most complicated Math problem of my life.
Here it is :
First we need to figure out how many strings of set [a-zA-Z0-9] (Which is 26 Small Letters, 26 Capital Letters, and 0 to 9 digits] Are possible to construct of length 32 characters.
Then we need to subtract these 3 out of our result.
1. Possible 32 character strings which has only small letters. [a-z]
2. Possible 32 character strings which has only big letters. [A-Z]
3. Possible 32 character strings which has only digits. [0-9]
Let me know if any questions.
• Do you mean #(strings with at least one of the three types of characters)? Commented Feb 5, 2017 at 0:16
• YES. Apology if I made the problem sound more complicated. Commented Feb 5, 2017 at 0:17
• Example of strings are : yXgZxJsaRG8o7EBMCXknL6XFW7V9Smmg , x8k9Ifp9ejbEYzyt2TPdkZfY1T5fZNJm , C958NV3IN3m18jVBgrtBkWdSigaDVZao Commented Feb 5, 2017 at 0:18
• This is a nice question. What sorts of approaches have you considered? Commented Feb 5, 2017 at 0:19
• @dxiv I'm only ok at maths till a certain level. And this question is like puzzle of my life I can't figure out till now. Commented Feb 5, 2017 at 0:27
Please let me know if I have grossly misinterpreted the problem.
It seems clear to me that we are allowed to repeat letters and numbers. Then, for each slot in our $32$-character string, we have $62$ choices, adding all of the options together ($26$ letters, both upper and lower case, and the $10$ numbers).
Since there are $32$ characters and repetition is allowed, we come to $62^{32} = 2272657884496751345355241563627544170162852933518655225856$ possibilities. Good lord. According to Wolfram Alpha, that is roughly 43 million times the number of possible chess positions. Good lord.
For strings made entirely of capital letters, we have $26^{32}=1901722457268488241418827816020396748021170176$ possibilities. This is also true for the lowercase letters.
Similarly, there are $10^{32}=100000000000000000000000000000000$ possible $32$-element strings of just digits.
So, your answer, apparently, is $62^{32}-2(26)^{32}-10^{32} = 2272657884492947900440704487144706514530812140022612885504$.
• Well counted. Commented Feb 5, 2017 at 3:03
• @CarstenS Ah! Ah! Ah! Commented Feb 5, 2017 at 4:40
Let $U,L,D$ be the sizes of the sets of strings containing only uppercase letters (26 usable characters), lowercase letters (26) or digits (10) respectively. Then the number of admissible strings is $$62^{32}-U-L-D=62^{32}-2\cdot26^{32}-10^{32}$$ $$=2.272\dots×10^{57}$$
• I got the impression the OP just wanted to exclude strings of only capitals, only lowercase, and only numbers. Commented Feb 5, 2017 at 0:34
• @TheCount Right, fixed that. All come to one answer. Commented Feb 5, 2017 at 0:38
• The alphabet size is 26 (a-z or A-Z, no punctuation or spaces included), not 36. Commented Feb 5, 2017 at 0:38
• @GreyMatters I used "alphabet" here in a computer science sense to mean "number of usable characters", but I've swapped the latter in. Commented Feb 5, 2017 at 0:40
• You and I still have different answers. Commented Feb 5, 2017 at 0:41
First, there are the following possibilities for strings:
• 32-character strings containing only small letters
• 32-character strings containing only capital letters
• 32-character strings containing only digits
• 32-character strings containing only small letters and capital letters
• 32-character strings containing only small letters and digits
• 32-character strings containing only capital letters and digits
• 32-character strings containing small letters, capital letters and digits
So, by removing the first 3 categories, what you'll end up with is the total number of 32-character strings which combine 2 or more of the categories. Make sure that's the answer you want.
26 small letters + 26 capital letters + 10 digits = 62 characters total. For the first character in a string, you obviously have 62 possibilities. Since the next character can be the same or different, you have 62 possibilities there as well, or $62^{2}$ possibilities for 2 characters. Adding a 3rd character gives $62^{3}$ possibilities. For 32 characters, then, there's $62^{32}$ possibilities.
There are 26 possible small letters, so by the same logic, you'll get $26^{32}$ possibilities for 32 character strings containing all small letters.
The same logic and math applies to capital letter-only strings, so there's another $26^{32}$ possibilities.
For 10 different digits, 0 through 9, this works out to $10^{32}$ possibilities.
The equation you want, therefore, is:
$$62^{32}-26^{32}-26^{32}-10^{32}$$
You can get the final total on, say, Wolfram|Alpha.
Again, what you're getting is the number of possible 32-character strings containing combinations of 2 or more of the "small letter", "capital letter" and "digit" categories. It's not only important to get the total, but to understand the meaning of the result, as well.
• I got the impression the OP just wanted to exclude strings of only capitals, only lowercase, and only numbers. Commented Feb 5, 2017 at 0:36
• Nice! We agree! I always love having multiple explanations. +1 Commented Feb 5, 2017 at 0:42
• All the answers posted at this point match, so we're all reading the question in the same way. Now the real question is whether this is actually the answer Virat Choudhary needs? Commented Feb 5, 2017 at 0:46
• It is. Now this problem is no longer complicated to me except the number of digits in the answer haha. Commented Feb 5, 2017 at 0:52
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How to change popular term checklist title when edit post?
I have a title_custom field in the article category. How can I show that field in place of the title in the category selection box when editing or adding posts? It’s in the function wp_popular_terms_checklist () Someone have anyway? Thanks very much!
Termination of term rewiting using strict partial order on subterms
Are there any good books, research reports, surveys, theses, or papers that display proof techniques, with clear proofs of termination of term rewriting problems that have the following form…?
Terms are represented by directed acyclic graphs where the terms are vertices with arcs labelled $$arg_{1}…arg_{n}$$ pointing to the immediate sub-terms. There would be additional equality arcs between vertices. Thinking of the transitive closure of the equality arcs as representing equivalence classes of vertices that are "merged", the $$arg$$ arcs in the graph form a lattice (because or the strict order on sub-terms, and some sub-terms might be shared). A rewrite rule would add extra arcs, such that existing partial order would be preserved and added to, so the rewrite rules would be constructing a series of partial orders (represented in the graph state at each step) $$p_{0} \subset … \subset p_{m}$$ more and more "constraining" the partial order relation between vertices until either the re-write rules find nothing to re-write or a rewrite would introduce a cycle (immediately detectable by a depth first search). I think this kind of termination proof is correct because we can say every step was a reduction under the partial order $$p_{m}$$ but I’d like a formal justification because I have worries about my not knowing $$p_{m}$$ before hand, only when it is constructed. And if the rewrite finds a cycle then that cycle was implicit from the beginning. Again I think that’s OK because my re-write rules are prove-ably LHS iff RHS so they transform the problem to an equivalent problem. I call this "construct a partial order or die trying." Is there a more formal name for this kind of proof?
Ideally the proof examples would be constructive and mathematically thorough. I see some papers that assume a lot of prior knowledge, probably because of brevity requirement, and not wanting to bore an expert audience. And others with "wordy" explanations, which are great to give intuitive understanding, but proofs should not depend on them.
Background
I once implemented a datatype representing arbitrary real numbers in Haskell. It labels every real numbers by having a Cauchy sequence converging to it. That will let $$\mathbb{R}$$ be in the usual topology. I also implemented addition, subtraction, multiplication, and division.
But my teacher said, "This doesn’t seem to be a good idea. Since comparison is undecidable here, this doesn’t look very practical. In particular, letting division by 0 to fall in an infinite loop doesn’t look good."
So I wanted my datatype to extend $$\mathbb{Q}$$. Since equality comparison of $$\mathbb{Q}$$ is decidable, $$\mathbb{Q}$$ is in discrete topology. That means a topology on $$\mathbb{R}$$ must be finer than the discrete topology on $$\mathbb{Q}$$.
But, I think I found that, even if I could implement such datatype, it will be impractical.
Proof, step 1
Let $$\mathbb{R}$$ be finer than $$\mathbb{Q}$$ in discrete topology. Then $$\{0\}$$ is open in $$\mathbb{R}$$. Assume $$+ : \mathbb{R}^2 → \mathbb{R}$$ is continuous. Then $$\{(x,-x): x \in \mathbb{R}\}$$ is open in $$\mathbb{R}^2$$. Since $$\mathbb{R}^2$$ is in product topology, $$\{(x,-x)\}$$ is a basis element of $$\mathbb{R}^2$$ for every $$x \in \mathbb{R}$$. It follows that $$\{x\}$$ is a basis element of $$\mathbb{R}$$ for every $$x \in \mathbb{R}$$. That is, $$\mathbb{R}$$ is in discrete topology.
Proof, step 2
Since $$\mathbb{R}$$ is in discrete topology, $$\mathbb{R}$$ is computably equality comparable. This is a contradiction, so $$+$$ is not continuous, and thus not computable.
Question
What is bugging me is the bolded text. It is well-known that every computable function is continuous (Weihrauch 2000, p. 6). Though the analytic definition and the topological definition of continuity coincide in functions from and to Euclidean spaces, $$\mathbb{R}$$ above is not a Euclidean space. So I’m unsure whether my proof is correct. What is the definition of "continuity" in computable analysis?
Context
We recently added a feature that used a library whose API we misunderstood. Long story short, if user A sends a request to our web application, the library caches some result, and that result may show in a response to user B’s request. Needless to say, this is a security bug, specifically, data from user A leaks to user B.
Although it is well-known that web application should be stateless, the long dependency graph of such application makes the likelihood of some downstream library (or its bad usage) accidentally leaking data between requests non-zero. I can imagine this bug is possible with a wide range of web frameworks and environments (e.g., Django, .NET, NodeJS, AWS Lambda), since they all reuse the application between request to avoid cold starts.
Questions
1. What is the proper term for data leaking server-side between HTTP requests, due to an honest developer mistake? Terms such as session hijacking and session fixation seem to refer exclusively to malicious attacks.
2. Are there tools and method to test for such mistakes or detect them in production?
Use Custom Post Type archive page for the taxonomies term archive page
In my wordpress theme I created a new custom post type ‘books’ with 2 taxonomies (‘series’, ‘genres’). when I visit the archive of the cpt ‘books’ (site.com/books), I list all the books.
I added a custom frontend filter to get in this archive page books by taxonomy terms by passing an argument to the url with the name of the taxonomy and the terme (like this: site.com/books/?genres=action). Like a book browser.
But wordpress is by default creating the link for my taxonomies like this (site.com/genres/action/) and i want it to be "redirected" to the books post type with the taxonomy argument (site.com/books/?genres=action).
Is it possible to achieve that ? Thank you
What’s the term for a hash sent early and plain text revealed later?
I think there is a known pattern where you post the hash of a document, e.g. on Twitter, in order to have its time registered. You could then later publish the document and have it accredited for the time of the hash.
I’m sure someone gave this procedure a name. What is that name?
I found trusted timestamping, but that is a thing for digital certificates, which do not come into play here.
Factor independent term from a summation?
Can I factor $$t^2$$ from $$\frac{1}{t}\sum_{n=1}^{n1}t^2Exp[-a n^2+bn]$$ to get
$$t\sum_{n=1}^{n1}Exp[-a n^2+bn]$$ using some command?
1/t Sum[t^2 Exp[-a n^2 + b n], {n, 1, n1}]
Generalization of The Term “Insider Threat”
A definition of an Insider Threat in enterprises/organizations context is: "A current or former employee or business associate who has access to sensitive information or privileged accounts within the network of an organization, and who misuses this access."
I would like to know if such a threat can be generalized in a broader context so I can say that: "An Insider Threat refers to any user or entity that misuses the delegated access by taking the privilege that it is already authenticated and authorized to the system. The misuse of delegated access can be unintentional such as program flaws and failure, or intentional such as user account compromise."
Is my generalization of the term "Insider Threat" correct?
If it is not, what term is used to designate the type of threat that I defined in my generalization (2nd paragraph)?
[ Politics ] Open Question : Why would anyone vote to give Trump another term when things have gotten so much worse under him already in less than four years?
Just look at the state of the country today: Massive protests in the streets. 40 million+ people out of work. 100,000+ people dead from coronavirus. Racial relations badly strained.
Why is the term “nation state” used to refer to a government-sponsored effort in infosec, and is it accurate?
I work in infosec and as such, have read many whitepapers and been to many conference talks. I hear all the time, especially in conversation and literature about malware, the term “nation state” used to refer to a government entity or government-sponsored activity. The term “state actor” is also used.
My question is, why? According to Wikipedia:
A nation state is a state in which a great majority shares the same culture and is conscious of it. The nation state is an ideal in which cultural boundaries match up with political boundaries.1 According to one definition, “a nation state is a sovereign state of which most of its subjects are united also by factors which defined a nation such as language or common descent.” It is a more precise concept than “country”, since a country does not need to have a predominant ethnic group.
According to Merriam-Webster, the definition is:
a form of political organization under which a relatively homogeneous people inhabits a sovereign state
Is there some inherent need for the government sponsor to be primarily of one ethnic background in infosec literature? I just don’t understand why this term in particular is so frequently used, when there are many forms of states, such as a federated state, multinational state, or even more general terms such as “government” or “country,” all of whom would likely be capable of and do participate in infosec activities.
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# Did our ancestors calculate the distance between Prithvi (Earth) and Surya (Sun) correctly?
In Hanuman Chalisa, it is said:
जुग सहस्र जोजन पर भानू। लील्यो ताहि मधुर फल जानू॥ १८ ॥
meaning:
The Sun is at thousand yojanas(a unit of measurement of distance), and thought of it as a sweet fruit.
As per calculations,
``````1 Yug = 12000 years
1 Sahastra = 1000
1 Yojan = 8 Miles
Yug x Sahastra x Yojan = par Bhanu
12000 x 1000 x 8 miles = 96000000 miles
1 mile = 1.6kms
96000000 miles = 96000000 x 1.6kms = 1536000000 kms to Sun
``````
NASA has said that, the above is the exact distance between Earth and Sun(during the Aphelion).
Does this prove that our ancestors calculated the distance between Earth and Sun centuries before scientists did it?
• Apr 13, 2016 at 18:07
• @Wally That is not debunking but BS, viewpoint of a so called skeptic who thinks he's speaking science but know neither Science nor Sanskrit! Jul 19, 2016 at 21:29
• @ABcDexter The skeptic presents mutiple views and presents all the numbers and language mis-interpretations. These people are picking one out of a million numbers that too years and time and language misinterpreted. Jul 20, 2016 at 6:45
• Look, being skeptical never means that you defy history, and it's because of the philosophy of religion that we(as in human beings) have had, which makes us logical enough to question. So, it's not good to not have an open mind, also being Scientific in approach doesn't imply that you or anyone start attacking any religion! Jul 20, 2016 at 6:49
• No. How did they convert yojana and yuga to modern metric system. There is no where in Hindu scripture where it says a Yojana is 8 miles. It's pure algebra by modern Hindu preachers
– Gap
Apr 29, 2017 at 22:40
Interesting Facts & Fascinating Story of Hanuman Chalisa!
Hanuman Chalisa has 40 Chaupais (Chalis=40 in Hindi) on Hanuman is a devotional hymn dedicated to Bajrang Bali by the Great Indian poet, philosopher and saint Shri Goswami Tulsidas. Born in the 16th century, Tulsidas authored Hanuman Chalisa in Awadhi language when he was quite young.
Hanuman Chalisa Tells Us The Distance Between Earth & Sun
When Hanuman was very young, he flew from Earth to the sky in the direction of the Sun to eat it, assuming it to be a ripe, luscious fruit. Tulsidas while stating this incident in the chalisa in simple languages gives the distance between Earth and the Sun.
The Line is –
“जुग सहस्र जोजन पर भानू। लील्यो ताहि मधुर फल जानू॥ १८ ॥”
Meaning:
Sun is at the distance of sahastra(thousand) yojan(an anstonomical unit of distance).
After certain intellectuals decoded this famous line of Hanuman Chalisa by Tulsidas they could find the distance of Earth, they found that tt is exactly the same as that discovered by scientist later.
In the year 1653, astronomer Christian Huygens with his guesswork estimated the distance from earth to the sun.
Now let’s see what the intellectuals (Garjajev’s Research Group) got after decoding the lines:
1 Yuga is 12000 years
1 Sahastra is 1000.
1 Yojan is 8 miles
Thus,
Yuga Shastra Yojan par Bhanu means
12000 X 1000 X 8 = 96000000 miles
and 1 mile is 1.6 km
So, 96000000 X 1.6 = 1536000000 km
This is the exact round figure distance of earth from Sun on 3rd July because studies mention the distance of 152,093,481 km from earth to sun during the aphelion (The period when earth is farthest from the Sun). This is very close to the current figure by NASA.
• How year * miles = miles ? What is per Bhanu? FYI, i'm not against Tulsidas or Hanuman but want to know the exact truth. But fact is Ancient India was more advanced than entire world due to words of infallible Vedas. Jul 19, 2016 at 14:46
• Welcome to Hinduism.SE. Please try to provide links for your research :) Jul 19, 2016 at 20:47
• @TheDestroyer, 'par bhanu पर भानु' could be translated as 'sun is at'. Even though the dimensionality problem remains. Jul 20, 2016 at 10:22
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A073961 Let R be the polynomial ring GF(2)[x]. Then a(n) = number of distinct products f*g with f,g in R and 1 <= deg(f),deg(g) <= n. 1
3, 18, 72, 262, 975, 3562, 13456, 50765, 194122, 745526, 2882670, 11173191, 43485970, 169656399, 663259282, 2598327983, 10190686903, 40038932993, 157431481559, 619871680780, 2442107519364, 9632769554849, 38008189079970, 150127212826428, 593141913076502 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS W. Edwin Clark computed the initial terms. LINKS Table of n, a(n) for n=1..25. FORMULA a(n) = A086908(n) - 1 - Sum_{i=0, n} A001037(i). - Andrew Howroyd, Jul 10 2018 EXAMPLE From Andrew Howroyd, Jul 10 2018: (Start) Case n=1: The following 3 polynomials can be represented: x^2 = x*x, x^2 + 1 = (x + 1)*(x+1), x^2 + x = x*(x + 1). (End) CROSSREFS Cf. A001037, A027424, A086908. Sequence in context: A174764 A114633 A135070 * A305623 A280804 A152897 Adjacent sequences: A073958 A073959 A073960 * A073962 A073963 A073964 KEYWORD nonn AUTHOR Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 13 2003 EXTENSIONS a(9)-a(25) from Andrew Howroyd, Jul 10 2018 STATUS approved
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Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
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Last modified September 27 14:11 EDT 2023. Contains 365711 sequences. (Running on oeis4.)
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# Pedal Coordinates
Pedal coordinates are tangential coordinates that describe the position of a point x on a pedal curve γ by two numbers: the distance from the origin r (the pedal point), and the distance from the origin to the tangent of γ at point x. The tangent of the curve depends on the line from the origin to the point.
The pedal equation is defined as [2]
f(r, p) = 0.
## History and Applications of Pedal Coordinates
The name pedal coordinates appears to be due to H.J. Purkiss, a Victorian-era student at the University of Cambridge, who proposed the name because “they are the polar coordinates of the foot of the perpendicular on the tangent” [3]. However, Purkiss didn’t discover the system. According to B. Pourciau [4] Newton uses pedal coordinates in Principia Mathematica Philosophica Naturalis to show the relationship between the inverse square law and a trajectory; the coordinate system is implied, not explicitly stated [5]. Pedal coordinates are more natural than Cartesian or polar coordinates in some settings, like the study of force problems of classical mechanics in the plane [1].
## References
Pedal curve image: Sam Derbyshire at English Wikipedia, CC BY-SA 3.0, via Wikimedia Commons
[1] Blaschke, P. PEDAL COORDINATES, DARK KEPLER AND OTHER FORCE PROBLEMS. Retrieved January 16, 2022 from: http://arxiv-export-lb.library.cornell.edu/pdf/1704.00897
[2] Yates, R. (1974). Curves and Their Properties. The National Council of Teachers of Mathematics. from: https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.853.7639&rep=rep1&type=pdf
[3] The Oxford, Cambridge, and Dublin Messenger of Mathematics, Volume 3. Macmillan and Company, 1866.
[4] Pourciau, B. Reading the Master: Newton and the Birth of Celestial Mechanics.
[5] Olivier Bruneau. ICT AND HISTORY OF MATHEMATICS: the case of the pedal curves from
17th-century to 19th-century. 6th European Summer University on the History and Epistemology in Mathematics Education, Jul 2010, Vienna, Austria. pp.363-370. ffhal-01179909f
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# AREA Units Conversionsquare-inches to decares
1 Square Inches
= 6.4516E-7 Decares
Category: area
Conversion: Square Inches to Decares
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Inches] symbol/abbrevation: (in2, sq in)
[Decares] symbol/abbrevation: (daa)
How to convert Square Inches to Decares (in2, sq in to daa)?
1 in2, sq in = 6.4516E-7 daa.
1 x 6.4516E-7 daa = 6.4516E-7 Decares.
Always check the results; rounding errors may occur.
Definition:
The decare (symbol daa) is derived from deka, the prefix for 10 and are, and is equal to 10 ares or 1000 square metres. It is used in Norway and in the former Ottoman areas o ..more definition+
In relation to the base unit of [area] => (square meters), 1 Square Inches (in2, sq in) is equal to 0.00064516 square-meters, while 1 Decares (daa) = 1000 square-meters.
1 Square Inches to common area units
1 in2, sq in = 0.00064516 square meters (m2, sq m)
1 in2, sq in = 6.4516 square centimeters (cm2, sq cm)
1 in2, sq in = 6.4516E-10 square kilometers (km2, sq km)
1 in2, sq in = 0.0069444474344208 square feet (ft2, sq ft)
1 in2, sq in = 1 square inches (in2, sq in)
1 in2, sq in = 0.0007716049382716 square yards (yd2, sq yd)
1 in2, sq in = 2.4909766863756E-10 square miles (mi2, sq mi)
1 in2, sq in = 1000000 square mils (sq mil)
1 in2, sq in = 6.4516E-8 hectares (ha)
1 in2, sq in = 1.5942236697094E-7 acres (ac)
Square Inchesto Decares (table conversion)
1 in2, sq in = 6.4516E-7 daa
2 in2, sq in = 1.29032E-6 daa
3 in2, sq in = 1.93548E-6 daa
4 in2, sq in = 2.58064E-6 daa
5 in2, sq in = 3.2258E-6 daa
6 in2, sq in = 3.87096E-6 daa
7 in2, sq in = 4.51612E-6 daa
8 in2, sq in = 5.16128E-6 daa
9 in2, sq in = 5.80644E-6 daa
10 in2, sq in = 6.4516E-6 daa
20 in2, sq in = 1.29032E-5 daa
30 in2, sq in = 1.93548E-5 daa
40 in2, sq in = 2.58064E-5 daa
50 in2, sq in = 3.2258E-5 daa
60 in2, sq in = 3.87096E-5 daa
70 in2, sq in = 4.51612E-5 daa
80 in2, sq in = 5.16128E-5 daa
90 in2, sq in = 5.80644E-5 daa
100 in2, sq in = 6.4516E-5 daa
200 in2, sq in = 0.000129032 daa
300 in2, sq in = 0.000193548 daa
400 in2, sq in = 0.000258064 daa
500 in2, sq in = 0.00032258 daa
600 in2, sq in = 0.000387096 daa
700 in2, sq in = 0.000451612 daa
800 in2, sq in = 0.000516128 daa
900 in2, sq in = 0.000580644 daa
1000 in2, sq in = 0.00064516 daa
2000 in2, sq in = 0.00129032 daa
4000 in2, sq in = 0.00258064 daa
5000 in2, sq in = 0.0032258 daa
7500 in2, sq in = 0.0048387 daa
10000 in2, sq in = 0.0064516 daa
25000 in2, sq in = 0.016129 daa
50000 in2, sq in = 0.032258 daa
100000 in2, sq in = 0.064516 daa
1000000 in2, sq in = 0.64516 daa
1000000000 in2, sq in = 645.16 daa
(Square Inches) to (Decares) conversions
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# Simulation Optimization Essay Sample
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• Category: risk
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## Simulation Optimization Essay Sample
Simulation Optimization is providing solutions to important practical problems previously beyond reach. This paper explores how new approaches are significantly expanding the power of Simulation Optimization for managing risk. Recent advances in Simulation Optimization technology are leading to new opportunities to solve problems more effectively. Specifically, in applications involving risk and uncertainty, Simulation Optimization surpasses the capabilities of other optimization methods, not only in the quality of solutions, but also in their interpretability and practicality. In this paper, we demonstrate the advantages of using a Simulation Optimization approach to tackle risky decisions, by showcasing the methodology on two popular applications from the areas of finance and business process design.
Keywords: optimization, simulation, portfolio selection, risk management.
1. Introduction
Whenever uncertainty exists, there is risk. Uncertainty is present when there is a possibility that the outcome of a particular event will deviate from what is expected. In some cases, we can use past experience and other information to try to estimate the probability of occurrence of different events. This allows us to estimate a probability distribution for all possible events. Risk can be defined as the probability of occurrence of an event that would have a negative effect on a goal. On the other hand, the probability of occurrence of an event that would have a positive impact is considered an opportunity (see Ref. 1 for a detailed discussion of risks and opportunities).
Therefore, the portion of the probability distribution that represents potentially harmful, or unwanted, outcomes is the focus of risk management. Risk management is the process that involves identifying, selecting and implementing measures that can be applied to mitigate risk in a particular situation.1 The objective of risk management, in this context, is to find the set of actions (i.e., investments, policies, resource configurations, etc.) to reduce the level of risk to acceptable levels. What constitutes an acceptable level will depend on the situation, the decision makers’ attitude towards risk, and the marginal rewards expected from taking on additional risk. In order to help risk managers achieve this objective, many techniques have been developed, both qualitative and quantitative.
Among quantitative techniques, optimization has a natural appeal because it is based on objective mathematical formulations that usually output an optimal solution (i.e. set of decisions) for mitigating risk. However, traditional optimization approaches are prone to serious limitations. In Section 2 of this paper, we briefly describe two prominent optimization techniques that are frequently used in risk management applications for their ability to handle uncertainty in the data; we then discuss the advantages and disadvantages of these methods. In Section 3, we discuss how Simulation Optimization can overcome the limitations of traditional optimization techniques, and we detail some innovative methods that make this a very useful, practical and intuitive approach for risk management. Section 4 illustrates the advantages of Simulation Optimization on two practical examples. Finally, in Section 5 we summarize our results and conclusions.
Very few situations in the real world are completely devoid of risk. In fact, a person would be hard-pressed to recall a single decision in their life that was completely risk-free. In the world of deterministic optimization, we often choose to “ignore” uncertainty in order to come up with a unique and objective solution to a problem. But in situations where uncertainty is at the core of the problem – as it is in risk management – a different strategy is required. In the field of optimization, there are various approaches designed to cope with uncertainty.2,3 In this context, the exact values of the parameters (e.g. the data) of the optimization problem are not known with absolute certainty, but may vary to a larger or lesser extent depending on the nature of the factors they represent.
In other words, there may be many possible “realizations” of the parameters, each of which is a possible scenario. Traditional scenario-based approaches to optimization, such as scenario optimization and robust optimization, are effective in finding a solution that is feasible for all the scenarios considered, and minimizing the deviation of the overall solution from the optimal solution for each scenario. These approaches, however, only consider a very small subset of possible scenarios, and the size and complexity of models they can handle are very limited.
1.1 Scenario Optimization
Dembo4 offers an approach to solving stochastic programs based on a method for solving deterministic scenario subproblems and combining the optimal scenario solutions into a single feasible decision. Imagine a situation in which we want to minimize the cost of producing a set J of finished goods. Each good j (j=1,…,n) has a per-unit production cost cj associated with it, as well as an associated utilization rate aij of resources for each finished good. In addition, the plant that produces the goods has a limited amount of each resource i (i=1,…,m), denoted by bi. We can formulate a deterministic mathematical program for a single scenario s (the scenario subproblem, or SP) as follows:
SP:
zs = minimize [pic](1)
Subject to:[pic]for i=1,…,m(2)
xj ≥ 0for j=1,…,n(3)
where cs, as and bs respectively represent the realization of the cost coefficient, the resource utilization and the resource availability data under scenario s. Consider, for example, a company that manufactures a certain type of Maple door. Depending on the weather in the region where the wood for the doors is obtained, the costs of raw materials and transportation will vary. The company is also considering whether to expand production capacity at the facility where doors are manufactured, so that a total of six scenarios must be considered.
The six possible scenarios and associated parameters for Maple doors are shown in Table 1. The first column corresponds to the particular scenario; Column 2 denotes whether the facility is at current or expanded capacity; Column 3 shows the probability of each capacity scenario; Column 4 denotes the weather (dry, normal or wet) for each scenario; Column 5 provides the probability for each weather instance; Column 6 denotes the probability for each scenario; Column 7 shows the cost associated with each scenario (L = low, M = medium, H = high); Column 8 denotes the utilization rate of the capacity (L = low, H = high); and Column 9 denotes the expected availability associated with each scenario.
Minimize [pic](4)
Subject to: xj ≥ 0 for j=1,…,n(5)
The purpose of this tracking model is to find a solution that is feasible under all the scenarios, and penalizes solutions that differ greatly from the optimal solution under each scenario. The two terms in the objective function are squared to ensure non-negativity.
More sophisticated tracking models can be used for various different purposes. In risk management, for instance, we may select a tracking model that is designed to penalize performance below a certain target level.
1.2.Robust Optimization
Robust optimization may be used when the parameters of the optimization problem are known only within a finite set of values. The robust optimization framework gets its name because it seeks to identify a robust decision – i.e. a solution that performs well across many possible scenarios. In order to measure the robustness of a given solution, different criteria may be used. Kouvelis and Yu identify three criteria: (1) Absolute robustness; (2) Robust deviation; and (3) Relative robustness5.
We illustrate the meaning and relevance of these criteria, by describing their robust optimization approach. Consider an optimization problem where the objective is to minimize a certain performance measure such as cost. Let S denote the set of possible data scenarios over the planning horizon of interest. Also, let X denote the set of decision variables, and P the set of input parameters of our decision model. Correspondingly, let Ps identify the value of the parameters belonging to scenario s, and let Fs identify the set of feasible solutions to scenario s. The optimal solution to a specific scenario s is then:
[pic](6)
We assume here that f is convex. The first criterion, absolute robustness, also known as “worst-case optimization,” seeks to find a solution that is feasible for all possible scenarios and optimal for the worst possible scenario. In other words, in a situation where the goal is to minimize the cost, the optimization procedure will seek the robust solution, zR, that minimizes the cost of the maximum-cost scenario. We can formulate this as an objective function of the form
[pic](7)
Variations to this basic framework have been proposed (see Ref. 5 for examples) to capture the risk-averse nature of decision-makers, by introducing higher moments of the distribution of zs in the optimization model, and implementing weights as penalty factors for infeasibility of the robust solution with respect to certain scenarios. The problem with both of these approaches, as with most traditional optimization techniques that attempt to deal with uncertainty, is their inability to handle a large number of possible scenarios. Thus, they often fail to consider events that, while unlikely, can be catastrophic. Recent approaches that use innovative Simulation Optimization techniques overcome these limitations by providing a practical, flexible framework for risk management and decision-making under uncertainty.
3. Simulation Optimization
Simulation Optimization can efficiently handle a much larger number of scenarios than traditional optimization approaches, as well as multiple sources and types of risk. Modern simulation optimization tools are designed to solve optimization problems of the form: MinimizeF(x)(Objective function)
Subject to:Ax < b (Constraints on input variables)
gl < G(x) < gu (Constraints on output measures)
l < x < u (Bounds),
where the vector x of decision variables includes variables that range over continuous values and variables that only take on discrete values (both integer values and values with arbitrary step sizes).7 The objective function F(x) is, typically, highly complex. Under the context of Simulation Optimization, F(x) could represent, for example, the expected value of the probability distribution of the throughput at a factory; the 5th percentile of the distribution of the net present value of a portfolio of investments; a measure of the likelihood that the cycle time of a process will be lower than a desired threshold value; etc. In general, F(x) represents an output performance measure obtained from the simulation, and it is a mapping from a set of values x to a real value. The constraints represented by inequality Ax ≤ b are usually linear (given that non-linearity in the model is embedded within the simulation itself), and both the coefficient matrix A and the right-hand-side values corresponding to vector b are known.
The constraints represented by inequalities of the form gl ≤ G(x) ≤ gu impose simple upper and/or lower bound requirements on an output function G(x) that can be linear or non-linear. The values of the bounds gl and gu are known constants. All decision variables x are bounded and some may be restricted to be discrete, as previously noted. Each evaluation of F(x) and G(x) requires an execution of a simulation of the system. By combining simulation and optimization, a powerful design tool results. Simulation enables fast, inexpensive and non-disruptive examination and testing of a large number of scenarios prior to actually implementing a particular decision in the “real” environment. As such, it is quickly becoming a very popular tool in industry for conducting detailed “what-if” analysis. Since simulation approximates reality, it also permits the inclusion of various sources of uncertainty and variability into forecasts that impact performance.
The need for optimization of simulation models arises when the analyst wants to find a set of model specifications (i.e., input parameters and/or structural assumptions) that leads to optimal performance. On one hand, the range of parameter values and the number of parameter combinations is too large for analysts to enumerate and test all possible scenarios, so they need a way to guide the search for good solutions. On the other hand, without simulation, many real world problems are too complex to be modeled by tractable mathematical formulations that are at the core of pure optimization methods like scenario optimization and robust optimization. This creates a conundrum; as shown above, pure optimization models alone are incapable of capturing all the complexities and dynamics of the system, so one must resort to simulation, which cannot easily find the best solutions. Simulation Optimization resolves this conundrum by combining both methods.
Optimizers designed for simulation embody the principle of separating the method from the model. In such a context, the optimization problem is defined outside the complex system. Therefore, the evaluator (i.e. the simulation model) can change and evolve to incorporate additional elements of the complex system, while the optimization routines remain the same. Hence, there is a complete separation between the model that represents the system and the procedure that is used to solve optimization problems defined within this model. The optimization procedure – usually based on metaheuristic search algorithms – uses the outputs from the system evaluator, which measures the merit of the inputs that were fed into the model. On the basis of both current and past evaluations, the method decides upon a new set of input values (Figure 1 shows the coordination between the optimization engine and the simulation model).
[pic]
Fig. 1: Coordination between the optimization engine and the simulation
Provided that a feasible solution exists, the optimization procedure ideally carries out a special search where the successively generated inputs produce varying evaluations, not all of them improving, but which over time provide a highly efficient trajectory to the globally best solutions. The process continues until an appropriate termination criterion is satisfied (usually based on the user’s preference for the amount of time devoted to the search). As stated before, the uncertainties and complexities modeled by the simulation are often such that the analyst has no idea about the shape of the response surface – i.e. the solution space. There exists no closed-form mathematical expression to represent the space, and there is no way to gauge whether the region being searched is smooth, discontinuous, etc.
While this is enough to make most traditional optimization algorithms fail, metaheuristic optimization approaches, such as tabu search5 and scatter search8, overcome this challenge by making use of adaptive memory techniques and population sampling methods that allow the search to be conducted on a wide area of the solution space, without getting stuck in local optima. The metaheuristic-based simulation optimization framework is also very flexible in terms of the performance measures the decision-maker wishes to evaluate. In fact, the only limitation is not on the side of the optimization engine, but on the simulation model’s ability to evaluate performance based on specified values for the decision variables. In order to provide in-depth insights into the use of simulation optimization in the context of risk-management, we present some practical applications through the use of illustrative examples.
4. Illustrative Examples
4.1.Selecting Risk-Efficient Project Portfolios
Companies in the Petroleum and Energy (P&E) Industry use project portfolio optimization to manage investments in exploration and production, as well as power plant acquisitions.9,10 Decision makers typically wish to maximize the return on invested capital, while controlling the exposure of their portfolio of projects to various risk factors that may ultimately result in financial losses. In this example, we look at a P&E company that has sixty-one potential projects in its investment funnel. For each project, the pro-forma revenues for a horizon of 10 to 20 periods (depending on the project) are given as probability distributions. To carry it out, each project requires an initial investment and a certain number of business development, engineering and earth sciences personnel. The company has a budget limit for its investment opportunities, and a limited number of personnel of each skill category. In addition, each project has a probability of success (POS) factor.
This factor has a value between 0 and 1, and affects the simulation as follows: let’s suppose that Project A has a POS = 0.6; therefore, during the simulation, we expect that we will be able to obtain the revenues from Project A in 60% of the trials, while in the remaining 40%, we will only incur the investment cost. The resulting probability distribution of results from simulating Project A would be similar to that shown in Figure 2, where about 40% of the trials would have negative returns (i.e. equal to the investment cost), and the remaining 60% would have returns resembling the shape of its revenue distribution (i.e. equal to the simulated revenues minus the investment cost).
[pic]
Fig. 2: sample probability distribution of returns for a single simulated project
Projects may start in different time periods, but there is a restricted window of opportunity of up to three years for each project. The company must select a set of projects to invest in that will best further its corporate goals. Probably, the best-known model for portfolio optimization is rooted in the work of Nobel laureate Harry Markowitz. Called the mean-variance model11, it is based on the assumption that the expected portfolio returns will be normally distributed. The model seeks to balance risk and return in a single objective function, as follows. Given a vector of portfolio returns r, and a covariance matrix Q of returns, then we can formulate the model as follows:
Maximize rT w – kwT Qw(8)
Subject to:(i ciwi = b(9)
w ( {0,1}(10)
where k represents a coefficient of the firm’s risk aversion, ci represents the initial investment in project i, wi is a binary variable representing the decision whether to invest in project i, and b is the available budget. We will use the mean-variance model as a base case for the purpose of comparing to other selected models of portfolio performance. To facilitate our analysis, we make use of the OptFolio® software that combines simulation and optimization into a single system specifically designed for portfolio optimization.12, 13 We examine three cases, including Value at Risk (VaR) minimization, to demonstrate the flexibility of this method to enable a variety of decision alternatives that significantly improve upon traditional mean-variance portfolio optimization, and illustrate the flexibility afforded by simulation optimization approaches in terms of controlling risk. The results also show the benefits of managing and efficiently allocating scarce resources like capital, personnel and time. The weighted average cost of capital, or annual discount rate, used for all cases is 12%.
Case 1: Mean-Variance Approach
In this first case, we implement the mean-variance portfolio selection method of Markowitz described above. The decision is to determine participation levels (0 or 1) in each project with the objective of maximizing the expected net present value (NPV) of the portfolio while keeping the standard deviation of the NPV below a specified threshold of \$140M. We denote the expected value of the NPV by (NPV, and the standard deviation of the NPV by (NPV. This case can be formulated as follows:
Maximize (NPV(objective function)(11)
Subject to:(NPV < \$140M(requirement)(12)
[pic](budget constraint)(13)
[pic](personnel constraints)(14)
All projects must start in year 1(15)
xi ( {0, 1} (i(binary decisions)(16)
The optimal portfolio has the following performance metrics:
(NPV = \$394M, (NPV = \$107M, P(5)NPV = \$176M, where P(5)NPV denotes the 5th percentile of the resulting NPV probability distribution (i.e. the probability of the NPV being lower than the P(5) value is 5%). The bound imposed on the standard deviation in Eq. (12) does not seem binding. However, due to the binary nature of the decision variables, no project additions are possible without violating the bound. Figure 2 shows a graph of the probability distribution of the NPV obtained from 1000 replications of this base model. The thin line represents the expected value.
[pic]
Fig. 2: distribution of returns for mean-variance approach
Case 2: Risk controlled by 5th Percentile
In the context of risk management, statistics such as variance or standard deviation of returns are not always easy to interpret, and there may be other measures that are more intuitive and useful. For example, it provides a clearer picture of the risk involved if we say: “there is a 5% chance that the portfolio return will be below some value X,” than to say that “the standard deviation is \$107M.” The former analysis can be easily implemented in a simulation optimization approach by imposing a requirement on the 5th percentile of the resulting distribution of returns, as we describe here. In Case 2, the decision is to determine participation levels (0, 1) in each project with the objective of maximizing the expected NPV of the portfolio, while keeping the 5th percentile of the NPV distribution above the value of the 5th percentile obtained in Case 1.
In this way, we seek to “move” the distribution of returns further to the right, so as to reduce the likelihood of undesired outcomes. This is achieved by imposing the requirement represented by Eq. (18) in the model below. In other words, we want to find the portfolio that produces the maximum average return, as long as no more than 5% of the trial observations fall below \$176M. In addition, we allow for delays in the start dates of projects, according to windows of opportunity defined for each project. In order to achieve this, in the simulation model we have created copies of each project that are shifted by one, two or three periods into the future (according to the windows of opportunity defined for each project). Mutual exclusivity clauses among these stages copies of a project ensure that only one start date is selected. For example, to represent the fact that Project A can start at time t = 0, 1 or 2, we include the following mutual exclusivity clause as a constraint:
Project A0 + Project A1 + Project A2 ( 1
The subscript following the project name corresponds to the allowed start dates for the project, and the constraint only allows at most one of these to be chosen.
Maximize (NPV(17)
Subject to:P(5)NPV > \$176M(requirement)(18)
[pic](budget constraint)(19)
[pic](personnel constraints)(20)
[pic](mutual exclusivity)(21)
xi ( {0, 1} (I(binary decisions)(22)
where mi denotes the set of mutually exclusive projects related to project i. In this case we have replaced the standard deviation with the 5th percentile as a measure of risk containment. The resulting portfolio has the following attributes:
(NPV = \$438M, (NPV = \$140M, P(5)NPV = \$241M
By using the 5th percentile instead of the standard deviation as a measure of risk, we are able to obtain an outcome that shifts the distribution of returns to the right, compared to Case 1, as shown in Figure 3.
[pic]
Fig. 3: distribution of returns for Case 2
This case clearly outperforms case 1. Although the distribution of returns exhibits a wider range (remember, we are not constraining the standard deviation here), not only do we obtain significantly better financial performance, but we also achieve a higher personnel utilization rate, and a more diverse portfolio.
Case 3: Probability-Maximizing and Value-at-Risk
In Case 3, the decision is to determine participation levels (0, 1) in each project with the objective of maximizing the probability of meeting or exceeding the mean NPV found in Case 1. This objective is expressed in Eq. (23) of the following model.
Maximize P(NPV ≥ \$394M)(23)
Subject to:[pic](budget constraint)(24)
[pic](personnel constraints)(25)
[pic](mutual exclusivity)(26)
xi ( {0, 1} (i(binary decisions)(27)
This case focuses on maximizing the chance of achieving a goal and essentially combines performance and risk containment into one metric. The probability in (23) is not known a priori, so we must rely on the simulation to obtain it. The resulting optimal solution yields a portfolio that has the following attributes:
(NPV = \$440M, (NPV = \$167M, P(5) = \$198M
Although this portfolio has a performance similar to the one in Case 2, it has a 70% chance of achieving or exceeding the NPV goal (whereas Case 1 had only a 50% chance). As can be seen in the graph of Figure 4, we have succeeded in shifting the probability distribution even further to the right, therefore increasing our chances of exceeding the returns obtained with the traditional Markowitz approach. In addition, in cases 2 and 3, we need not make any assumption about the distribution of expected returns.
[pic]
Fig. 4: distribution of returns for Case 3
As a related corollary to this last case, we can conduct an interesting analysis that addresses Value-at-Risk (VaR). In traditional (securities) portfolio management, VaR is defined as the worst expected loss under normal market conditions over a specific time interval and at a given confidence level. In other words, VaR measures how much the investor can lose, with probability = (, over a certain time horizon.14 In the case of project portfolios, we can define VaR as the probability that the NPV of the portfolio will fall below a specified value. Going back to our present case, the manager may want to limit the probability of incurring negative returns. In this example, we formulate the problem in a slightly different way: we still want to maximize the expected return, but we limit the probability that we incur a loss to ( = 1% by using the requirement shown in Eq. (29) as follows:
Maximize (NPV(28)
Subject to:P(NPV < 0) ≤ 1%(requirement)(29)
[pic](budget constraint)(30)
[pic](personnel constraints)(31)
[pic](mutual exclusivity)(32)
xi ( {0, 1} (i(binary decisions)(33)
The portfolio performance under this scenario is:
(NPV = \$411M, (NPV = \$159M, P(5) = \$195M
The results from the VaR model turn out to be slightly inferior to the case where the probability was maximized. This is not a surprise, since the focus of VaR is to limit the probability of downside risk, whereas before, the goal was to maximize the probability of obtaining a high expected return. However, this last analysis could prove valuable for a manager who wants to limit the VaR of the selected portfolio. As shown here, for this particular set of projects, a very good portfolio – in financial terms – can still be selected with that objective in mind.
4.2.Risk Management in Business Process Design
Very common measures of process performance are the cycle time, a.k.a turnaround time, the throughput, and the operational cost. For our present example, we consider a process manager at a hospital emergency room (ER). Typically, emergency patients that arrive at the ER present different levels of criticality. In our example, we consider 2 levels: Level 1 patients are very critical, and require immediate treatment; Level 2 patients are not as critical, and must undergo an assessment by a triage nurse before being assigned to an ER room.
Usually, a patient will have a choice in the care provider he or she prefers, so there is an inherent risk of lost business related to the quality of service provided. For instance, if patients must spend a very long time in the ER, there is a risk that patients will prefer to seek care at another facility on subsequent, follow-up visits. Figure 5 shows a high-level flowchart of the process for Level 1 patients (the process for Level 2 patients differs only in that the “Fill out registration” activity is done before the “Transfer to room” activity, during the triage assessment.)
[pic]
Fig. 5: process flowchart for Level 1 patients
The current operation consists of 7 nurses, 3 physicians, 4 patient care technicians (PCTs), 4 administrative clerks and 20 ER rooms. This current configuration has a total operating cost (i.e. wages, supplies, ER rooms’ costs, etc.) of \$52,600 per 100 hours of operation, and the average time a Level 1 patient spends at the ER (i.e. cycle time) has been estimated at 1.98 hours. Hospital management has just issued the new operational budget for the coming year, and has allocated to the ER a maximum operational cost of \$40,000 per 100 hours of operation.
The manager has three weeks to make changes to the process to ensure compliance with the budget, without deteriorating the current service levels. The process manager has stated his goals as: to minimize the average cycle time for Level 1 patients, while ensuring that the operational cost is below \$40,000 per 100 hours of operation. Since arrival times and service times in the process are stochastic, we use a simulation model to simulate the ER operation, and OptQuest to find the best configuration of resources in order to achieve the manager’s goals. We obtain the following results: • Operational Cost = \$36,200 (3 nurses, 3 physicians, 1 PCT, 2 clerks, 12 ER rooms) • Average Cycle Time = 2.08 hours
The configuration of resources above (shown in parentheses) results in the lowest possible cycle time given the new operational budget for the ER. Obviously, in order to improve the service level at the current operational budget, it is necessary to re-design the process itself. The new process proposed for Level 1 patients is depicted in Figure 6. In the proposed process, the “Receive treatment” and “Fill out registration” activities are now done in parallel, instead of in sequence. The simulation of this new process, with the resource configuration found earlier, reduces the average cycle time from 2.08 hours to 1.98 hours.
[pic]
Fig. 6: redesigned process for Level 1 patients
When designing business processes, however, it is very important to set goals that are not subject to the Law of Averages – that is, the goals should be set so that they accommodate a large percentage of the demand for the product or service that the process is intended to deliver. In the above example, it is possible that the probability distribution of cycle time turns out to be highly skewed to the right, so that even if the average cycle is under 2 hours, there could be a very large number of patients who will spend a much longer time in the ER.
It would be better to restate the manager’s performance goals as: ensure that at least 95% of Level 1 patients will spend no more than 2 hours in the ER while ensuring that the operational cost is under \$40,000. This gives a clear idea of the service level to which the process manager aspires. If we re-optimize the configuration of resources with the new goal, we obtain the following results: • Operational Cost = \$31,800 (4 nurses, 2 physicians, 2 PCTs, 2 clerks, 9 ER rooms) • Average Cycle Time = 1.94 hours
• 95th Percentile of Cycle Time = 1.99 hours
Through the use of simulation optimization, we have obtained a process design that complies with the new budget requirements as well as with an improved service level. If implemented correctly, we can be sure that 95% of the critical patients in the ER will be either released or admitted into the hospital for further treatment in less than 2 hours. Thus, the risk of having unsatisfied patients is minimized.
5. Results and Conclusions
Practically every real-world situation involves uncertainty and risk, creating a need for optimization methods that can handle uncertainty in model data and input parameters. We have briefly described two popular methods, scenario optimization and robust optimization, that seek to overcome limitations of classical optimization approaches for dealing with uncertainty, and which undertake to find high-quality solutions that are feasible under as many scenarios as possible. However, these methods are unable to handle problems involving moderately large numbers of decision variables and constraints, or involving significant degrees of uncertainty and complexity.
In these cases, simulation optimization is becoming the method of choice. The combination of simulation and optimization affords all the flexibility of the simulation engine in terms of defining a variety of performance measures and risk profiles, as desired by the decision maker. In addition, as we demonstrate through two practical examples, modern optimization engines can enforce requirements on one or more outputs from the simulation, a feature that scenario-based methods cannot handle. This affords the user different alternatives for controlling risk, while ensuring that the performance of the system is optimized.
The combination of simulation and optimization creates a tool for decision making that is fast (a simulation runs at a small fraction of real-time, and the optimization guides the search for good solutions without the need to enumerate all possibilities), inexpensive and non-disruptive (solutions can be evaluated without the need to stop the normal operation of the business, as opposed to pilot projects which can also be quite expensive). Finally, simulation optimization produces results that can be conveyed and grasped in an intuitive manner, providing the user with an especially useful and easy-to-use tool for identifying improved business decisions under risk and uncertainty.
References
1. D. Vose, Risk Analysis: A Quantitative Guide, (John Wiley and Sons, Chichester, 2000). 2. M. Fukushima, How to deal with uncertainty in optimization – some recent attempts, International Journal of Information Technology & Decision Making, 5.4 (2006), 623 – 637. 3. H. Eskandari and L. Rabelo, Handling uncertainty in the analytic hierarchy process: a stochastic approach, International Journal of Information Technology & Decision Making, 6.1 (2007), 177 – 189. 4. R. Dembo, Scenario Optimization, Annals of Operations Research 30 (1991), 63 – 80. 5. P. Kouvelis and G. Yu, Robust Discrete Optimization and Its Applications,(Kluwer: Dordrecht, Netherlands, 1997), 8 – 29. 6. J. P. Kelly, Simulation Optimization is Evolving, INFORMS Journal of Computing 14.3 (2002), 223 – 225. 7. F. Glover and M. Laguna, Tabu Search, ( Kluwer: Norwell, MA, 1997). 8. F. Glover, M. Laguna and R. Martí, Fundamentals of scatter search and path relinking, Control and Cybernetics
29.3 (2000), 653 – 684. 9. W. J. Haskett, Optimal appraisal well location through efficient uncertainty reduction and value of information techniques, in Proceedings of the Society of Petroleum Engineers Annual Technical Conference and Exhibition, (Denver, CO, 2003). 10. W. J. Haskett, M. Better and J. April, Practical optimization: dealing with the realities of decision management, in Proceedings of the Society of Petroleum Engineers Annual Technical Conference and Exhibition, (Houston, TX, 2004). 11. H. Markowitz, Portfolio selection, Journal of Finance 7.1 (1952), 77 – 91. 12. J. April, F. Glover and J. P. Kelly, Portfolio Optimization for Capital Investment Projects, in Proceedings of the 2002 Winter Simulation Conference, (eds.) S. Chick, T. Sanchez, D. Ferrin and D. Morrice, (2002), 1546 – 1554. 13. J. April, F. Glover and J. P. Kelly, Optfolio – A Simulation Optimization System for Project Portfolio Planning, in Proceedings of the 2003 Winter Simulation Conference, (eds.) S. Chick, T. Sanchez, D. Ferrin and D. Morrice, (2003), 301 – 309. 14. S. Benninga, and Z. Wiener, Value-at-Risk (VaR), Mathematica in Education and Research 7.4 (1998), 1 – 7.
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[1] Published in the International Journal of Information Technology & Decision Making, Vol 7, No 4 (2008) 571-587.
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Understanding Statistical Significance in Algorithmic Trading
Introduction ::
In the world of algorithmic trading, understanding the concept of statistical significance is paramount. It allows traders to differentiate between reliable trading strategies and mere chance. In this article, we will explore the fundamental principles and techniques involved in assessing statistical significance in algo trading.
The Basics: Normal Distribution and Standard Deviation
At the core of statistical significance lies the normal distribution and standard deviation. The normal distribution, often resembling a bell-shaped curve, represents a random process such as “white noise.” Within this distribution, the standard deviation defines a range that encompasses a significant portion of the observed samples or values. Typically, about 68% of the values fall within one standard deviation, referred to as 1 sigma.
Understanding the 68-95-99 Rule
The 68-95-99 rule, widely recognized in empirical sciences, provides further insight. According to this rule, nearly all values (around 99.7%) are expected to lie within three standard deviations of the mean. This allows researchers and traders to treat a probability of 99.7% as a near certainty, making values outside this range potential signals of statistical significance.
The Z-Score: Measuring Confidence
To quantify the significance of an observed deviation from the mean, traders often rely on the z-score. The z-score determines the number of standard deviations a value is away from the mean. A higher z-score indicates a greater level of confidence that the observed result is not due to chance alone.
Calculating Confidence Levels
By using the z-score, traders can calculate confidence levels for their trading strategy results. For instance, a strategy result lying 2.58 sigmas away from the mean corresponds to a confidence level of approximately 99%. This implies a very low probability that the result is merely due to chance.
Determining Statistical Significance in Algo Trading
Determining the statistical significance of trading results is crucial. Traders need to ascertain whether their strategy’s performance is a result of skill or simply random chance. Conducting extensive backtesting, observing multiple trades, and calculating z-scores can help in this process. A higher number of trades increases confidence in the strategy’s effectiveness, reducing the influence of chance occurrences.
Conclusion:
Statistical significance plays a pivotal role in the evaluation of algo trading strategies. Understanding concepts like the normal distribution, standard deviation, and the z-score empowers traders to discern reliable signals from noise. By conducting thorough backtesting and analyzing confidence levels, traders can gain a deeper understanding of the statistical significance of their trading strategies. This knowledge aids in making informed decisions and achieving consistent results in the dynamic world of algo trading.
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# Problems with Random.Range
Hello, I’m using `Random.Range` to roll a 1d6 dice but it always seems to return the same number? Also (according to the Unity handbook) if the range is between two ints, and it rolls maximum, you will instead get the minimum. I think I’ve noticed this behaviour before, but it absolutely baffles me why this would be the case. Is there any way round it without converting from floats? Adding +1 to the maximum is not acceptable as that means there’s still a greater chance to roll a one than a six.
That doesn’t seem likely. When something used as often as random numbers doesn’t seem to work I would suggest that a developer check their code. Write a simple test that proves or disproves your theory.
The following says nothing about rolling the maximum. It is explaining that the max is “exclusive” i.e. that value is never returned. If you pass two identical integers then the min value is returned. What else could it do it only has one number in the range?
Note max is exclusive. Random.Range(0, 10) can return a value between 0 and 9. Return min if max equals min. The Random.Range distribution is uniform. Range is a Random Number Generator.
1 Like
I think this contradicts what you say. If the max is 10, the max it can actually return is 9. Therefore it never returns the maximum? This seems to be confirmed by the test, though I retract what I said about defaulting to one.
Yes the max value is never returned unless you use the same value for min and max.
(so Random.Range(1,1) will return 1).
In any case, what is the problem exactly?
Either use Random.Range(0,6) + 1, or Random.Range(1,7) to return the value for a six sided die.
Thanks. I’ve decided to do that. My issue is that I don’t think that’s how it should work and I have a whole host of `Random.Range` instances to do that for across about fifty files, and I’m not sure where they all are.
I mean, you can search within all your files. You could also make your own random range function that uses Random.Range and does the +1, so you can use values that make sense to you when you code.
Find them and remove them. Replace them with a proper API, something like:
``````int Roll1d6()
{
return Random.Range( 1, 7);
}
``````
A more-advanced API might do something like:
``````int EvaluateDice( string dice)
{
/// parse and roll the dice string
}
``````
where dice could be something complicated like `1d6` or `2d6+4` or `1d20+1d4` or whatever you happen to want.
There are libraries to do this already, but it’s so easy to write, I would just do it yourself. The benefits will begin to accumulate as soon as you start down this path.
Also, if in the future you ever need to debug your code with extremely rare dice rolls, just change that one function to return those numbers in order!
You should study a bit about languages in between Unity. The term “exclusive” means excluding so it is documented as being “less than” the maximum. Languages typically define the range as “exclusive” or “inclusive”. It works as designed.
1 Like
Seriously, you are not sure where they all are? Visual Studio will list them for you and you can jump to each one. And if read and understand AcidArrow’s response. You can wrap any library functionality in a function of your own. Add it to a static class and it is available anywhere. And if you find some other random library you prefer to use you change the implementation in that function and nowhere else.
Can we agree when you posted “it returns the same number” that it does not in fact return the same number?
I can tell you where most of them are, but there is always an outside chance I used it
It returned 5 about twenty times in a row. I don’t know why it started or stopped doing that. It could have been pure chance.
Ah yes, I had a heating system like that once.
Thanks. My current setup works around a similar system to the second one, but probably not as neat.
Random.Range reacts differently whether it is an int or a float so you need to check the doc each time to see if it is max inclusive or exclusive.
If you want some good algorithms there are CMWC and Mersenne Twister, the second one has some kind of copyright though.
I don’t see the point in a long discussion here. It’s simple, if you have d6 then the Range is 1 to 7 and the integer result is {1, 2, 3, 4, 5, 6}.
Range is not a dice simulator, it’s an RNG. So wrap it in a dice class and you’re done.
``````public class Dice
{
int sides = 6;
public int Roll()
{
return Random.Range(1, (sides + 1));
}
}
``````
In mathematical notation
• minInt <= randomInt < maxInt
• minF <= randomF <= maxF
It all depends on what system you have. In games, the range 0-5 (numbers on dice) is referred to as z6, and the range 1-6 (numbers on dice) as d6.
Edit:
Complementing the previous topic. Range probably requires this notation for integers, because the draw is done on a float anyway.
So how do you ensure an equal chance of rolling d6 on a float?
int 1 : {1.0f - 1.999999};
int 2 : {2.0f - 2.999999};
int 3 : {3.0f - 3.999999};
int 4 : {4.0f - 4.999999};
int 5 : {5.0f - 5.999999};
int 6 : {6.0f - 6.999999};
int 7 : {end of range}
If you add the +1 after the result your range is actually going to be 2-6.
When it comes to the int version of Range, the low end is inclusive. The high end is exclusive.
So, 0-6 returns 0,1,2,3,4,5. Thus adding 1 gives you 1,2,3,4,5,6.
1 Like
It is so odd that these types of discussions can continue for decades. If the random number generator that is included with Unity isn’t very random would someone generate data that shows the distribution is not normal? It would pretty much put an end to the discussion and would serve as evidence that could be shared with the Unity dev team.
1 Like
Well, Unity uses an xorshift variant for ages now (actually xorshift128) which is a quite cheap to compute but also quite good PRNG. That information had been shared by andeeeee a (former?) employee at Unity Technologies.
Depending on your exact usecase, if you want / need long term consistency, you may want to use your own implementation. I actually implemented an xorshift64 variant over here. So the internal state is only 64 bits, though for most applications that’s usually good enough.
Random generators have lots of uses but to get a random number when the user interacts you can as well take Time.deltaTime with some kind of modulo.
While this is true, this would not give you a good distribution and would change heavily depending on the users machine. Cryptographically secure random number generators usually “accumulate” entropy from various sources of the system (maybe audio hardware, mouse movement, CPU thread switching, RTC, using a PRNG to read / index RAM or files, …) over time. The difficult thing is usually how to properly map it to the desired range and still aim for a uniform distribution. A lot sources have a strong bias that is difficult to get rid of. So unless you studied those subjects indepth, I would not recommend to implement that yourself. You’re most likely make it worse ^^.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2015-04-24 23:18:27
niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,062
Let P(x) be a quadratic polynomial with real coefficients such that for all real x the relation 2(1+P(x))=P(x-1)+P(x+1) holds.
P(0)=8 and P(2)=32.
If the range of P(x) is [m,∞), then the value of m is?
friendship is tan 90°.
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## #2 2015-04-24 23:31:21
bob bundy
Registered: 2010-06-20
Posts: 8,206
hi Niharika,
You could proceed like this:
Use the 'P' recurrence relationship to work out P(1).
Then call the quadratic ax^2 + bx + c, substitute in the known values when x = 0, 1 and 2 and solve for a, b, and c.
Once you know the quadratic you'll see it has a positive x^2 so it is 'U' shaped rather than the other way up. So it has a minimum value. The minimum for a quadratic is always at -b/(2a) and all you need is the 'y' coordinate at that minimum point.
Hope that helps,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #3 2015-04-25 00:33:22
Olinguito
Member
Registered: 2014-08-12
Posts: 649
straightaway.
Bassaricyon neblina
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## #4 2015-05-01 15:09:12
niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,062
Here is one more question on quadratic equations.
Assume that p is a real number.Find the possible values of p in order to have real solutions for
.
I shifted x^1/3 to RHS and cubed both sides and tried to simplify it to a 2 degree equation so that I could apply D≥0 for real roots, but was unsuccessful in making the equation quadratic.
Pls help.
friendship is tan 90°.
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## #5 2015-05-01 17:07:04
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi;
I tried the substitution x = y^3 and that found the value of the discriminant for p but I am not sure about that idea.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #6 2015-05-01 20:10:06
bob bundy
Registered: 2010-06-20
Posts: 8,206
hi Niharika
I started by having a look at a graph. By using 'a' instead of 'p' you can use the slider to try varying 'a'.
You'll see they all look similar with the intersection moving right as 'a' gets bigger.
Here's the graph when a = 2.
So then I did this:
where y = x^(1/3)
So
We need y ≥ 0 so
That seems to agree with the graphical results.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #7 2015-05-02 03:54:25
niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,062
Well the answer given is p≥ -1/4.
friendship is tan 90°.
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## #8 2015-05-02 04:37:48
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
That is the answer I got but I do not like it.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #9 2015-05-02 05:09:49
bob bundy
Registered: 2010-06-20
Posts: 8,206
That appears to come from the discriminant of the y quadratic. But you still have to find x values from those ys.
If p = -0.25 then y = -0.5 and so x = -0.7937.
In the original equation the LHS = 0.10969 which is not 1. So that is not a solution.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #10 2015-05-02 23:06:06
Olinguito
Member
Registered: 2014-08-12
Posts: 649
bob bundy wrote:
If p = -0.25 then y = -0.5 and so x = -0.7937.
Bassaricyon neblina
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## #11 2015-05-02 23:18:22
Olinguito
Member
Registered: 2014-08-12
Posts: 649
niharika_kumar wrote:
.
Last edited by Olinguito (2015-05-04 00:09:46)
Bassaricyon neblina
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## #12 2015-05-05 00:08:47
bob bundy
Registered: 2010-06-20
Posts: 8,206
hi
Many thanks Olinguito, for finding my error. I had done cube root of y to get x, instead of cube y to get x.
I was still puzzled about why my graph didn't show any negative x values. I experimented with the grapher for some time but couldn't get a reliable graph. As negative values will have a real cube root there should be a graph for negative x. It seems to be the way the grapher has been programmed.
So I used Excel (sorry folks ) to generate values, checked a few by long hand, and when satisfied it was working out ok I made a scatter graph of the points. That's the best graph I think, for this data.
Here's a screen shot of two graphs with p = -0.24 and -0.26 and then the critical one with p = -0.25.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #13 2015-05-05 10:34:53
Olinguito
Member
Registered: 2014-08-12
Posts: 649
You are taking values of x and p independently of each other – but they are not independent of each other! x and p are related by
In other words your <cube root c> minus <cube root a> must be equal to 1 – but as you can see they are not.
Last edited by Olinguito (2015-05-05 14:08:21)
Bassaricyon neblina
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## #14 2015-05-05 10:57:52
Olinguito
Member
Registered: 2014-08-12
Posts: 649
PS:
so maybe this is the graph you want to plot: http://www.wolframalpha.com/input/?i=pl … a=%5E_Real
Last edited by Olinguito (2015-05-05 14:09:18)
Bassaricyon neblina
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## #15 2015-05-05 21:14:24
bob bundy
Registered: 2010-06-20
Posts: 8,206
hi Olinguito
I was deliberately taking x and p as independent; looking at the graph for a particular p, and finding which x (if any) makes the expression come to 1. I explored a whole family of graphs with varying p and could see that as p approached -0.25 from above the graph dropped lower. At p = -0.24 the graph just crosses the line y = 1. At p = -0.25 it just touches y = 1; and at p = -0.26 the curve fails to rise high enough.
I realise that isn't a proof, but I like to get a visual demonstration of a result. Before your proof others were unsure if p = -0.25 was the correct result. I did the graphs to show it is plausible. I'm following the bobbym's mantra to try it with numbers .
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #16 2015-05-06 07:18:55
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
bob bundy wrote:
Before your proof others were unsure if p = -0.25 was the correct result.
I am sorry for my comment without an explanation but I was unsure of the answer I got. For one thing, I do not like the way the problem is worded.
Find the possible values of p in order to have real solutions for
I am bothered by the word "possible."
Synonyms for possible: conceivable, plausible, imaginable, thinkable, believable, likely, potential, probable, credible, tenable, odds-on
Holmes wrote:
I am accustomed to have mystery at one end of my cases, but to have it at both ends is too confusing.
I am wary of problems that have any ambiguity in the phrasing. Anyway, that led me to try to find the real roots of that equation when p = - 1 / 4 or ( - 1 / 4 ) <= p < 0. I was unable to find any real solutions. Wolfram can not either unless you make a choice between principal value of the root or real value of the root. One produces an answer, the other does not.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #17 2015-05-07 01:15:53
niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,062
well I came up with the same solution yesterday as olinguito.
It was quietly easy.
Thanks for helping me out.
friendship is tan 90°.
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## #18 2015-05-07 03:32:11
bob bundy
Registered: 2010-06-20
Posts: 8,206
bobbym wrote:
( - 1 / 4 ) <= p < 0. I was unable to find any real solutions.
Does that mean that my graph for p = -0.24 in post 12 is wrong?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #19 2015-05-07 09:29:58
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi;
I truly do not know, that is why I said I did not like my answer.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #20 2015-05-07 10:23:12
bob bundy
Registered: 2010-06-20
Posts: 8,206
hi bobbym,
If p = -0.24 and x = -0.064
x + 3p + 1 = - 0.064 - 0.72 + 1 = 0.216 so cube root = 0.6
and cube root of x = - 0.4 so LH expression = 0.6 -- 0.4 = 1
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #21 2015-05-07 12:23:13
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi Bob;
It depends on how you interpret the cube root. Wolfram believes there are two interpretations. A principal cube root and a real valued one, just like with square roots. It is my understanding that the default one is the principal cube root. Using that with your p and q will return a complex number and not 0.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #22 2015-05-07 15:06:38
Olinguito
Member
Registered: 2014-08-12
Posts: 649
When you allow complex values, then every real number will have exactly three cube roots: one of them is real and the other two conjugate complex numbers. If you want only real solutions, you should tell Wolfram explicitly – otherwise it will give you its "principal value", which is not always the real root.
Last edited by Olinguito (2015-05-07 15:10:32)
Bassaricyon neblina
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## #23 2015-05-07 15:18:10
Olinguito
Member
Registered: 2014-08-12
Posts: 649
Wolfram always returns a complex number as the principal value of the cube root of a negative real number. Presumably this is the complex number in the first quadrant of the Argand diagram.
Last edited by Olinguito (2015-05-07 15:27:54)
Bassaricyon neblina
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## #24 2015-05-07 20:11:31
Olinguito
Member
Registered: 2014-08-12
Posts: 649
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# HW4 - e-y 2 H n y H m y(c Using these results compute the...
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Physics 731 Assignment #4, due Monday, October 5 1. Determine the (normalized) odd parity eigenfunctions and allowed energies for a particle bound in the finite square well: V = 0 , | x | < a V 0 , | x | > a, in which V 0 > 0 . Discuss the limiting behavior as V 0 0 and V 0 → ∞ . 2. Find accurate numerical values for the bound state energy eigenvalues of a particle in the above finite square well potential, in which ± 2 mV 0 a 2 ¯ h 2 ! 1 2 = 2 . Do this (a) numerically, and (b) graphically (with reasonable precision). 3. (a) Use the Hermite generating function, g ( y,t ) = e - t 2 +2 ty = X n =0 t n n ! H n ( y ) , to prove the following expressions: H n ( y ) = e y 2 / 2 ² y - d dy ³ n e - y 2 / 2 H 0 n ( y ) = 2 nH n - 1 ( y ) H n +1 ( y ) = 2 yH n ( y ) - 2 nH n - 1 ( y ) , (b) and to evaluate Z -∞
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Unformatted text preview: e-y 2 H n ( y ) H m ( y ) . (c) Using these results, compute the matrix elements of X and P between two energy eigenfunctions of the one-dimensional simple harmonic oscillator: h n | X | m i and h n | P | m i . 4. Calculate the probability that a particle in the ground state of the one-dimensional simple harmonic oscillator is farther from the origin than the classical turning points (where E = V ). 5. Evaluate both sides of the uncertainty relation for the n th energy eigenstate of the one-dimensional simple harmonic oscillator. 1...
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## optics
• ### Demo Highlight: Laser and Double Slit
One popular demonstration in our collection for introducing concepts of wave optics is M1-11: Laser Interference: Fixed Double Slits.
Collimated light waves come from the laser and pass through a pair of narrow slits in the slide; the light passes through and then projects on the distant screen. But light travels as an electromagnetic wave, so when the light comes out of the two slits, it forms two wavefronts, just like ripples from two stones dropped in a pond. These two wavefronts can interfere with each other, as we can model with this pair of overlapping concentric circles. Where two peaks or two valleys of the wave pattern line up, they add together, interfering constructively; when a peak and a valley overlap, they cancel out, interfering destructively. The same happens with light waves; the light from the two slits overlaps, and creates a pattern of bright spots (constructive interference) and dark spots (destructive interference).
When using this in class, we can adjust the slide to use different sets of slits, with different slit widths and different spacing between the slits. This is a good opportunity to challenge students to predict how changing these two variables will change the resulting interference pattern.
The spacing between the bright and dark fringes ultimately depends on three things: the distance between the slits and the screen, the wavelength of the light, and the spacing between the two slits. The first is probably obvious from your everyday experience – if you step farther away from the point that light spreads out from, it spreads out more!
Likewise, if you increase the wavelength, the space between the peaks of the waves gets larger, so it’s not a surprise that the spaces between their overlaps would get bigger, too. The effect of the slit spacing, though, takes a moment to think about. If the slits move closer together, the two wavefronts are more and more similar; so the differences between them, the points where they fully cancel out, are farther apart. So increasing the slit spacing decreases the spacing of the fringes, and decreasing the slit spacing increases the spacing of the fringes.
We can see this modeled in a ripple tank simulation here in the Physlet Physics collection at AAPT’s compadre.orghttps://www.compadre.org/Physlets/optics/prob37_7.cfm Use your mouse to measure the positions of the peaks relative to the double slit at the base of the image.
To experiment with this at home, check out this PhET Simulation at the University of Coloradohttps://phet.colorado.edu/sims/cheerpj/quantum-wave-interference/latest/quantum-wave-interference.html
Use the button on the right of the simulation screen to activate the double slit barrier. You will then be able to simulate precisely this experiment!
Light of a particular frequency is released from a source, passes through a double slit, and then is projected on a virtual “screen.” So you can see both the interference pattern as it is formed through space, and the final pattern that you can detect at a distance.
Carrying out the computations for this simulation is processor intensive, so it may run slowly; and as you can see in the image, the resolution is limited. There are limits to how well software can simulate reality!
But at the same time, we can use the simulator to try things we can’t easily do in the laboratory. The slider at the bottom will let you change the laser’s frequency, variable along the full visible spectrum and a bit beyond (fun challenge: is there anything remarkable about the spectrum selection here?).
The sliders at the right will let you change some of the physical parameters of the barrier. You can modify the width of the slits, or the separation distance between the two slits. The “Vertical position” slider lets you adjust the distance between the slits and the screen.
So here we can see the value of combining both real-world experiments and simulations – each alone is useful in learning, and each has its benefits and limitations, but the combination lets us see things we cannot with either one or the other on its own.
• ### Demo Highlight: Polarizers and Light Source
This week we’re exploring the physics of polarized light! We have several demonstrations of polarization in our collection; two of the most popular are perhaps the most straightforward: M7-03, which consists of two polarizing filters (or polarizers) and a light source; and M7-07, which adds a third polarizer.
Light is an electromagnetic wave, made up of oscillating electric and magnetic fields. We call a wave polarized when this oscillation has a particular orientation as the wave travels through space. The direction of the electric field defines the direction of polarization of the wave.
You can see two polarizers in action in this video starring Prof. Manuel Franco Sevilla.
A polarizer like this, also called a polarizing filter, passes only light of a given linear polarization. So it acts as a filter; if the first one is polarized vertically, it will block any horizontally polarized component of the light, and pass only the vertically polarized components. When the two polarizers are in line (which is to say that their axes of polarization are aligned), the second polarizer has very little effect on the light passing through. The first polarizer creates linearly polarized light; the second one, with the same polarization, passes nearly all the light that came through the first one. If we rotate the second polarizer, though, the axis of polarization, the direction in which it requires light to be polarized in order to pass through, rotates. So when the second polarizer is out of line with the first polarizer, it is only passing whatever component of the light from the first polarizer is also in line with the second one. As they rotate farther apart, that component is reduced. Once the two polarizers are fully 90 degrees apart, they no longer have any component in common, so together they pass no light at all! If the first one is polarized entirely vertically, and the second is polarized entirely horizontally, they are perpendicular.
Which is an important aspect of physics that this demo shows: that linearly polarized light can be treated as having separable components, just like we can separate the component vectors of linear motion of an object in space, and a polarizing filter passes only light components parallel to its polarization. So if we add a third polarizer, canted with respect to the other two, it can pass components parallel to its axis of polarization, however we choose to orient that. This can have some interesting results, as we see in the next video, starring Dan Horstman.
The passage of light depends on the orientation of the current wave’s polarization and the filter it encounters – so adding the third filter actually could allow more light to pass!
• ### Demonstration Highlight: Cosmic Ray Detector
Demonstration P4-04, the Cosmic Ray Detector, has recently been upgraded. PhD student Liz Friedman shows it in action in this video.
When energetic particles from space hit the upper atmosphere, they create a cascade of particles in the air around us. Even a single cosmic ray might create a whole shower of particles in our atmosphere. They’re invisible, but with the right apparatus we can detect them. Each of the two dark blocks in this device is a scintillator. When a particle passes through one scintillator, it makes a tiny spark of light, which is picked up by a sensitive photodetector. By checking for correlations between the two paddles, we can spot which sparks are being caused by particles passing straight through from space.
These particles stream around and through us all the time, harmlessly; but it’s pretty amazing that we can build a tabletop device that can measure them!
You can read more about Liz’s adventures in physics, hunting neutrinos in Antarctica, in last summer’s Odyssey magazine https://cmns.umd.edu/news-events/features/4642
• ### Demonstration Highlight: Focusing Heat
Welcome back! This week we’re experimenting with optics, as graduate student Naren Manjunath introduces us to demonstration L3-18, in which we focus light with a parabolic mirror. Check out his video below:
This demonstration uses one of the old overhead transparency projectors that focuses the light by a large parabolic mirror under the platform. These used to be standard issue in every classroom; but in the era of digital presentations, they only see use on special occasions.
The mirror focuses essentially all of the light striking it, both visible light and invisible (to our eyes) infrared light, to a single point. All of the energy carried by those light waves arrives at that point. When we place a piece of paper at that point, it absorbs that energy and heats up rapidly, bursting into flame.
Many old fashioned projectors like this would have a “heat filter” built in, a piece of glass treated to be transparent to visible light but reflect infrared light. This was to prevent exactly this from happening, since you don’t want your materials bursting into flame in the middle of class! Even some recent projectors have had similar filters, although modern light sources (such as LEDs) that produce light only in the visible wavelengths desired make them unnecessary.
• ### Demonstration Highlight: Gravitational Lensing Model
In astronomy, gravitational lensing is the phenomenon whereby gravitational forces around a mass bend light in a way similar to a conventional refracting lens does. When a large mass lies between an observer and the light source they're observing, sometimes that mass can bend the incoming light, causing the source to appear in a different location, or even in multiple locations at once. This can even allow an observer to see a light source that would otherwise be unobservable due to being directly behind another object.
We have a model of this in our collection, as demonstration E1-21, a glass lens that is specially shaped to produce a similar effect to gravitational lensing. Light is bent more the closer it is to the lens' center axis. As a light source moves behind the lens, you can see the source appear to be displaced, or even see one source appear to become several, or become a ring of light around the center of the lens. All of these phenomena can be seen from gravitational lensing in space as well.
In this drawing, you can see a cross section of part of the lens. The changing curvature produces the gravity-like effect of increasing refraction towards the center.
Try experimenting with this simulation https://slowe.github.io/LensToy/ to see it in action in a starfield!
• ### Demonstration Highlight: Laser Waterfall
This week’s Demonstration of the Week is L5-11: The Laser Waterfall. Check it out in this video below, presented by physics student Ela Rockafellow, a leader of our Society of Physics Students.
The laser here is illustrating internal reflection: depending on the index of refraction of the water and the angle the light hits it at, more light can be reflected back and forth within the stream of water than passes through it. At a certain point, it exhibits total internal reflection, where essentially all of the light is traveling along the stream rather than heading straight out the side.
You can experiment with this phenomenon in this TIR simulation from Boston University. You can change the index of refraction of the outside medium (usually about 1 for air) and the interior (about 1.333 for water, around 1.5 for most glass, for example. Try different combinations and see whether the light stays in the stream or passes outside!
This is the same physics that lets fibreoptic cables carry signals over long distances – the light stays within the cable, traveling with very little loss, and allowing us to operate much of modern communications technology.
• ### Demonstration Highlight: More Fun with Polarization
Earlier this year, we took a look at new videos of our popular demonstrations of the polarization of light, demos M7-03 and M7-07. This week, we’re returning to the topic to check out some simulations that let you try this at home!
The first simulation, by Tom Walsh at the oPhysics site, lets you model a wave as it passes through a series of polarizing slits. You can independently adjust the angle of up to three such slit-filters, and see how the resulting wave responds. Experiment with it at https://www.ophysics.com/l3.html.
The second simulation, created by Andrew Duffy and hosted by Boston University, shows a graph of light intensity as it passes through a series of polarizing filters. Again, you can independently vary the angle of each of three filters, and now you can see how this changes the intensity of the light after each. Try it at http://physics.bu.edu/~duffy/HTML5/polarized_light.html.
Speaking of trying things at home, this isn’t a purely academic question – this is how polarized sunglasses cut the glare from sunlight reflecting off the road without preventing you from seeing where you’re going! Try rotating a pair of polarized sunglasses and see how their effect changes with angle. It may look something like the animation below. We do this in the classroom, too – check out demonstration M7-18
• ### Demonstration Highlight: Plane Mirror
Today we’re looking at a simple but powerful demonstration of optics, Demo L2-01: Optical Board and Plane Mirror.
A bright white light source is directed through a baffle with several slits, producing a set of rays. Lenses are used to collimate these rays, and they are then reflected off of a long plane mirror. If the lenses are adjusted such that the incoming rays are approximately parallel, the reflected rays will be as well.
Unlike light diffracted through a lens or prism, reflected light from a surface is unaffected by the frequency of the light. The reflection off of a flat mirror is dependent only on the angle the light strikes at. This demonstration shows that the angle of incidence is equal to the angle of reflection.
You can experiment with this at home – just find any flat mirror (there’s often one on the wall above the sink), shine a light on it with a flashlight or phone, and measure how the reflected light moves when you move the light.
To see this in mathematical detail, check out this simulation at oPhysics: https://www.ophysics.com/l9.html . You can see how light rays from the tips of an object will reflect off a surface. Try dragging the object (a big blue arrow) around to see how the reflection moves. The simulation can also trace out virtual rays – the light rays “behind” the mirror that aren’t really there, but are the geometric extensions of the reflected rays. Following the virtual rays lets you predict where you eye would “see” the reflected image of the object being.
• ### Demonstration Highlight: Reflecting Telescope Models
The most popular design of telescopes for astronomical research is the reflecting telescope. First developed in the 17th century, the typical reflecting telescope uses a curved primary mirror to focus incoming light, and a secondary mirror to direct that light to an eyepiece or sensor. There are many variations on the design, but the underlying principle is the same: light is focused largely by reflection, rather than refraction as in a lens-based Galilean refracting telescope, which allows them to avoid both the chromatic aberration common to lenses and the weight required to create very large ones. Reflecting telescopes have a long history in astronomy and astrophysics, from William and Caroline Herschel to the Hubble Space Telescope and beyond.
We have two models in our collection of how reflecting telescopes work. Demonstration L7-14 models the behaviour of light in a reflecting telescope on our optical board, which uses real optical elements to create a viewable two-dimensional ray diagram.
We also have a static model, demonstration E2-54, which shows the construction of a typical reflecting telescope with strings to represent the paths of light rays through the device. The two are best used in combination to show students how this system lets us observe distant objects.
You can experiment with this at home as well, with this simulation from JavaLab. You can adjust the angle of the incoming light and see how it reflects off the primary mirror and forms an image at the secondary mirror, and use an eyepiece lens to focus it on an observer. Try it out at https://javalab.org/en/newtonian_reflector_en/
• ### Demonstration Highlight: Refraction
Welcome back! This week, we’re taking a lot a three different demonstrations that are very valuable for introducing students to optical refraction. Refraction is the process by which light bends as it passes from one medium to another. Consider demonstration L4-03, with a metal rod sitting in a small tank of water; depending on the angle you view it from, the rod may seem to bend or break at the surface; there may even appear to be more than one of it, if you look in through the corner of the tank! And in demonstration L4-06, you can see a laser beam enter a similarly-sized tank of water, changing direction as it passes through the water’s surface.
The degree to which light bends depends on the difference in the index of refraction of each medium, which relates to the relative speeds of wave transmission in the medium. Prisms, lenses, and many other optical devices rely on carefully chosen angles and indices of refraction to create optical effects.
This public domain animation from Wikipedia does an excellent job of illustrating the effect. As the waves pass from one medium to another, the change in propagation speed induces a change in angle, as seen here.
You can see this happening on a larger scale in demonstration L4-01. Three rays of light are directed through a heavy slab of transparent plastic. The index of refraction of the plastic is much greater than that of air, so the light bends. In the photo below, you can see the beams change direction as they enter and exit the slab.
You can try it for yourself at home with this simulation in the PhET Interactive Simulations Collection: https://phet.colorado.edu/sims/html/bending-light/latest/bending-light_en.html
• ### Demonstration Highlight: Tyndall's Experiment
Today we’re featuring an interesting experiment that explores the classic question of springtime: Why Is The Sky Blue?
Demonstration M7-31: Tyndall’s Experiment uses a chemical reaction to simulate light scattering through Earth’s atmosphere, but in a tiny container. Solutions of sodium thiosulfate and hydrogen chloride are dissolved in water, then mixed together in a glass tank in front of a light source. These chemicals react and form new particles. The resulting particles form a colloid, a liquid with particles suspended throughout, giving it optical properties that let it simulate our much thinner atmosphere on a smaller scale. Over the course of several minutes, as more and more particles form, we see more and more light scattered out the sides of the tank, while less and less passes straight through. As it goes through this transition, the light coming out the end changes from nearly white, to orange, to red, perhaps finally vanishing entirely.
In the atmosphere, light scatters off of molecules and other tiny particles in the air. The angle at which light scatters depends on the wavelength of the light; shorter wavelengths scatter farther than longer ones. Visible light consists of a spectrum of various wavelengths, with blue light having shorter wavelengths and red light having longer wavelengths. This means that blue light scatters more than red.
This property in the atmosphere was written about by physicists John William Strutt, Lord Rayleigh, and is commonly called Rayleigh Scattering as a result; but the experiment to observe similar effects in a liquid was developed by physicist John Tyndall, hence the name of the demonstration. John Tyndall was well known in his time for his interest in education, and gave public lectures with demonstrations, much like we do today! He also carried out important research on the effect of carbon dioxide in the atmosphere, what we now know as the greenhouse effect.
As the Earth turns, the angle we see the sun at changes. Early and late in the day, we’re seeing the sun through the atmosphere at a different angle, and looking through more atmosphere as a result. So at those times we mainly see red light; the shorter blue wavelengths have been scattered away. At midday, with the sun shining directly down on us, we see more blue!
A fascinating bit of physics, and also very beautiful.
• ### Demonstration Highlight: Visible and Invisible Spectra
A recurring favourite optics demonstration in many of our classes is N1-05 Spectra: Visible and Invisible. This seemingly simple setup can show us some important truths about electromagnetic radiation.
A carbon arc lamp is used to create a bright, broad-spectrum white light. This is an example of what is known (confusingly) as blackbody radiation, the light that an object emits due to its temperature. Technically, a hot object radiates light across many frequencies, but what we think of as its “color” is made up of the frequency ranges with the greatest intensity, which depends on the object’s temperature. This we see here a bright blue-white light, with high emission across all the visible frequencies.
Lenses right next to the source focus this light onto a narrow slit, which then passes a narrow beam of light, focused by an additional lens, to a prism. The prism refracts the light at different angles depending on its frequency. So projected onto the wall we will see, rather than a spot of bright white light, a spectrum of all the colors making up the light.
But here’s what’s interesting about this carbon arc lamp. Not all of the light is in that visible range! We have a fluorescent screen, which glows in the visible light range when it absorbs higher-frequency ultraviolet light; using this, we can see that there are bright bands of ultraviolet light off beyond the blue end of the visible spectrum on the wall.
So does this mean there’s something off beyond the red end as well? To check this, we have a thermopile, a horn containing a series of sensors that sense when they get warm. Using this, connected to an audio oscillator that changes pitch when the thermopile senses heat, we can scan across the wall… and indeed, we can hear the pitch change when the horn is in the dark area past the red end of the spectrum. There is infrared light hiding here, frequencies too low for us to see!
You can vary the temperature of your source and see how that changes not only the intensity of light, but its color – or, more accurately, its distribution of color. A light source can radiate light across a broad range of frequencies, which may be centered within, above, or below the range we can see. Try it out for yourself, compare the spectra of a household lightbulb to the Sun or another star, and see if you can guess the temperature of the arc lamp we use here!
• ### Quantum Demonstration and Simulation: The Hydrogen Atom
We love our demonstrations, but there are some things you can’t easily demonstrate in the classroom, either because the physics isn’t compatible with that environment, or because the scale is beyond what we can practically see. This is where simulations can be valuable, in letting us go beyond what we can do on the tabletop and look inside the black boxes.
The quantum nature of the hydrogen atom is a good example. We can demonstrate the emission spectrum of hydrogen with the Balmer Series demonstration P3-51, and we have simple models of electron orbitals for more complex atoms, but how can we look at the structure of the hydrogen atom itself?
Here are some simulations available for looking inside our smallest atom.
• ### Space News: The Nancy Grace Roman Space Telescope
NASA announced last month that their upcoming infrared observatory project had been named the Nancy Grace Roman Space Telescope. Today, we’re going to take a brief look at this project, Dr. Roman, and the role of physics in astronomy.
Nancy Grace Roman (1925-2018) was the first Chief of Astronomy for NASA, and the first woman ever to hold an executive-level position there. A graduate of Western High School in Baltimore, and later of Swarthmore College and the University of Chicago, she began her career in astronomy research specializing in the emission spectra of stars.
Spectroscopy is the study of the spectrum of light, the individual frequencies and colors that make up the light we see. Every element and compound emits its own distinctive pattern of frequencies of light, based on the structure and energy of the electrons within the atom. By analyzing the light from stars, we can use these distinctive patterns as a kind of fingerprint to identify the chemical makeup of distant stars and planets. Our demonstration collection at UMD Physics has many demonstrations about spectra and spectroscopy; be sure to click here and check them out!
Not all light is visible to our eyes; lower frequency light is below the range that we can see, in infrared wavelengths. Space telescopes are often valuable for observing this invisible light without interference from our atmosphere, just as they are in the visible spectrum.
Despite her skill as a researcher, the widespread discrimination against women in the sciences made it difficult for Dr. Roman to advance her career in academia. Eventually, she moved to working in government instead, first joining the Naval Research Laboatory and then NASA. She essentially created the astronomical science program at NASA, plotting its course for decades to come. She was a key player in the development of many of NASA’s research satellites, including Uhuru, the first X-ray astronomy satellite; and she was a leader in the creation of the Hubble Space Telescope. She advocated that NASA science should be for everyone, and ensured that their research and data were publicly available.
Dr. Roman retired from NASA, and began a second career in scientific computing. She learned programming at Montgomery College and went on to work as a contractor specializing in scientific data management, eventually returning to NASA as a contractor to manage the Astronomical Data Center at Goddard Space Flight Center.
Dr. Roman worked hard to inspire more women to become scientists and leaders in science, and we can hope to follow in her footsteps.
• ### STEM News Tip: Raman Spectroscopy for COVID-19 Testing
At the APS March Meeting, Prof Miguel José Yacamán of Northern Arizona University reported on their NSF-funded research to develop a new method of rapid testing for COVID-19 and other coronaviruses by using surface-enhanced Raman spectroscopy.
• ### STEM News Tip: Titan and its Atmosphere on APOD
Today's Astronomy Picture of the Day shows seven different views of Saturn's moon Titan. The surface of Titan is ahrd to see at most visible wavelengths, due to the extreme scattering of light due to particles in the atmosphere. But scattering is wavelength-dependent; anyone who has seen our demonstration M7-31: Tyndall's Experiment can report that you can see very different scattering of light at different frequencies - this is what gives us our rich red sunsets and the brilliant blue sky overhead at noon. To see the surface of Titan, astronomers can use even longer infrared wavelengths, to penetrate the clouds and see what's happening down there.
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Expected Value (EV) Calculations
# Tricks For Calculating Your EV On The Fly
## For some, calculating the EV on the fly is almost second nature, but for most of us, need lots and lots of practice to master those calculations in the midst of the action.
In the beginning, poker (and Hold'em in particular) seemed like an easy game for all of us. We only had to know when to hold’em and when to fold’em right? As time passed though, and as we gathered more and more experience, we suddenly realized Holdem is much more than meets the eye. We learned that Holdem is actually a very easy game to learn and a very hard game to master. To be the best we had to go in-depth, create profitable strategies, think of ways to play profitable both preflop and postflop, put our foes on ranges , and yes, calculate our Expected Value or EV on the go among many other things. So we started searching the Internet and well... here we are!
Don’t worry though, we know what you are going through. All of us have been in your spot. For some, calculating the EV on the fly is almost second nature, but for most of us, we needed lots and lots of practice to master those calculations in the midst of the action. You will probably need practice too, but before hitting the tables, you may be able to jump ahead by reading this piece of poker strategy letting all this valuable info sink in. After that, you can resume your play and start putting into practice what you’ve learned. Are you ready? Let’s begin then.
## The Rule Of 2 And 4
And let’s just start with the easiest mental shortcut you will ever find in poker. Forget about charts and complex formulas, all you ever need for this trick to work is some basic math. The trick is commonly referred to as the Rule of 2 and 4.
To apply the Rule of 2 and 4, you must know your outs . For example, if you have a flush draw, you will have 9 outs to make an actual flush - the total of 13 cards of one suit minus the four cards already on the board and in your hand. If you have an open-ended straight draw, you will have 8 outs. If you have a gutshot, you will have only 4 outs to make a straight. If you have an overcard to the board that can actually turn your hand into the best one, then you will have 3 outs. With all these outs, you can calculate how often you improve your hand in a very simple way.
So let’s say you’re on the flop and you have a flush draw.
You can multiply the number of outs (9) by four and get 36 which is actually a very close approximation of the hand equity.
36% of the time you will make a flush on the next two streets. When you have an open-ender, you will have some 32% (8 x 4) equity while with your sole overcard, you will have approximately 12% (3 x 4).
If you’re on the turn and the flush or straight draw is NOT complete, you can find out what’s your equity by multiplying the number of outs by 2 and after that adding 2 to your result. So with a FD you have around 20% equity - (9 x 2) + 2 - while with a SD you have around 18% equity - (8 x 2) + 2.
Ok that sounds easy enough you may say, but what does this rule have to do with the actual Expected Value? If you know your hand equity and also the pot odds/equity, you can easily find out if a play is +EV or -EV on the spot.
So let’s say you have a flush draw on the flop, you are against one opponent and the pot is \$10 before the flop. Your opponent goes ahead and bets \$10 making the pot \$20. To see the turn, you need to invest \$10 having a 2-to-1 pot odds. In terms of percentages, 2-to-1 is around 33% - basically your \$10 / (\$20 pot + \$10 you are willing to invest). What does this mean? Your hand equity is greater than the pot equity therefore calling is +EV. If you have the nut flush draw, your Ace might actually be good too so you may have 12 outs instead of 9 which would mean around 48% equity, basically a flip and a great spot to get it all in on some occasions.
To summarize this part,
it is +EV to call ONLY when your hand equity is greater than the pot equity, otherwise the call is -EV. Easy as 1,2,3.
## How To Use The Rule Of 2 And 4 Properly
A word of caution though before actually applying this rule at the poker tables. Let’s return to the previous example with the flush draw on the flop and let’s say you did call thinking it was indeed +EV. The turn bricks, the pot is \$30 and villain bets big once more, \$30. This time, you once again get 2-to-1 pot odds or 33% but as you apply the Rule of 2, you realize you have only 20% hand equity so you correctly fold. Yet something is wrong and you can feel it. You actually misinterpreted the Rule of 2 and 4. But how?
If you read the situation carefully, you would realize you actually saw only one street and the Rule of 4 refers to your equity on two streets. Wouldn’t it have been better to use the Rule of 2 instead? Of course it would, it is actually advisable to use more the Rule of 2 rather than the Rule of 4 even if you’re on the flop. And villain’s line confirms it. If you had used the Rule of 2 on the flop, you would have noticed it is actually -EV to call that \$10 as you had 20% NOT 36% hand equity. That’s why you need to use the Rule of 4 more carefully and mostly rely on the Rule of 2, even if you have two more streets to go. Always think of your foe’s line and play before labeling a call +EV. Does he second-barrel often? What does his bet say? Does he have a strong enough hand to continue his aggression on the turn or is he just cbetting only to let the hand go on the next streets?
But what about the implied odds? Wouldn’t the implied odds justify our call? Well, the implied odds could justify almost any call in No Limit Texas Holdem. The reality, though, contradicts this assumption. Truth be told, implied odds are mostly overrated, especially by the poker newbies. Three cards of the same suit on the board can kill the action in many situation so don’t expect getting that much value out of your made flush. On the other hand, hidden straights have much more value and can pay off a well-timed - even slightly -EV - call. And yes, don’t forget about reverse implied odds: with the nut flush draw, you may be thinking you have 12 outs, yet if the opponent has a set, you actually have 7 or 8 good outs and 4 or 5 dirty outs that can only get you in a lot of trouble and lose a lot of money.
## Best Poker Equity Tools and Completing the Puzzle
Easy enough but remember, that’s only the beginning. From this point on, things can only get tougher and it’s only up to you to find the missing links and complete the puzzle.
Yes, poker and Holdem specifically isn’t only about draws and +EV calls. Many more concepts are involved like ranges and fold equity that can make the EV calculations much more difficult. Nevertheless, you can still do it by keeping up your play and thinking about the game away from the tables.
One very efficient way to master EV on the fly is to use the best poker equity tools available regularly like the ‘grandfather’ of them all PokerStove or Equilab. And let’s not forget the poker tracking software like Holdem Manager and Poker Tracker. Start by playing with them, find the most common patterns and memorize them. For example,
Pocket Aces are an 85% favorite versus a random hand preflop, an overpair vs. underpair preflop is 80% dog, while a pocket pair vs. a hand with an overcard and undercard to that pair is best around 70% of the time.
Ace-King versus any pocket pair is basically a flip pre while Ace-King versus a lowe Ace is many times over 70% favorite. Most importantly make a habit of thinking in terms of ranges and hand combinations, NOT sole hands.
If you play tournament poker, try out the best ICM tools out there like ICMIZER and Holdem Resources. Analyze as many hands as possible and learn the ranges to push/fold with in the most common situations. Yes, it will be hard at first, but as you will use the software more and more, you will slowly memorize all those ranges - e.g. find the hands are profitable to shove with (+EV) in almost any shortstack situation and so on.
Remember, you don’t need to calculate your exact EV at the poker tables in order to be profitable, you only have to know if your call/raise/fold has a positive expectation or a negative one. That’s the real secret to be ahead of your competition in any No Limit Holdem game, cash or tournament!
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# calc
Find the exact area of the surface obtained by rotating the curve about the x-axis.
9x = y2 + 36, 4 ≤ x ≤ 8
I keep on getting 4pi (3^(3/2) +1)
so I tried to put the answer in that way and also 24.78pi
But it is wrong
1. 👍
2. 👎
3. 👁
1. 9x = y^2 + 36
I assume you just mean the part above the x-axis, since the curve is symmetric about that axis. So, using discs, the volume is
v = ∫[4,8] πr^2 dx
where r^2 = y^2 = 9x-36
v = π∫[4,8] 9x-36 dx
= π(9/2 x^2 - 36x) [4,8]
= 72π
Using shells, we have
v = ∫[0,6] 2πrh dy
where r = y and h = (8-x) = 8-(y^2+36)/9)
v = 2π∫[0,6] y(8-(y^2+36)/9) dy
2π(2y^2 - y^4/36) [0,6]
= 72π
1. 👍
2. 👎
2. I am just suppose to find the surface area. We are studying arc length
1. 👍
2. 👎
3. Oops. Oh well. The observation about symmetry still holds. The surface area is just the sum of circumferences of many circles of radius y. So,
dS = 2πy ds
S = 2π∫[4,8] y√(1+y'^2) dx
9x = y^2+36, so y = 3√(x-4)
18 = 2yy'
y' = 9/y = 9/√(9x-36)
so, y'^2 = 81/(9x-36) = 9/(x-4)
S = 2π∫[4,8] (3√(x-4))√((x+5)/(x-4)) dx
= 6π∫[4,8] √(x+5) dx
= 4π(x+5)^(3/2) [4,8]
= 79.49 π
Better double-check my math.
1. 👍
2. 👎
4. I redid the problem using polar coordinates, where I set the (0,0,0) at the base of the paraboloid section, so that
z = 4 - (x^2+y^2)/9
That means that we have only to evaluate
S = 1/9 ∫[0,2π]∫[0,6] r√(81+4r^2) dr dθ
and we get a nice tidy 49π
Not sure where I went wrong using x = g(y) and rectangular coordinates. You can see a discussion of the coordinate change at
1. 👍
2. 👎
## Similar Questions
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# 20.2: Standard Entropy
Difficulty Level: At Grade Created by: CK-12
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#### How much energy is available?
As scientists explore energy supplies, geothermal sources look very appealing. The natural geysers that exist in some parts of the world could possibly be harnessed to provide power for many purposes. The change in energy content and the release of energy caused by steam condensing to liquid can help fill some of our growing energy needs.
### Standard Entropy
All molecular motion ceases at absolute zero (0 K). Therefore, the entropy of a pure crystalline substance at absolute zero is defined to be equal to zero. As the temperature of the substance increases, its entropy increases because of an increase in molecular motion. The absolute or standard entropy of substances can be measured. The symbol for entropy is S\begin{align*}S\end{align*} and the standard entropy of a substance is given by the symbol S\begin{align*}S^{\circ}\end{align*}, indicating that the standard entropy is determined under standard conditions. The units for entropy are J/K • mol. Standard entropies for a few substances are shown in Table below:
Substance S∘(J/K⋅mol)\begin{align*}S^\circ(\text{J/K} \cdot \text{mol})\end{align*} H2(g) 131.0 O2(g) 205.0 H2O(l) 69.9 H2O(g) 188.7 C(graphite) 5.69 C(diamond) 2.4
The knowledge of the absolute entropies of substances allows us to calculate the entropy change (ΔS)\begin{align*}(\Delta S^\circ)\end{align*} for a reaction. For example, the entropy change for the vaporization of water can be found as follows:
ΔS=S(H2O(g))S(H2O(l))=188.7 J/Kmol69.9 J/Kmol=118.8 J/Kmol\begin{align*}\Delta S^\circ &=S^\circ(\text{H}_2\text{O}(g)) - S^\circ(\text{H}_2\text{O}(l)) \\ &=188.7 \ \text{J/K} \cdot \text{mol} - 69.9 \ \text{J/K} \cdot \text{mol}=118.8 \ \text{J/K} \cdot \text{mol}\end{align*}
The entropy change for the vaporization of water is positive because the gas state has higher entropy than the liquid state.
In general, the entropy change for a reaction can be determined if the standard entropies of each substance are known. The equation below can be applied.
ΔS=nS(products)nS(reactants)\begin{align*}\Delta S^\circ=\sum nS^\circ(\text{products}) - \sum nS^\circ (\text{reactants})\end{align*}
The standard entropy change is equal to the sum of all the standard entropies of the products minus the sum of all the standard entropies of the reactants. The symbol “n\begin{align*}n\end{align*}” signifies that each entropy must first be multiplied by its coefficient in the balanced equation. The entropy change for the formation of liquid water from gaseous hydrogen and oxygen can be calculated using this equation:
2H2(g)+O2(g)2H2O(l)ΔS=2(69.9)[2(131.0)+1(205.0)]=327 J/Kmol\begin{align*}& 2\text{H}_2(g)+\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \\ & \Delta S^\circ=2(69.9) - [2(131.0)+1(205.0)]=-327 \ \text{J/K} \cdot \text{mol}\end{align*}
The entropy change for this reaction is highly negative because three gaseous molecules are being converted into two liquid molecules. According to the drive towards higher entropy, the formation of water from hydrogen and oxygen is an unfavorable reaction. In this case, the reaction is highly exothermic and the drive towards a decrease in energy allows the reaction to occur.
### Summary
• Calculations of change in entropy using standard entropy are described.
### Review
1. When is the entropy of any material at its lowest?
2. In the reaction involving the formation of water from hydrogen and oxygen, why is the entropy value negative?
3. Why would diamond have a lower standard entropy value than graphite?
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Phys205A_Lecture32
# Physics for Scientists & Engineers with Modern Physics (4th Edition)
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Lecture 32 Midterm 4 Wednesday Nov 19 Topics: Rotation Center of mass Angular momentum Gravitation Chapter 13 Universal Gravitation Homework 10 Chapter 11 (angular momentum) Q 9,13 P 21,29 Chapter 13 Q 2,10 P 33
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Newton’s Law of Universal Gravitation z Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the distance between them z G is the universal gravitational constant and equals 6.673 x 10 -11 N m 2 / kg 2 12 2 g mm FG r =
Law of Gravitation, cont z This is an example of an inverse square law z The magnitude of the force varies as the inverse square of the separation of the particles z The law can also be expressed in vector form z The negative sign indicates an attractive force 12 12 12 2 ˆ mm G r =− Fr r
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Notation z is the force exerted by particle 1 on particle 2 z The negative sign in the vector form of the equation indicates that particle 2 is attracted toward particle 1 z is the force exerted by particle 2 on particle 1 12 F r 21 F r
More About Forces z z The forces form a Newton’s Third Law action-reaction pair z Gravitation is a field force that always exists between two particles, regardless of the medium between them z The force decreases rapidly as distance increases z A consequence of the inverse square law 12 21 =− FF rr
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Gravitational Force Due to a Distribution of Mass z The gravitational force exerted by a finite- size, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center z The force exerted by the Earth on a particle of mass m near the surface of the Earth is 2 E g E Mm FG R =
G vs. g z Always distinguish between G and g z G is the universal gravitational constant z It is the same everywhere z g is the acceleration due to gravity z g = 9.80 m/s 2 at the surface of the Earth z g will vary by location
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Finding g from G z The magnitude of the force acting on an object of mass m in freefall near the Earth’s surface is mg z This can be set equal to the force of universal gravitation acting on the object 2 2 E E E E Mm mg G R M gG R = =
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# If a square region has area x, what is the length of its diagonal in
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If a square region has area x, what is the length of its diagonal in terms of x?
(A) √x
(B) √(2x)
(C) 2√x
(D) x√2
(E) 2x
[Reveal] Spoiler: OA
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Kudos [?]: 264 [0], given: 551
If a square region has area x, what is the length of its diagonal in [#permalink]
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27 Sep 2017, 05:56
Bunuel wrote:
If a square region has area x, what is the length of its diagonal in terms of x?
(A) √x
(B) √(2x)
(C) 2√x
(D) x√2
(E) 2x
Because $$s^2$$ = area of a square:
$$s^2 = x$$
$$s = \sqrt{x}$$
A square's diagonal* is given by $$s\sqrt{2}$$
$$\sqrt{x} * \sqrt{2} = \sqrt{2x}$$
*OR use Pythagorean theorem, where hypotenuse = diagonal = d
$$(\sqrt{x})^2 + (\sqrt{x})^2 = d^2$$
$$2x = d^2$$
$$\sqrt{2x} = \sqrt{d^2}$$
$$d =\sqrt{2x}$$
Kudos [?]: 264 [0], given: 551
If a square region has area x, what is the length of its diagonal in [#permalink] 27 Sep 2017, 05:56
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