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USU Junior Contest October'2003
After a success of the previous Vasechkin’s program that allowed to calculate the results of the elections in cause of two days Artemy Sidorovich was placed at the head of the department. At the
moment Artemy Sidorovich prepares a task for his subordinate — programmer Petechkin. The task is to write a very useful function that would ease the life of all the department programmers. For each
integer from 0 to M the function would calculate how many times this number appears in the N-element array. Artemy Sidorovich deems that the function should work as follows (the sample code for N =
3, M = 1):
│ C │ Pascal │
│ if (arr[0]==0) ++count[0]; │ if arr[0]=0 then count[0] := count[0] + 1; │
│ if (arr[0]==1) ++count[1]; │ if arr[0]=1 then count[1] := count[1] + 1; │
│ if (arr[1]==0) ++count[0]; │ if arr[1]=0 then count[0] := count[0] + 1; │
│ if (arr[1]==1) ++count[1]; │ if arr[1]=1 then count[1] := count[1] + 1; │
│ if (arr[2]==0) ++count[0]; │ if arr[2]=0 then count[0] := count[0] + 1; │
│ if (arr[2]==1) ++count[1]; │ if arr[2]=1 then count[1] := count[1] + 1; │
Artemy Sidorovich wants to estimate the time that Petechkin will need to execute the task. We know that Petechkin needs one second to write a line of the code (he’s fast, isn’t he?). Artemy
Sidorovich doesn’t know exactly bounds for M and N. Your task is to write program that would calculate a number of seconds that Petechkin will write the code.
The only line contains integers N (0 ≤ N ≤ 40000) and M (0 ≤ M ≤ 40000).
Output an amount of seconds that Petechkin needs to write the program.
Problem Author: Den Raskovalov
Problem Source: Open collegiate programming contest for high school children of the Sverdlovsk region, October 11, 2003
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The Weekly Challenge - Perl & Raku
| Day 9 | Day 10 | Day 11 |
The gift is presented by Bruce Gray. Today he is talking about his solution to “The Weekly Challenge - 154”. This is re-produced for Advent Calendar 2022 from the original post by him.
In which we search for a needle in a lendee
(or maybe a chatchka in a haystack),
and delight in some lazy CPAN comfort.
Find all permutations missing from a list.
- I have snipped most of the task permutations to make the code fit better in the blog post.
- I don't want to write my own permutation code again.
- Raku has built-in .permutations method, and (-) set-difference operator.
- Perl has several CPAN modules for permutations; List::Permutor is the first one Google returned, and its ->next method allowed me to write a loop that was different than my Raku solution.
- Python has Sets, and the itertools library handles permutations. itertools has lots of features that I miss when working outside of Raku, but they cannot be combined as nicely as Raku's, due to the basic(underlying(Python(function(call(syntax())))).
The partial list of juggled letters is in @in.
my @in = <PELR PREL ...snip...
Take the first word; we could have used any of them.
With no argument, .comb splits into a list of single characters.
.permutations gives all the possible rearrangements of those characters.
The ». makes a hyper method call that will be run on each item in the list of permuted characters, joining them back into words.
my @all = @in[0].comb.permutations».join;
When we do a set operation on a List, it is automatically converted to a Set.
(-) is the ASCII version of the set difference operator; it returns a Set of items present in the left-hand Set that are absent from the right-hand Set.
Iterating over the resulting Set gives us Pair objects, where the .value is always True, and the .key is the part we are interested in.
say .key for @all (-) @in;
The Perl version of Raku’s Set is a hash,
initialized via my %h = map { $_ => 1 } @stuff;.
use List::Permutor;
my @in = qw<PELR PREL ...snip...>;
my %in_set = map { $_ => 1 } @in;
my $permutor = List::Permutor->new( split '', $in[0] );
while ( my @letters = $permutor->next() ) {
my $word = join '', @letters;
say $word if not $in_set{$word};
The Python code mirrors the Perl solution. When given only a single string, list breaks it into characters.
In hindsight, I really should have named the last variable word instead of s.
from itertools import permutations
input_words = "PELR PREL ...snip...".split()
input_set = set(input_words)
for i in permutations(list(input_words[0])):
s = "".join(i)
if s not in input_set:
Compute the first 10 distinct prime Padovan Numbers.
I don’t want to write my own is_prime() code again.
There were several ways to write the code block. I chose one that highlights $c being requested then deliberately unused.
.squish only suppresses consecutive duplicates, so it works efficiently with lazy lists.
constant @Padovan = 1, 1, 1, { sink $^c; $^a + $^b } ... *;
say @Padovan.grep(&is-prime).squish.head(10);
I was very happy to discover List::Lazy, which make easy both the generation of the Padovan Numbers, and filtering them for primes.
It was going to be more trouble than it was worth to make a version of Raku’s .squish that would work with List::Lazy, so I used the foreknowledge that there is exactly one duplicate, and just called
uniq on the 10+1 numbers returned by ->next.
use List::Util qw<uniq head>;
use List::Lazy qw<lazy_list>;
use ntheory qw<is_prime>;
my $Padovan = lazy_list {
push @$_, $_->[-2] + $_->[-3];
shift @$_;
} [1, 1, 1];
my $prime_pad = $Padovan->grep( sub { is_prime($_) } );
say join ', ', uniq $prime_pad->next(11);
I am pleased with the brevity of the Padovan() generator.
head() is from an itertools recipe; it is not my own code.
Comparing the final print line to the second line of the Raku code makes me wish for Raku’s flexibility of choosing method calls vs function calls.
from sympy import isprime
from itertools import islice
def Padovan():
p = [1, 1, 1]
while True:
p.append(p[-2] + p[-3])
yield p.pop(0)
def squish(a):
last = None
for i in a:
if i != last:
yield i
last = i
def head(n, iterable):
return list(islice(iterable, n))
print(head(10, squish(filter(isprime, Padovan()))))
Primes, Primes, everywhere a Prime.
Making all the Integers, alone or combined
Sieve N, to the square root.
Can't you see its Prime?
If you have any suggestion then please do share with us perlweeklychallenge@yahoo.com.
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Chapter 4 Review | Introduction to Electricity, Magnetism, and Circuits | Textbooks
Chapter 4 Review
Key Terms
amount of charge stored per unit volt
device that stores electrical charge and electrical energy
insulating material used to fill the space between two plates
dielectric breakdown
phenomenon that occurs when an insulator becomes a conductor in a strong electrical field
dielectric constant
factor by which capacitance increases when a dielectric is inserted between the plates of a capacitor
dielectric strength
critical electrical field strength above which molecules in insulator begin to break down and the insulator starts to conduct
energy density
energy stored in a capacitor divided by the volume between the plates
induced electric-dipole moment
dipole moment that a nonpolar molecule may acquire when it is placed in an electrical field
induced electrical field
electrical field in the dielectric due to the presence of induced charges
induced surface charges
charges that occur on a dielectric surface due to its polarization
parallel combination
components in a circuit arranged with one side of each component connected to one side of the circuit and the other sides of the components connected to the other side of the circuit
parallel-plate capacitor
system of two identical parallel conducting plates separated by a distance
series combination
components in a circuit arranged in a row one after the other in a circuit
Key Equations
Capacitance of a parallel-plate capacitor
Capacitance of a vacuum spherical capacitor
Capacitance of a vacuum cylindrical capacitor
Capacitance of a series combination
Capacitance of a parallel combination
Energy density
Energy stored in a capacitor
Capacitance of a capacitor with dielectric
Energy stored in an isolated capacitor with
Dielectric constant
Induced electrical field in a dielectric
4.1 Capacitors and Capacitance
• A capacitor is a device that stores an electrical charge and electrical energy. The amount of charge a vacuum capacitor can store depends on two major factors: the voltage applied and the
capacitor’s physical characteristics, such as its size and geometry.
• The capacitance of a capacitor is a parameter that tells us how much charge can be stored in the capacitor per unit potential difference between its plates. Capacitance of a system of conductors
depends only on the geometry of their arrangement and physical properties of the insulating material that fills the space between the conductors. The unit of capacitance is the farad, where
4.2 Capacitors in Series and in Parallel
• When several capacitors are connected in a series combination, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.
• When several capacitors are connected in a parallel combination, the equivalent capacitance is the sum of the individual capacitances.
• When a network of capacitors contains a combination of series and parallel connections, we identify the series and parallel networks, and compute their equivalent capacitances step by step until
the entire network becomes reduced to one equivalent capacitance.
4.3 Energy Stored in a Capacitor
• Capacitors are used to supply energy to a variety of devices, including defibrillators, microelectronics such as calculators, and flash lamps.
• The energy stored in a capacitor is the work required to charge the capacitor, beginning with no charge on its plates. The energy is stored in the electrical field in the space between the
capacitor plates. It depends on the amount of electrical charge on the plates and on the potential difference between the plates.
• The energy stored in a capacitor network is the sum of the energies stored on individual capacitors in the network. It can be computed as the energy stored in the equivalent capacitor of the
4.4 Capacitor with a Dielectric
• The capacitance of an empty capacitor is increased by a factor of
• Each dielectric material has its specific dielectric constant.
• The energy stored in an empty isolated capacitor is decreased by a factor of
4.5 Molecular Model of a Dielectric
• When a dielectric is inserted between the plates of a capacitor, equal and opposite surface charge is induced on the two faces of the dielectric. The induced surface charge produces an induced
electrical field that opposes the field of the free charge on the capacitor plates.
• The dielectric constant of a material is the ratio of the electrical field in vacuum to the net electrical field in the material. A capacitor filled with dielectric has a larger capacitance than
an empty capacitor.
• The dielectric strength of an insulator represents a critical value of electrical field at which the molecules in an insulating material start to become ionized. When this happens, the material
can conduct and dielectric breakdown is observed.
Answers to Check Your Understanding
4.4 a.
4.5 a.
4.6 a.
4.7 a.
4.9 a.
Conceptual Questions
4.1 Capacitors and Capacitance
1. Does the capacitance of a device depend on the applied voltage? Does the capacitance of a device depend on the charge residing on it?
2. Would you place the plates of a parallel-plate capacitor closer together or farther apart to increase their capacitance?
3. The value of the capacitance is zero if the plates are not charged. True or false?
4. If the plates of a capacitor have different areas, will they acquire the same charge when the capacitor is connected across a battery?
5. Does the capacitance of a spherical capacitor depend on which sphere is charged positively or negatively?
4.2 Capacitors in Series and in Parallel
6. If you wish to store a large amount of charge in a capacitor bank, would you connect capacitors in series or in parallel? Explain.
7. What is the maximum capacitance you can get by connecting three
capacitors? What is the minimum capacitance?
4.3 Energy Stored in a Capacitor
8. If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.
4.4 Capacitor with a Dielectric
9. Discuss what would happen if a conducting slab rather than a dielectric were inserted into the gap between the capacitor plates.
10. Discuss how the energy stored in an empty but charged capacitor changes when a dielectric is inserted if (a) the capacitor is isolated so that its charge does not change; (b) the capacitor
remains connected to a battery so that the potential difference between its plates does not change.
4.5 Molecular Model of a Dielectric
11. Distinguish between dielectric strength and dielectric constant.
12. Water is a good solvent because it has a high dielectric constant. Explain.
13. Water has a high dielectric constant. Explain why it is then not used as a dielectric material in capacitors.
14. Elaborate on why molecules in a dielectric material experience net forces on them in a non-uniform electrical field but not in a uniform field.
15. Explain why the dielectric constant of a substance containing permanent molecular electric dipoles decreases with increasing temperature.
16. Give a reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. How does a dielectric material also allow a greater
voltage to be applied to a capacitor? (The dielectric thus increases
and permits a greater
17. Elaborate on the way in which the polar character of water molecules helps to explain water’s relatively large dielectric constant.
18. Sparks will occur between the plates of an air-filled capacitor at a lower voltage when the air is humid than when it is dry. Discuss why, considering the polar character of water molecules.
4.1 Capacitors and Capacitance
19. What charge is stored in a
capacitor when
is applied to it?
20. Find the charge stored when
is applied to an
21. Calculate the voltage applied to a
capacitor when it holds
of charge.
22. What voltage must be applied to an
capacitor to store
of charge?
23. What capacitance is needed to store
of charge at a voltage of
24. What is the capacitance of a large Van de Graaff generator’s terminal, given that it stores
of charge at a voltage of
25. The plates of an empty parallel-plate capacitor of capacitance
apart. What is the area of each plate?
26. A
vacuum capacitor has a plate area of
What is the separation between its plates?
27. A set of parallel plates has a capacitance of
How much charge must be added to the plates to increase the potential difference between them by
28. Consider Earth to be a spherical conductor of radius
and calculate its capacitance.
29. If the capacitance per unit length of a cylindrical capacitor is
what is the ratio of the radii of the two cylinders?
30. An empty parallel-plate capacitor has a capacitance of
How much charge must leak off its plates before the voltage across them is reduced by
4.2 Capacitors in Series and in Parallel
31. A
is connected in series with an
capacitor and a
potential difference is applied across the pair. (a) What is the charge on each capacitor? (b) What is the voltage across each capacitor?
32. Three capacitors, with capacitances of
respectively, are connected in parallel. A 500-V potential difference is applied across the combination. Determine the voltage across each capacitor and the charge on each capacitor.
33. Find the total capacitance of this combination of series and parallel capacitors shown below.
34. Suppose you need a capacitor bank with a total capacitance of
but you have only
capacitors at your disposal. What is the smallest number of capacitors you could connect together to achieve your goal, and how would you connect them?
35. What total capacitances can you make by connecting a
and a
36. Find the equivalent capacitance of the combination of series and parallel capacitors shown below.
37. Find the net capacitance of the combination of series and parallel capacitors shown below.
38. A
capacitor is charged to a potential difference of
Its terminals are then connected to those of an uncharged
capacitor. Calculate: (a) the original charge on the
capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.
39. A
capacitor and a
capacitor are connected in series across a
potential. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor and the voltage across each
4.3 Energy Stored in a Capacitor
40. How much energy is stored in an
capacitor whose plates are at a potential difference of
41. A capacitor has a charge of
when connected to a
battery. How much energy is stored in this capacitor?
42. How much energy is stored in the electrical field of a metal sphere of radius
that is kept at a
43. (a) What is the energy stored in the
capacitor of a heart defibrillator charged to
? (b) Find the amount of the stored charge.
44. In open-heart surgery, a much smaller amount of energy will defibrillate the heart. (a) What voltage is applied to the
capacitor of a heart defibrillator that stores
of energy? (b) Find the amount of the stored charge.
45. A
capacitor is used in conjunction with a dc motor. How much energy is stored in it when
is applied?
46. Suppose you have a
battery, a
capacitor, and a
capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series. (b) Do the same for a parallel connection.
47. An anxious physicist worries that the two metal shelves of a wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. (a) What is the
capacitance of the empty shelves if they have area
and are
apart? (b) What is the voltage between them if opposite charges of magnitude
are placed on them? (c) To show that this voltage poses a small hazard, calculate the energy stored. (d) The actual shelves have an area
times smaller than these hypothetical shelves. Are his fears justified?
48. A parallel-plate capacitor is made of two square plates
on a side and
apart. The capacitor is connected to a
battery. With the battery still connected, the plates are pulled apart to a separation of
. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating the plates?
49. Suppose that the capacitance of a variable capacitor can be manually changed from
by turning a dial, connected to one set of plates by a shaft, from
With the dial set at
(corresponding to
), the capacitor is connected to a
source. After charging, the capacitor is disconnected from the source, and the dial is turned to
If friction is negligible, how much work is required to turn the dial from
8.4 Capacitor with a Dielectric
50. Show that for a given dielectric material, the maximum energy a parallel-plate capacitor can store is directly proportional to the volume of dielectric.
51. An air-filled capacitor is made from two flat parallel plates
apart. The inside area of each plate is
(a) What is the capacitance of this set of plates? (b) If the region between the plates is filled with a material whose dielectric constant is
what is the new capacitance?
52. A capacitor is made from two concentric spheres, one with radius
the other with radius
(a) What is the capacitance of this set of conductors? (b) If the region between the conductors is filled with a material whose dielectric constant is
what is the capacitance of the system?
53. A parallel-plate capacitor has charge of magnitude
on each plate and capacitance
when there is air between the plates. The plates are separated by
. With the charge on the plates kept constant, a dielectric with
is inserted between the plates, completely filling the volume between the plates. (a) What is the potential difference between the plates of the capacitor, before and after the dielectric has been
inserted? (b) What is the electrical field at the point midway between the plates before and after the dielectric is inserted?
54. Some cell walls in the human body have a layer of negative charge on the inside surface. Suppose that the surface charge densities are
the cell wall is
thick, and the cell wall material has a dielectric constant of
(a) Find the magnitude of the electric field in the wall between two charge layers. (b) Find the potential difference between the inside and the outside of the cell. Which is at higher potential? (c)
A typical cell in the human body has volume
Estimate the total electrical field energy stored in the wall of a cell of this size when assuming that the cell is spherical. (Hint: Calculate the volume of the cell wall.)
55. A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge
leaving the plates. (a) A voltmeter reads
when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads
. What is the dielectric constant of the material? (b) What will the voltmeter read if the dielectric is now pulled away out so it fills only one-third of the space between the plates?
8.5 Molecular Model of a Dielectric
56. Two flat plates containing equal and opposite charges are separated by material
thick with a dielectric constant of
If the electrical field in the dielectric is
what are (a) the charge density on the capacitor plates, and (b) the induced charge density on the surfaces of the dielectric?
57. For a Teflon™-filled, parallel-plate capacitor, the area of the plate is
and the spacing between the plates is
If the capacitor is connected to a
battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.
58. Find the capacitance of a parallel-plate capacitor having plates with a surface area of
and separated by
of Teflon™.
59. (a) What is the capacitance of a parallel-plate capacitor with plates of area
that are separated by
of neoprene rubber? (b) What charge does it hold when
is applied to it?
60. Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electrical field is
When the space is filled with dielectric, the electrical field is
(a) What is the surface charge density on each surface of the dielectric? (b) What is the dielectric constant?
61. The dielectric to be used in a parallel-plate capacitor has a dielectric constant of
and a dielectric strength of
The capacitor has to have a capacitance of
and must be able to withstand a maximum potential difference
What is the minimum area the plates of the capacitor may have?
62. When a
air capacitor is connected to a power supply, the energy stored in the capacitor is
While the capacitor is connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by
(a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?
63. A parallel-plate capacitor has square plates that are
on each side and
apart. The space between the plates is completely filled with two square slabs of dielectric, each
on a side and
thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is
find how much electrical energy can be stored in this capacitor.
Additional Problems
64. A capacitor is made from two flat parallel plates placed
apart. When a charge of
is placed on the plates the potential difference between them is
. (a) What is the capacitance of the plates? (b) What is the area of each plate? (c) What is the charge on the plates when the potential difference between them is
? (d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed
65. An air-filled (empty) parallel-plate capacitor is made from two square plates that are
on each side and
apart. The capacitor is connected to a
battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of
(a) What is the capacitance of this new capacitor? (b) What is the charge on each plate? (c) What is the electrical field between the plates?
66. Suppose that the capacitance of a variable capacitor can be manually changed from
by turning a dial connected to one set of plates by a shaft, from
With the dial set at
(corresponding to
), the capacitor is connected to a
source. After charging, the capacitor is disconnected from the source, and the dial is turned to
(a) What is the charge on the capacitor? (b) What is the voltage across the capacitor when the dial is set to
67. Earth can be considered as a spherical capacitor with two plates, where the negative plate is the surface of Earth and the positive plate is the bottom of the ionosphere, which is located at an
altitude of approximately
The potential difference between Earth’s surface and the ionosphere is about
(a) Calculate the capacitance of this system. (b) Find the total charge on this capacitor. (c) Find the energy stored in this system.
68. A
capacitor and a
capacitor are connected in parallel across a
supply line. (a) Find the charge on each capacitor and voltage across each. (b) The charged capacitors are disconnected from the line and from each other. They are then reconnected to each other with
terminals of unlike sign together. Find the final charge on each capacitor and the voltage across each.
69. Three capacitors having capacitances of
respectively, are connected in series across a
potential difference. (a) What is the charge on the
capacitor? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged
plates connected together. What is the voltage across each capacitor in the parallel combination?
70. A parallel-plate capacitor with capacitance
is charged with a
battery, after which the battery is disconnected. Determine the minimum work required to increase the separation between the plates by a factor of
71. (a) How much energy is stored in the electrical fields in the capacitors (in total) shown below? (b) Is this energy equal to the work done by the
source in charging the capacitors?
72. Three capacitors having capacitances
are connected in series across a
potential difference. (a) What is the total energy stored in all three capacitors? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then
reconnected in parallel with each other with the positively charged plates connected together. What is the total energy now stored in the capacitors?
73. (a) An
capacitor is connected in parallel to another capacitor, producing a total capacitance of
What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
74. (a) On a particular day, it takes
of electrical energy to start a truck’s engine. Calculate the capacitance of a capacitor that could store that amount of energy at
(b) What is unreasonable about this result? (c) Which assumptions are responsible?
75. (a) A certain parallel-plate capacitor has plates of area
separated by
of nylon, and stores
of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent?
76. A prankster applies
to an
capacitor and then tosses it to an unsuspecting victim. The victim’s finger is burned by the discharge of the capacitor through
of flesh. Estimate, what is the temperature increase of the flesh? Is it reasonable to assume that no thermodynamic phase change happened?
Challenge Problems
77. A spherical capacitor is formed from two concentric spherical conducting spheres separated by vacuum. The inner sphere has radius
and the outer sphere has radius
A potential difference of
is applied to the capacitor. (a) What is the capacitance of the capacitor? (b) What is the magnitude of the electrical field at
just outside the inner sphere? (c) What is the magnitude of the electrical field at
just inside the outer sphere? (d) For a parallel-plate capacitor the electrical field is uniform in the region between the plates, except near the edges of the plates. Is this also true for a
spherical capacitor?
78. The network of capacitors shown below are all uncharged when a
potential is applied between points
with the switch
open. (a) What is the potential difference
? (b) What is the potential at point
after the switch is closed? (c) How much charge flows through the switch after it is closed?
79. Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for
fraction of a second with an average light power output of
(a) If the conversion of electrical energy to light is
efficient (because the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of
when the stored energy equals the value stored in part (a). What is the capacitance?
80. A spherical capacitor is formed from two concentric spherical conducting shells separated by a vacuum. The inner sphere has radius
and the outer sphere has radius
A potential difference of
is applied to the capacitor. (a) What is the energy density at
just outside the inner sphere? (b) What is the energy density at
just inside the outer sphere? (c) For the parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for the
spherical capacitor?
81. A metal plate of thickness
is held in place between two capacitor plates by plastic pegs, as shown below. The effect of the pegs on the capacitance is negligible. The area of each capacitor plate and the area of the top and
bottom surfaces of the inserted plate are all
What is the capacitance of this system?
82. A parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is
and separation between plates is
show that the capacitance is given by
83. A parallel-plate capacitor is filled with two dielectrics, as shown below. Show that the capacitance is given by
84. A capacitor has parallel plates of area
separated by
The space between the plates is filled with polystyrene. (a) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown. (b) When the voltage equals the value found in
part (a), find the surface charge density on the surface of the dielectric.
Candela Citations
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Clustering – The Dan MacKinlay stable of variably-well-consider’d enterprises
May 23, 2015 — June 7, 2016
feature construction
sparser than thou
Getting a bunch of data points and approximating them (in some sense) by their membership (possibly fuzzy) in some groups or regions of feature space. Quantization, in other words.
For certain definitions, this can be the same thing as non-negative and/or low-rank matrix factorisations if you use mixture models, and is only really different in emphasis from dimensionality
reduction. If you start with a list of features and then think about “distances” between observations, you have just implicitly intuited a weighted graph from your hitherto non-graphy data and are
now looking at a networks problem.
If you care about clustering as such, spectral clustering feels like a nice entry point, maybe via Chris Ding’s tutorial on spectral clustering.
• CONCOR induces a cute similarity measure.
• MCL: Markov Cluster Algorithm, a fast and scalable unsupervised cluster algorithm for graphs (also known as networks) based on simulation of (stochastic) flow in graphs.
There are many useful tricks in here, e.g. Belkin and Niyogi (2003) shows how to use a graph Laplacian (possibly a contrived or arbitrary one) to construct “natural” Euclidean coordinates for your
data, such that nodes that have much traffic between them in the Laplacian representation have a small Euclidean distance (The “Urban Traffic Planner Fantasy Transformation”) Quickly gives you a
similarity measure on non-Euclidean data. Questions: Under which metrics is it equivalent to multidimensional scaling? Is it worthwhile going the other way and constructing density estimates from
induced flow graphs?
1 Clustering as matrix factorization
If I know me, I might be looking at this page trying to remember which papers situate k-means-type clustering in matrix factorization literature.
The single-serve paper doing that is Bauckhage (2015), but there are broader versions (Singh and Gordon 2008; Türkmen 2015), some computer science connections in Mixon, Villar, and Ward (2016), and
an older one in Zass and Shashua (2005).
Further things I might discuss here are the graph-flow/Laplacian notions of clustering and the density/centroids approach. I will discuss that under mixture models
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Dice efficiency in Ashes Reborn
Often times when people ask about how to win in Ashes Reborn, experienced players will tell them, “Use your dice more efficiently than your opponent.” However, because units and spells can exist on
the board from round to round, it can be difficult to easily identify what constitutes an efficient use of dice.
Understanding dice efficiency is further complicated by the fact that you have to consider outcomes vis-à-vis your opponent; whenever you’re talking about dice efficiency, it’s with relation to how
your opponent has spent their dice.
Lastly, before I get into the nitty gritty of evaluating outcomes for their dice efficiency, I want to also mention that while efficiently dealing damage is extremely important, it’s not the only
thing that will win games. Smart sequencing, tempo plays, and gaining higher utility from your cards compared to your opponent all play a part, as well. This is simply one piece of the puzzle.
Damage vs. utility
Cards in Ashes tend to do one (or more) of three things:
1. Deal damage (attack values on units, direct damage)
2. Prevent or mitigate taking damage (life value on units, healing, destruction effects)
3. Offer a utility effect (adjust dice, manipulate exhaustion, etc.)
There are a lot of different utility effects, and they can be very difficult to evaluate from the standpoint of dice efficiency. Utility effects often show their worth through play, and sometimes
only if you use the correct play line (or situation) for them. As a result, while I can help coach you through evaluating dice efficiency, learning which utility effects you need and prefer will
require playing the game. Smart deck building and play can allow you to use cards the community generally considers inefficient to great effect by compounding utility effects.
(Incidentally, if you’re ever wondering why “decks full of units” are so popular in Ashes: it’s because units often do all three of the things above! They deal damage by attacking or countering,
prevent damage to your Phoenixborn by blocking or encouraging your opponent to attack them, and usually have some utility effect.)
With that out of the way, let’s take a look at the starting point for evaluating your damage-to-dice efficiency!
Base damage output
The starting point for calculating dice efficiency is to look at your base damage output. This is situational, but at the most simplistic you can boil it down to “how many wounds—or wound
equivalents—does this card cause compared to how many dice it costs?” For instance:
• Frost Bite is a Ready Spell that deals 1 damage for 1 die; this is a 1-to-1 ratio
• Final Cry is a spell that deals 2 damage to your opponent for 1 die; this is a 2-to-1 ratio
This actually illustrates the full range of base damage output in Ashes! (Some cards have ratios below 1-to-1, but they typically have some utility effect that complicates calculating their actual
There are also some cards like Summon Frostback Bear that have a “book tax”: a play cost that is effectively amortized across the total number of conjurations you summon all game. In this instance,
if you only summon 1 Frostback Bear, it costs 3 dice (2 damage to 3 dice). But if you summon two, they effectively cost 2.5 dice each, and so on. Since the book tax typically only impacts your First
Five, most people round it to zero for subsequent summons—so a Frostback Bear effectively costs 2 dice, for your standard base damage output of 1 damage to 1 die.
However, base damage output is merely a starting place! To calculate your actual dice efficiency, you have to look at outcomes.
Calculating dice efficiency through outcomes
To calculate the dice efficiency of a card, you need to consider its total outcome: that is, how many wounds it dealt and was dealt until it was destroyed. Note that there’s a difference between
wounds and damage in Ashes! Base damage is how much damage the unit is capable of outputting in a simple attack to the Phoenixborn compared to how much dice you spent. Dice efficiency is more about
how many wounds the unit actually places, though.
For instance, say I summon a Hammer Knight. Its base damage-to-dice ratio is 3-to-3. However, if you respond by playing Sword of Virtue to destroy my Hammer Knight before I have a chance to attack
with it, then I have spent 3 dice to deal 0 wounds, and you have spent 2 dice to effectively deal 4 wounds (since that’s how much damage the Hammer Knight would normally take to destroy).
That scenario is pretty easy to intuit the efficiency (“I spent 3 dice, you spent 2 dice, and we’re back where we started, so you were more efficient.”). Things start to get complicated when both
players are dealing damage, however.
For a second scenario, say I have a Hammer Knight, and you have a ready Frostback Bear and an exhausted Mist Spirit that attacked on a previous turn (this is the first round, so the Frostback Bear
costs 3, including the book tax). I attack the Frostback Bear and deal it 3 damage to destroy it, while it deals 2 counter damage back. I then use the Hammer Knight’s Aftershock ability to deal 1
damage to the Mist Spirit. In this instance, I have spent 3 dice for 4 wounds, while you have spent 4 dice for 3 wounds (two from the Bear’s counter, and 1 from the initial attack from the Mist
Spirit). My efficiency is slightly better, but more importantly we are not done with the Hammer Knight’s outcome, because the Hammer Knight is still in play. For instance, you might use Aradel’s
Water Blast ability to deal 2 more damage to the Knight, killing it. That makes the final outcome 4 wounds to 3 dice (1.33) for me and 5 wounds for 4 dice for you (1.25): my efficiency was slightly
better, because I have a slightly higher ratio. If you subtract the two ratios, you end up with 0.08; so you could say that in that exchange I was ahead by about a tenth of a wound.
The reason that knights are so popular, however, is because that minor efficiency improvement is usually the floor for Knights (barring hard removal, as described above). If you don’t have Water
Blast (or an equivalent way to kill the Knight) and the round ends, then the outcome is a lot worse for you because the Knight’s recover 2 value clears off your two wounds and I get to use my Knight
For argument’s sake, let’s say that happens and you attack the Knight with a Frostback Bear, then use Water Blast to kill it. At this point, my efficiency is 7 wounds for 3 dice (4 in first round, 3
in counter damage to kill the Bear this round), or 2.33. Your dice efficiency is 5 wounds for 7 dice (1 in first round from Mist Spirit, since the Bear’s wounds were wiped out by Recovery; then 4
from the Bear and Water Blast this round), or 0.71. Subtract those two numbers and you get 1.62: I was ahead by over one and a half wounds! That sort of thing adds up, because there’s only so many
wounds you can soak up with your dice (and available conjurations or units from hand) before I start converting that damage into damage on your Phoenixborn.
Messing with your opponent’s outcomes
In the examples above, all damage being dealt was the perfect amount to destroy a unit (or not). But much of the time that won’t be the case. You can increase your dice efficiency by ensuring that
your units output as close to their full damage as possible, while your opponent’s units waste their potential damage output.
For instance, if I play a Hammer Knight, and you attack it with two Shadow Spirits (across subsequent turns), my efficiency is 2 wounds for 3 dice (each Shadow Spirit only has 1 life and the Hammer
Knight remains ready after countering)—0.67—while your efficiency is 4 wounds for 2 dice—2. That’s a difference of 1.33 wounds in your favor!
By making smart choices about which units to block or guard and which to attack, you can maximize the wound output from your units and minimize the output from your opponent’s units to increase your
relative dice efficiency.
Why dice efficiency matters, even when it’s from wounds dealt to units
Ultimately, the only damage that matters is damage dealt to your opponent’s Phoenixborn, but because dice, cards, and the number of units available to you are finite resources, considering the dice
efficiency with which your deck can handle various scenarios is important. There is an opportunity cost to playing and attacking with units, which is one of the reasons Alert knights are so played so
widely. Although they have a relatively high dice cost, they make up for it by potentially killing off a bunch of your opponent’s units (and, in severely disadvantageous matchups, ultimately swinging
to face, as well). Additionally, decks can only put so much attack, damage, and life on the board each round, and efficiently dealing with what your opponent has played can allow you to build up very
big dice efficiency differentials simply when your units persist to a new round and swing again.
Practical applications for calculating dice efficiency
Exactly calculating your dice efficiency in the middle of a game of Ashes like I’ve done in the examples above isn’t a useful endeavor. However, considering dice efficiency can be very important
during deck construction, before games when you know your opponent’s list (to determine your ideal play lines), and after games (to understand where your play lines or deck building choices might
need to change to improve your outcomes).
For instance, at the time of this writing I just finished playing a Noah deck in the 2021 Shufflebus 5 tournament which fielded mostly several 2/1 conjurations that cost 1 die each. Doing some simple
efficiency outcome calculations, I can determine what books are optimal to lock down with Noah’s Shadow Target ability to avoid inefficient trades. I faced a deck that was running Summon Turtle Guard
and Summon Ruby Cobra. From an efficiency standpoint:
• If they attack a 2/1 with the Ruby Cobra, they spend 1 die for 1 damage (and a mill, which is a utility effect that is difficult to value under this framework) and I spend 1 die for 2 damage. We
basically break even there, so there’s not much reason to worry about locking down that book, and if I attack it I kill it and leave an exhausted 2/1 unit behind (which requires them to expend
more resources to destroy, improving my efficiency).
• Turtle Guard is less simple, because it has Recover 1 and is effectively immune to damage while exhausted. So they play it for 2 dice (1 for the book tax), I attack with a 2/1, then the round
turns over and I attack it with a second 2/1 to kill it. That’s 3 wounds for 2 dice for me (1.33) and 2 wounds for 2 dice for them (1). On paper that looks to be slightly in my favor, but because
Turtle Guard has Unit Guard that means I don’t get to decide where the damage goes (they effectively get 2 free guard actions, which is a big deal if I need to efficiently deal damage).
Subsequent Turtle Guards only cost 1 die, too, so the same pattern repeated in the second round would mean their efficiency is 2 wounds for 1 die, and mine would be 3 for 2 (plus all the same
efficiency costs). That means locking down Turtle Guard with Noah’s ability was a high priority for me.
Those particular examples are kind of obvious, but hopefully illustrate the concept. You can also consider the opposite: why did my opponent choose those two books?
• Ruby Cobra is a 1/2 unit on attack, or a 0/2 unit on defense. For decks that aren’t running 2/1 units (my deck is an outlier in that regard; in the online meta as of this writing it’s an unusual
statline to see), that means that a likely outcome for the Cobra is to deal 1 wound for 1 die (a low baseline damage, but consider it also has a utility effect), but then require 1-2 dice spent
by your opponent to kill it (or it might soak up a Knight swing, causing them to waste a potential wound).
• Turtle Guard is a 2/3 unit that can’t attack, which effectively costs 1 die (disregarding the book tax). That makes it very difficult to kill by anything except Knights, with whom it trades
beautifully (it deals damage equal to half a typical Knight’s health, making it much easier to efficiently kill the Knight).
These less specific “good enough” calculations are typically how most players think about dice efficiency. Tracing specific, full outcomes is often too difficult, very specific to individual
match-ups, and is complicated by the fact that dice efficiency is a constantly evolving thing; in a way, the true “outcome” would have to be tracking efficiency from the very start to the very end of
the game, because it’s very common for highly efficient outcomes to be turned on their head (for instance, perhaps I efficiently kill a Hammer Knight with my 2/1 units only to a have my opponent play
a second Hammer Knight that wrecks me with Aftershock damage and survives to the next round). Examining specific outcomes can hopefully help lead to a more general understanding of efficiency,
Dice efficiency isn’t everything
I mentioned it earlier, but it bears re-iteration: dice efficiency isn’t everything! Simply collecting all the most efficient units in a single deck won’t necessarily win you games; timing, smart
play, and exploiting utility effects that work well together are all incredibly important parts of Ashes, as well. However, gaining an understanding of what constitutes dice efficiency will
definitely help improve your ability to construct decks and make smart choices in game, so it’s worth thinking about.
Good luck and have fun!
2 responses to “Dice efficiency in Ashes Reborn”
1. Dear, sharp articles here! thanks for it.
About this one, all crystal clear until you introduced values like 2/1, 0/1, 2/3. What does it mean?
Btw, wondering if you’d like to write on Red Rains solo-coop mode soon…
Best ashes!
□ Ack, I completely missed that you’d posted this comment! Many apologies. Values like “2/1” mean the unit has 2 attack and 1 life.
Leave a response
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TANCET 2014 DS 70
TANCET 2014 DS 70: Linear Equations
Directions for Data Sufficiency Question
The question is followed by two statements labeled (1) and (2) in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question.
Using the data given in the problem plus your knowledge of mathematics and everyday facts, choose the answer as:
1. Choice 1 if statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
2. Choice 2 if statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
3. Choice 3 if both the statements (1) and (2) TOGETHER are sufficient, but NEITHER statement alone is sufficient.
4. Choice 4 if each statement ALONE is sufficient.
5. Choice 5 if statements (1) and (2) TOGETHER are not sufficient, and additional data is needed.
What is the sum of x, y and z?
1. Statement 1: 2x + y + 3z = 45
2. Statement 2: x + 2y = 30
Correct Answer Choice (3). Statements (1) and (2) TOGETHER are sufficient.
Explanatory Answer - step by step
• What should we know from the Question Stem?
Before evaluating the two statements, answer the following questions to get clarity on when the data is sufficient.
What kind of an answer will the question fetch?
The question is "What is the sum of x, y and z?"
The answer to the question should be a number. For e.g., 65.
When is the data sufficient?
If we are able to come up with a UNIQUE value for the sum of x, y, and z, the data is sufficient.
If we are not able to come up with a UNIQUE value for (x + y + z), the data is NOT sufficient.
What is the approach?
Use the statements independently first and check whether by multiplying or dividing the equation by a number will result in (x + y + z). If it does, data is sufficient.
Else combine the two statements. Try and add, subtract or manipulate the equations in some way to arrive at x + y + z. If you could find a unique value data is sufficient. Else, data is NOT
• Statement (1) ALONE
2x + y + 3z = 45
We will not be able to get the value of x + y + z with the above equation.
Statement (1) ALONE is NOT sufficient.
If statement (1) ALONE is NOT sufficient, we can eliminate choices 1 and 4.
Choices narrow down to 2, 3, or 5.
• Statement (2) ALONE
x + 2y = 30.
At least the first statement had all 3 variables in it. The second statement has only 2 variables in it.
No, we will not be able to find the value of x + y + z using statement 2
Statement (2) ALONE is NOT sufficient.
If statement (2) ALONE is NOT sufficient, we can eliminate choice 2 as well.
Choices narrow down to 3, or 5.
• Statements Together
2x + y + 3z = 45
x + 2y = 30
Though at first sight it appears as if we have only two equations and 3 variables, a bit of tweaking is likely to give us the answer.
Add the two equations. {2x + y + 3z = 45} + {x + 2y = 30}
= 3x + 3y + 3z = 75.
If 3x + 3y + 3z = 75, {x + y + z} = 25.
Using statements (1) and (2) together we were able to determine a unique value for the sum of x, y and z.
Hence, choice (3) is the answer.
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Using Transformations to Determine Similarity
Question Video: Using Transformations to Determine Similarity Mathematics
The triangle π ΄π ΅π Ά has been transformed onto triangle π ΄β ²π ΅β ²π Άβ ², which has then been transformed onto triangle π ΄β ³π ΅β ³π Άβ ³. Describe the single transformation that maps π
΄π ΅π Ά onto π ΄β ²π ΅β ²π Άβ ². Describe the single transformation that maps π ΄β ²π ΅β ²π Άβ ² onto π ΄β ³π ΅β ³π Άβ ³. Hence, are triangles π ΄π ΅π Ά and π ΄β ³π ΅β ³π Άβ ³
Video Transcript
The triangle π ΄π ΅π Ά has been transformed onto triangle π ΄ prime π ΅ prime π Ά prime, which has then been transformed onto triangle π ΄ double prime π ΅ double prime π Ά double prime.
Describe the single transformation that maps π ΄π ΅π Ά onto π ΄ prime π ΅ prime π Ά prime. Describe the single transformation that maps π ΄ prime π ΅ prime π Ά prime onto π ΄ double prime π
΅ double prime π Ά double prime. Hence, are triangles π ΄π ΅π Ά and π ΄ double prime π ΅ double prime π Ά double prime similar?
Weβ ll begin with the first part of this question. That asks us to describe the single transformation that maps π ΄π ΅π Ά onto π ΄ prime π ΅ prime π Ά prime. π ΄π ΅π Ά is this small
triangle at the center of our drawing. Then, π ΄ prime π ΅ prime π Ά prime is the larger one that sits around it. Now, we need to be really careful when describing the transformation. The keyword
in this question is the word β single.β Weβ re going to describe exactly one transformation rather than a series of these.
So, letβ s recall the transformations that we need to know. These are rotations, reflections, translations, and dilations. We recall that a rotation, the key is the letter T here, turns the shape.
We have reflections, and the β flβ in this word remind us that we flip the shape. When we translate, the β slβ reminds us to slide the shape. And finally, the dilation, sometimes called an
enlargement, makes a shape larger, thatβ s the l, or smaller.
So, letβ s see whatβ s happened to transform π ΄π ΅π Ά onto π ΄ prime π ΅ prime π Ά prime. We should quite quickly notice that the image π ΄ prime π ΅ prime π Ά prime is larger than the
original shape. Thatβ s an indication to us that π ΄π ΅π Ά has been dilated. Thatβ s not enough, though. Weβ re going to need to describe two more things. We need to give a center of
enlargement or dilation, and we need to give a scale factor. To find a scale factor for enlargement, we divide a dimension on the new shape by the corresponding dimension on the old shape.
And so, to find the scale factor of our dilation or enlargement, weβ re going to divide the length of the line segment π ΄ prime π ΅ prime by the length of the line segment π ΄π ΅. Line segment
π ΄ prime π ΅ prime is 12 units long, whereas the line segment π ΄π ΅ is four. So, the scale factor of enlargement or dilation, which Iβ ve shortened to s.f., is 12 divided by four, which is
equal to three.
So, we have the type of transformation and the scale factor. We need to decide where the center of enlargement lies. To do this, we join each corresponding vertex by a ray. So, for example, we can
join vertex π ΄ prime and π ΄. Similarly, weβ ll join vertex π Ά prime and π Ά. And then, we join the vertices π ΅ prime and π ΅. The point at which these rays meet is the center of
enlargement or dilation. And we can see that happens at point π ·. And so, the single transformation that maps triangle π ΄π ΅π Ά onto π ΄ prime π ΅ prime π Ά prime is a dilation from point π
· by a scale factor of three.
We now move on to question two. And that says, describe the single transformation that maps π ΄ prime π ΅ prime π Ά prime onto π ΄ double prime π ΅ double prime π Ά double prime. Weβ ve
already identified that π ΄ prime π ΅ prime π Ά prime is the enlargement of π ΄π ΅π Ά. And π ΄ double prime π ΅ double prime π Ά double prime is this triangle here. So, how have we got from
π ΄ prime π ΅ prime π Ά prime onto its image?
It appears that these triangles are the same size. And so we can disregard the dilation at this point. A translation involves a slide of the shape. When we slide the shape, they end up in the same
orientation. And these ones are clearly not in the same orientation; one appears to be upside down. So, weβ ll disregard the translation. So, we have two options. We have a reflection, thatβ s a
flip of the shape, and a rotation, thatβ s a turn.
Well, in fact, we see that if we turn or rotate π ΄ prime π ΅ prime π Ά prime, we end up in the same orientation as π ΄ double prime π ΅ double prime π Ά double prime. And so, the shape has
been rotated. There are two more things we need to decide. We need to decide the angle of rotation and the point about which the shape is rotated. If we clear the annotations off of our diagram, we
see thereβ s only really one point about which the shape can have rotated. Thatβ s the point π ·.
Then, weβ re going to join one of our vertices to this point. Weβ ll begin by joining π ΄ prime to point π ·. And then, weβ re going to join the rotated vertex, thatβ s π ΄ double prime, to
point π ·. We can now see that this line has rotated 180 degrees. And we can, therefore, say that the single transformation that maps π ΄ prime π ΅ prime π Ά prime onto π ΄ double prime π ΅
double prime π Ά double prime is a rotation by 180 degrees about point π ·.
Finally, weβ re asked, hence are triangles π ΄π ΅π Ά and π ΄ double prime π ΅ double prime π Ά double prime similar? The word β henceβ indicates to us that we need to use what weβ ve just
done. And so, what weβ re really asking is if we start off with a shape, then dilate it, and then rotate it, will we end up with two shapes that are similar?
Well, for two shapes to be similar, one must be an enlargement or a dilation of the other. We initially showed that π ΄ prime π ΅ prime π Ά prime is a dilation of π ΄π ΅π Ά. So, we can
certainly say that π ΄π ΅π Ά and π ΄ prime π ΅ prime π Ά prime are similar.
But what about when we rotate a shape? Well, when we rotate a shape, it does change the orientation, but it otherwise does not change the size of that shape. And we can, therefore, say that π ΄
prime π ΅ prime π Ά prime and π ΄ double prime π ΅ double prime π Ά double prime must be congruent. Theyβ re actually exactly the same. And so, if π ΄π ΅π Ά is similar to π ΄ prime π ΅
prime π Ά prime, but actually π ΄ prime π ΅ prime π Ά prime is congruent to π ΄ double prime π ΅ double prime π Ά double prime, then this in turn means that π ΄π ΅π Ά and π ΄ double prime
π ΅ double prime π Ά double prime must in fact be similar. And so, the answer is yes.
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A Game of Wacamole in my classroom - Riley Math Education
On Wednesday, I had a double period with one 6th grade class and I got lazy. I decided to let them work in their Eureka Workbooks on a lesson about solving Problems with Percents. I decided to let
them work at their tables since they would be doing the lesson in their workbooks. They began working and I began circulating through the room following my normal route and not chasing hands.
They stayed at their tables in groups. Even though the lesson claimed to be about solving real-world problems, it was still abstract and didn’t give much opportunity for creativity in the math.
Within a very few minutes, it became obvious that they did not have a conceptual understanding of “percent” and also did not feel the need to think very much since they had the workbooks.
It became a game of Wacamole. I would arrive at a table and spend a minute getting them back on task and try to guide them without teaching too much. Then I would move to the next group and notice
that I was leaving frustrated and confused students in my wake.
Most of them raised their hands for help at first. When I didn’t get there quick enough they got off task and started fooling around. I began moving faster from group to group creating even more
frustration and probably wearing ruts in the carpeting. The room became louder and more off task. I actually resorted to writing names on the board and threatening loss of recess.
Then I stopped and reflected. These were the same kids who worked so well the previous day when I gave them better activities and had them work at white boards. I realized it was not them and it was
not me, It was the lesson.
I stopped the class, got their attention (which took some effort) and had a heart-to-heart with them. I apologized for the way the class had gone. I congratulated them for being excellent students
and explained that I had given them a bad task that did not allow them to explore the concept of percent. These kids needed a more rich and concrete task.
Percent lessons are notorious for being formulaic and rote. It is a very simple small idea that is used for convenience, but it has opportunity to get them thinking deeply about what is meant by
“percent”. (Literally “Per One Hundred”)
I did not use this task the next period. I heard one of students talking about black holes from their science class. I decided to flex my Physics muscles and have a class discussion. We talked about
the nature of black holes which evolved into the nature of reality. One of them mentioned that we all see colors differently. This led to the proposition that perhaps we all have the same favorite
color but we describe it differently. We never touched on percent, but I did get to stop them at the end and reveal to them that the class had been a math discussion and by learning more math, they
would be able to have a better understanding of such abstract concepts. They left smiling. I took a nap.
Two Days Later............
After I wrote this, I realized I did something that is a pet peeve of mine. I presented a problem without suggesting a real solution. So I needed to put together some activities that do a good job of
getting kids thinking about percent. Here is what I did with the same class two day’s later:
The image at right is from the Mindset Mathematics Series Grade 5.
I have used this with fifth-grade students for estimation. For This activity, I wanted something less challenging so they could get a more accurate percent estimate in less time.
I posted this on the screen and asked:
"What percentage of this parking lot is full?"
They asked me:
How many cars are there: I Don’t Know?
How many parking spots are there: I Don’t Know?
Does it matter what color they are? No
Do we include the parking spots on the sides? Up to you
What about the car driving on the road? Up to you
The kids were running back and forth counting cars and parking spots and came up with some good estimations of the percentage. I didn’t give them enough time to count every car and parking spot.
I asked them:
What is your estimate? 35%
Are there 35 cars? No
Then what do you mean by 35%? [silence]
Someone needs to explain what 35% means. Take a few minutes in your groups and think about it.
I also asked them (as I circulated from group to group):
If I gave you some more time, would you be able to come up with an exact percentage? Yes
Then I asked them to look at the bulletin board next to the door and asked them: What percentage of the bulletin board is yellow?
How do we do that?
Well, is it more than half or less than half? Less than half.
So it is less than 50%? Yes
How could you get a more accurate answer? [awkward pause followed by whispering in their groups followed by kids going up to measure the yellow spots]
In each class, the first thing I noticed was that they were improvising a unit of area. In the photo above, the unit of area is his calculator.
My question:
Can you get an exact answer? No
Then I stood between the screen and the bulletin board and pointed out that in the first case, they are determining a percent of something that is countable. In the second case, the areas are
difficult to count, but you can still talk about percent of the bulletin board.
They stayed on task and thought about each question and came up with others. An excellent class.
The next day, I drew some random shapes on graph paper and asked the kids to shade a specific percentage. I assigned each group a different percentage and let them work on the problem. After they
were done (about ten minutes), I collected their work and showed each of them on the document camera from smallest to largest percentage.
My goal was to give them another concrete example of percent with areas. I also wanted them to think about visual examples of percent area as they saw them on the screen.
The concrete activities went much better. The kids stayed on task and later said they enjoyed the activity.
8 Comments
• HI MR. Riley, long time learner, fist time commenter. YOU ARE AMAZING
□ Thank you, Charlie. I really enjoyed having you in class these last three weeks. I will visit when I am in town.
Mr Riley
• Hello Mr. Riley, It is nice seeing our pictures up there. 🙂
• Hey Mr Really I am Happy to see me in a website Thanks (Nicholas Kurz)
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Absolute Value Equations
Absolute Value Equations
If you're trying to solve an equation containing a variable that's trapped inside absolute value bars, you'll need to slightly modify your technique. Here's why: Such equations may, in fact, have two
answers, instead of just one! That might be initially shocking and exciting (like finding out that while you thought you had one secret admirer, you actually have two), once you figure it all out in
the end, everything makes sense. (It must be because you are one irresistible hunk of burnin' love.)
Here's what to do if you encounter an equation whose poor, defenseless variable is trapped in absolute value bars:
1. Isolate the absolute value expression. Just like you isolated the variables before, this time isolate that entire expression that falls between the absolute value bars. Follow the same steps as
before start by adding or subtracting things out of the way and finish by eliminating a coefficient, if the expression has one.
2. Create two new equations. Here's the tricky part. You're actually going to design two completely separate equations from the original one. The first equation should look just like the original,
just without the bars on it. The second should look just like the first, only take the opposite of the right side of the equation. This might sound tricky, but trust me, it's easy.
3. Solve the new equations to get your answer(s). Both of the solutions you get are answers to the original absolute value equation.
You've Got Problems
You're creating two separate equations because absolute values change two different values (any number and its opposite) into the same thing. If you don't understand what I mean, check out the end of
To remind myself that absolute value equations require two separate parts, I sometimes imagine that those absolute value bars are little bars of dynamite that blow that original equation in half,
creating two distinct pieces. Now that you get the idea, let me show you how to handle the explosives correctly.
Example 3: Solve the equation 4|2x - 3| + 1 = 21.
Solution: Start by isolating the absolute value quantity on the left. To do so, first subtract 1 from both sides.
To complete the isolation process, divide both sides by 4.
• ^4|2x - 3| [4] = ^20 [4]
• |2x - 3| = 5
Now that only the absolute values remain on the left side, it's time to create two new equations. The first looks just like the above equation (without bars attached); its sister equation is an exact
replica, except its right side will be the opposite of its sibling's right side (it'll have -5 instead of 5).
Solve those equations separately.
2x - 3 = 5 2x - 3 = -5
2x = 8 2x = -2
x = 4 x = -1
You've Got Problems
Problem 4: Solve the equation |x - 5| - 6 = 4.
There you go; the answers are -1 and 4. Do you find two answers hard to swallow? Watch what happens when I check them both in the original equation.
4|2(4) - 3| + 1 = 21 4|2(-1)-1|+1=21
4|5| + 1 = 21 4|-5| + 1 = 21
4(5) + 1 = 21 4(5) + 1 = 21
20 + 1 = 21 20 + 1 =21
Notice that the contents of the absolute values are opposites, so once the absolute value is taken, you end up getting the same results.
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with
Alpha Books, a member of Penguin Group (USA) Inc.
You can purchase this book at Amazon.com and Barnes & Noble.
Here are the facts and trivia that people are buzzing about.
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IrisCTF 2024 Solution Guide - What the Beep (Forensics)
Jan 8, 2024
CTF challenges can be intimidating for beginners, especially those without much technical background. This is a step-by-step guide to the IrisCTF 2024 challenge “What the Beep”, aiming to show both
the thought process and the details of how I solved it, and how you can too, even without a lot of prior knowledge.
Skills Required
• Googling
• Use of web applications
• Some algebra
• Basic Python scripting
First Look
Challenge Description
A strange beep sound was heard across a part of the San Joaquin Valley. We have the records from some audio volume meters at various locations nearby that picked up this event. It’s understood
that the original sound was about 140 dB at the source, but can you find out where it originated from?
Let’s download the attached file and see what it has…
“Wait, the file has a weird extension .tar.gz. How do I even open it?”
Whenever you encounter something you’ve never seen before, just look it up.
Ok, let’s see what we get inside the folder.
The names of the HTML files resemble GPS coordinates, and if we open one of them in a browser it shows a graph as follows:
As the x and y axis have units of time (\(\text{s}\)) and decibel (\(\text{dB}\)) respectively, it’s reasonable to assume that the graph shows the sound level at the corresponding time, in which
there is a peak in loudness for about 2 seconds, at around \(50 \text{dB}\). That matches the “loud beep” description, and the attached audio file:
Let’s wrap up what information we have so far:
• A recording of a loud beep
• Four pairs of GPS coordinates, each with a graph of sound level over time
From the above, we can infer that the challenge is about finding the location of the sound, given the intensity of the sound recorded at different locations.
The Approach
To clear things up, let’s sketch a diagram of the situation:
Diagrams are always helpful, especially when you are stuck.
If we worked out \(r_A\) to \(r_D\), the distance between the sound source and each of the four locations, the location of the source could be easily found by drawing four circles around \(A\) to \(D
\) with radii \(r_A\) to \(r_D\) respectively, and look for their intersection:
This graphical method is simple and intuitive, as it avoids solving for the coordinates of the source by hand, but it’s not very accurate. However it suits our purpose as we only need to find the
approximate location of the sound source.
Physics Comes In Handy
Now that we only have the sound intensity, we need a way to work out the distance between the source and each location. We all know we can judge a distance simply by the loudness of a sound, but how
do we actually calculate it?
That might sound familiar to you if you took high school physics, as it’s the inverse square law:
\[\left(\frac{r_2}{r_1}\right)^2 = \frac{I_1}{I_2}\]
where \(I_1\) is the intensity at distance \(r_1\), and \(r_2\) is the distance at which \(I_2\) is measured. Note how the fraction on the right hand side is inverted, as it involves an inversely
proportional relationship.
The equation relates the intensity of sound at two different points with their distances to the source. In our case, \(I_2\) and \(r_2\) represents the intensity and distance at \(A\) to \(D\), and \
(I_1\) and \(r_1\) is what we get from a known reference point. Where can we find such a reference point?
Perfect! Now just set \(I_1\) to \(140 \text{dB}\), and \(r_1\) to \(1 \ \text{ft}\)… wait. That’s not how it works. The decibel scale is logarithmic and relative to a threshold \(I_0\), defined by:
\[ n = 10 \log_{10} \left(\frac{I}{I_0}\right) \] where \(n\) is the sound intensity in \(\text{dB}\). so we need to convert it to a linear scale first: \[\frac{I}{I_0} = 10^{n/10}\]
Luckily for us, we’re only interested in the ratio of the sound intensity at different locations, so we can just ignore the \(I_0\) term and work it out directly: \[\frac{I_1}{I_2} = \cfrac{\cfrac
{I_1}{I_0}}{\cfrac{I_2}{I_0}} = \cfrac{10^{{n_1}/{10}}}{10^{{n_2}/{10}}} = 10^{(n_1 - n_2)/10} \]
Intuitively, an increase of \(10 \text{dB}\) means multiplying the intensity by 10. So to find out the ratio between two different intensities \({I_1}\) and \({I_2}\), we figure out how many
times we have to add \(10\) to get from \(n_2\) to \(n_1\), then raise \(10\) to that power.
To sum up, if we were to find distance \(r_A\): \[r_A = r_1 \sqrt{10^{(n_1 - n_A)/10}} = (1 \ \text{ft}) \cdot \sqrt{10^{(140 - n_A)/10}}\]
where \(n_A\) is the intensity at \(A\), measured in \(\text{dB}\).
Obtaining the Data
Now our task is to extract the exact sound level from the graphs. But here’s a small problem: we get little bumps on the curve during the 2-second beep:
There’s not only one number, but a range of numbers. So we have to approximate the sound level instead, as accurately as possible. The best way I can think of is to take the average of the data
And rather than copying off the values one by one with the mouse, did you notice how the graphs are interactive? The actual data must be stored somewhere in the HTML file, and we can inspect the page
to find out where:
Note how the data is stored in the form of JavaScript arrays. We can just copy the array and paste it into a text editor, and treat it like Python lists:
data_a = [49.4862687673082, 49.11758306247154, 49.35891737763439, 49.60312825002279, 49.28094240869986, 49.33179344636332, 49.77218278810612, 49.33157050295794, 49.954163292100134, 49.46399894576454, 49.75225513933776, 49.51956668498062, 49.72709095876894, 49.08380931815951, 49.80535732712877, 49.466366411374636, 49.272738443513475, 49.537197963188916, 49.42320370510891, 49.324671083447626, 49.54118211146326, 49.49531460381351, 49.976621634496894, 49.3893728063094, 49.921942150468716, 49.19386224160513, 49.36279881936855, 49.3589213415847, 49.56066713691474, 49.12186675176159, 49.98362703411643, 49.52541697547485, 49.35868710209489, 49.43923653057155, 49.98372751405347, 49.28405742162781, 49.401207574823644, 49.01614674667963, 49.13219547793331, 49.64847624718398, 49.498071028322336, 49.334549685095496, 49.458331325541025, 49.16635204000964, 49.2845016923542, 49.04043406000734, 49.911997928476055, 49.522384277676096, 49.63639242519472, 49.5321507012455, 49.580222157005686, 49.462799630990716, 49.15286264591634, 49.5636105290103, 49.24446101814839, 49.17815265294301, 49.277052087309045, 49.34785136315813, 49.2099358209713, 49.18130715442975, 49.81637365701671, 49.58976121006631, 49.26447327335997, 49.07489408373105, 49.10738248956828, 49.82935754558414, 49.7076592827515, 49.56229242462191, 49.67051905124946, 49.042312629812045, 49.561770092276326, 49.66475069029362, 49.858494354189034, 49.048272583835754, 49.9132487579282, 49.71779824360189, 49.79452312717411, 49.50065500658594, 49.84834211295007, 49.220394666568154, 49.66254149768159, 49.83640438670091, 49.10061336144564, 49.42849201280895, 49.646915124964735, 49.78950547033567, 49.4929685819846, 49.73705541695538, 49.22359955303929, 49.79862536749438, 49.652865340678765, 49.066372510572236, 49.19935726756466, 49.12145308818689, 49.438711940843866, 49.004099870912, 49.502682207162174, 49.293165246893956, 49.112557507785844, 49.544065615449895,
49.663019552626615, 49.46132309525862, 49.28867460561771, 49.04716798758809, 49.35484951313734, 49.37733257790768, 49.84822901146003, 49.81145708386574, 49.88943707227456, 49.8760994755179, 48.99312519891967, 49.48120050185433, 49.52537947160789, 49.90610721676662, 49.914515091218576, 49.416331830579196, 49.348693840298814, 49.545231061032965, 49.114561921757456, 49.026512427769646, 49.14711681989299, 49.77105573603577, 49.536523596883534, 49.60021444492543, 49.60605081543197, 49.64891471841797, 49.4600478177719, 49.977585356380565, 49.64790786367474, 49.05339723365599, 49.776801982915465, 49.345914994020035, 49.460170041286936, 49.458597510753314, 49.404020334240904, 49.80110610568392, 49.92580226915928, 49.892740295161424, 49.32725220374159, 49.14689359709007, 49.272749076596895, 49.77397438750593, 49.092844212042216, 49.29759302412302, 49.44741729129354, 49.41099308272467, 49.079825857328835, 49.386676414641016, 49.10972967558096, 49.043040950598254, 49.11993424749808, 48.99318353627927, 49.10530939042136, 49.21146252088831, 49.15074800916907, 49.61678542581952, 49.35038069687581, 49.03078805691796, 49.6258955230806, 49.63094191644237, 49.47515877815869, 49.26668175133948, 49.31472885646965, 49.640732134272305, 49.228802255830445, 49.59159486655283, 49.06310688917667, 49.49737416549353, 49.97771220245058, 49.636406874411215, 49.173004449388536, 49.33266160439406, 49.014913705550754, 49.40246086899188, 49.329105230947825, 49.86540375321602, 49.843613528504285, 49.8657191427391, 49.219049611824765, 49.57851447868852, 49.98526126350094, 49.94445651733331, 49.45467432812931, 49.76296756625244, 49.7935233989949, 49.83262284962718, 49.93995818802516, 49.67306551801829, 49.54206148681675, 49.31504612258427, 49.94311396484975, 49.94210883611035, 49.11875893744185, 49.46127955948076, 49.7357488917027, 49.878480551044404, 49.897911048398925, 49.58542961303054, 49.59108511515634, 49.31169509848393]
I only extracted the numbers close to 50, since that’s the relevant part of the graph. Now we can work out the distance \(r_A\) in feet:
import math
avg_n_a = sum(data_a) / len(data_a)
r_a = 1 * math.sqrt(10 ** ((140 - avg_n_a) / 10))
That would give us 33573.94482372866 feet (around 10 km), which looks reasonable.
Same goes for \(r_B\), \(r_C\) and \(r_D\). Try working them out yourself!
Work the smart way, not the hard way. Perhaps let your machine do the job for you?
Putting It All Together
Now that we have the coordinates and distances, all that’s left is to draw the circles on a map… is there any convenient map application that allows us to draw circles?
There we are (I have no idea why this exists). Let’s try to use it, pasting in the coordinates and distances…
Ooh, there’s an intersection! But one of the circles looks off. What went wrong? Let’s click on the circle around \(A\):
The coordinates shown for the circle is entirely different from what we entered (37.185287, -120.292548)! How do we manually correct it?
We see a generated URL for the created map, in a text box below:
which contains the wrong coordinates of the circle. Now let’s do a little hack, replacing the numbers with the correct ones:
and finally pasting it into the address bar:
Voila! The circles now (nearly) intersect at a single point. Pick a point best approximating the location of the sound source, and we’re done!
Where do we submit our answer? Let’s look at the challenge description again:
“What’s the answer checker service? And what about nc what-the-beep.chal.irisc.tf 10500?”
I’ll give you a hint: nc is short for Netcat.
… no trick questions here, don’t worry. Just follow the instructions and the sacred line of text you’ve been craving for shall reveal itself:
If something is not working right, don’t complain. Hack your way around it.
TL;DR Solution
1. Extract the average sound intensity from the graphs, for each location
2. Use the inverse square law to calculate the distance between the sound source and each location
3. Draw circles on a map with the coordinates and distances
4. Find the intersection of the circles, and submit the coordinates to the answer checker
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Last change on this file since 11551 was 11551, checked in by nicolasmartin, 5 years ago
Bugfix: no index entries in math env
File size: 42.4 KB
1 \documentclass[../main/NEMO_manual]{subfiles}
3 \begin{document}
4 % ================================================================
5 % Chapter — Lateral Boundary Condition (LBC)
6 % ================================================================
7 \chapter{Lateral Boundary Condition (LBC)}
8 \label{chap:LBC}
10 \chaptertoc
12 \newpage
14 %gm% add here introduction to this chapter
16 % ================================================================
17 % Boundary Condition at the Coast
18 % ================================================================
19 \section[Boundary condition at the coast (\texttt{rn\_shlat})]
20 {Boundary condition at the coast (\protect\np{rn\_shlat})}
21 \label{sec:LBC_coast}
22 %--------------------------------------------namlbc-------------------------------------------------------
24 \nlst{namlbc}
25 %--------------------------------------------------------------------------------------------------------------
27 %The lateral ocean boundary conditions contiguous to coastlines are Neumann conditions for heat and salt
28 %(no flux across boundaries) and Dirichlet conditions for momentum (ranging from free-slip to "strong" no-slip).
29 %They are handled automatically by the mask system (see \autoref{subsec:DOM_msk}).
31 %OPA allows land and topography grid points in the computational domain due to the presence of continents or islands,
32 %and includes the use of a full or partial step representation of bottom topography.
33 %The computation is performed over the whole domain, \ie\ we do not try to restrict the computation to ocean-only points.
34 %This choice has two motivations.
35 %Firstly, working on ocean only grid points overloads the code and harms the code readability.
36 %Secondly, and more importantly, it drastically reduces the vector portion of the computation,
37 %leading to a dramatic increase of CPU time requirement on vector computers.
38 %The current section describes how the masking affects the computation of the various terms of the equations
39 %with respect to the boundary condition at solid walls.
40 %The process of defining which areas are to be masked is described in \autoref{subsec:DOM_msk}.
42 Options are defined through the \nam{lbc} namelist variables.
43 The discrete representation of a domain with complex boundaries (coastlines and bottom topography) leads to
44 arrays that include large portions where a computation is not required as the model variables remain at zero.
45 Nevertheless, vectorial supercomputers are far more efficient when computing over a whole array,
46 and the readability of a code is greatly improved when boundary conditions are applied in
47 an automatic way rather than by a specific computation before or after each computational loop.
48 An efficient way to work over the whole domain while specifying the boundary conditions,
49 is to use multiplication by mask arrays in the computation.
50 A mask array is a matrix whose elements are $1$ in the ocean domain and $0$ elsewhere.
51 A simple multiplication of a variable by its own mask ensures that it will remain zero over land areas.
52 Since most of the boundary conditions consist of a zero flux across the solid boundaries,
53 they can be simply applied by multiplying variables by the correct mask arrays,
54 \ie\ the mask array of the grid point where the flux is evaluated.
55 For example, the heat flux in the \textbf{i}-direction is evaluated at $u$-points.
56 Evaluating this quantity as,
58 \[
59 % \label{eq:LBC_aaaa}
60 \frac{A^{lT} }{e_1 }\frac{\partial T}{\partial i}\equiv \frac{A_u^{lT}
61 }{e_{1u} } \; \delta_{i+1 / 2} \left[ T \right]\;\;mask_u
62 \]
63 (where mask$_{u}$ is the mask array at a $u$-point) ensures that the heat flux is zero inside land and
64 at the boundaries, since mask$_{u}$ is zero at solid boundaries which in this case are defined at $u$-points
65 (normal velocity $u$ remains zero at the coast) (\autoref{fig:LBC_uv}).
67 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
68 \begin{figure}[!t]
69 \begin{center}
70 \includegraphics[width=\textwidth]{Fig_LBC_uv}
71 \caption{
72 \protect\label{fig:LBC_uv}
73 Lateral boundary (thick line) at T-level.
74 The velocity normal to the boundary is set to zero.
75 }
76 \end{center}
77 \end{figure}
78 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
80 For momentum the situation is a bit more complex as two boundary conditions must be provided along the coast
81 (one each for the normal and tangential velocities).
82 The boundary of the ocean in the C-grid is defined by the velocity-faces.
83 For example, at a given $T$-level,
84 the lateral boundary (a coastline or an intersection with the bottom topography) is made of
85 segments joining $f$-points, and normal velocity points are located between two $f-$points (\autoref{fig:LBC_uv}).
86 The boundary condition on the normal velocity (no flux through solid boundaries)
87 can thus be easily implemented using the mask system.
88 The boundary condition on the tangential velocity requires a more specific treatment.
89 This boundary condition influences the relative vorticity and momentum diffusive trends,
90 and is required in order to compute the vorticity at the coast.
91 Four different types of lateral boundary condition are available,
92 controlled by the value of the \np{rn\_shlat} namelist parameter
93 (The value of the mask$_{f}$ array along the coastline is set equal to this parameter).
94 These are:
96 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
97 \begin{figure}[!p]
98 \begin{center}
99 \includegraphics[width=\textwidth]{Fig_LBC_shlat}
100 \caption{
101 \protect\label{fig:LBC_shlat}
102 lateral boundary condition
103 (a) free-slip ($rn\_shlat=0$);
104 (b) no-slip ($rn\_shlat=2$);
105 (c) "partial" free-slip ($0<rn\_shlat<2$) and
106 (d) "strong" no-slip ($2<rn\_shlat$).
107 Implied "ghost" velocity inside land area is display in grey.
108 }
109 \end{center}
110 \end{figure}
111 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
113 \begin{description}
115 \item[free-slip boundary condition (\np{rn\_shlat}\forcode{=0}):] the tangential velocity at
116 the coastline is equal to the offshore velocity,
117 \ie\ the normal derivative of the tangential velocity is zero at the coast,
118 so the vorticity: mask$_{f}$ array is set to zero inside the land and just at the coast
119 (\autoref{fig:LBC_shlat}-a).
121 \item[no-slip boundary condition (\np{rn\_shlat}\forcode{=2}):] the tangential velocity vanishes at the coastline.
122 Assuming that the tangential velocity decreases linearly from
123 the closest ocean velocity grid point to the coastline,
124 the normal derivative is evaluated as if the velocities at the closest land velocity gridpoint and
125 the closest ocean velocity gridpoint were of the same magnitude but in the opposite direction
126 (\autoref{fig:LBC_shlat}-b).
127 Therefore, the vorticity along the coastlines is given by:
129 \[
130 \zeta \equiv 2 \left(\delta_{i+1/2} \left[e_{2v} v \right] - \delta_{j+1/2} \left[e_{1u} u \right] \right) / \left(e_{1f} e_{2f} \right) \ ,
131 \]
132 where $u$ and $v$ are masked fields.
133 Setting the mask$_{f}$ array to $2$ along the coastline provides a vorticity field computed with
134 the no-slip boundary condition, simply by multiplying it by the mask$_{f}$ :
135 \[
136 % \label{eq:LBC_bbbb}
137 \zeta \equiv \frac{1}{e_{1f} {\kern 1pt}e_{2f} }\left( {\delta_{i+1/2}
138 \left[ {e_{2v} \,v} \right]-\delta_{j+1/2} \left[ {e_{1u} \,u} \right]}
139 \right)\;\mbox{mask}_f
140 \]
142 \item["partial" free-slip boundary condition (0$<$\np{rn\_shlat}$<$2):] the tangential velocity at
143 the coastline is smaller than the offshore velocity, \ie\ there is a lateral friction but
144 not strong enough to make the tangential velocity at the coast vanish (\autoref{fig:LBC_shlat}-c).
145 This can be selected by providing a value of mask$_{f}$ strictly inbetween $0$ and $2$.
147 \item["strong" no-slip boundary condition (2$<$\np{rn\_shlat}):] the viscous boundary layer is assumed to
148 be smaller than half the grid size (\autoref{fig:LBC_shlat}-d).
149 The friction is thus larger than in the no-slip case.
151 \end{description}
153 Note that when the bottom topography is entirely represented by the $s$-coordinates (pure $s$-coordinate),
154 the lateral boundary condition on tangential velocity is of much less importance as
155 it is only applied next to the coast where the minimum water depth can be quite shallow.
158 % ================================================================
159 % Boundary Condition around the Model Domain
160 % ================================================================
161 \section[Model domain boundary condition (\texttt{jperio})]
162 {Model domain boundary condition (\protect\jp{jperio})}
163 \label{sec:LBC_jperio}
165 At the model domain boundaries several choices are offered:
166 closed, cyclic east-west, cyclic north-south, a north-fold, and combination closed-north fold or
167 bi-cyclic east-west and north-fold.
168 The north-fold boundary condition is associated with the 3-pole ORCA mesh.
170 % -------------------------------------------------------------------------------------------------------------
171 % Closed, cyclic (\jp{jperio}\forcode{ = 0..2})
172 % -------------------------------------------------------------------------------------------------------------
173 \subsection[Closed, cyclic (\forcode{jperio=[0127]})]
174 {Closed, cyclic (\protect\jp{jperio}\forcode{=[0127]})}
175 \label{subsec:LBC_jperio012}
177 The choice of closed or cyclic model domain boundary condition is made by
178 setting \jp{jperio} to 0, 1, 2 or 7 in namelist \nam{cfg}.
179 Each time such a boundary condition is needed, it is set by a call to routine \mdl{lbclnk}.
180 The computation of momentum and tracer trends proceeds from $i=2$ to $i=jpi-1$ and from $j=2$ to $j=jpj-1$,
181 \ie\ in the model interior.
182 To choose a lateral model boundary condition is to specify the first and last rows and columns of
183 the model variables.
185 \begin{description}
187 \item[For closed boundary (\jp{jperio}\forcode{=0})],
188 solid walls are imposed at all model boundaries:
189 first and last rows and columns are set to zero.
191 \item[For cyclic east-west boundary (\jp{jperio}\forcode{=1})],
192 first and last rows are set to zero (closed) whilst the first column is set to
193 the value of the last-but-one column and the last column to the value of the second one
194 (\autoref{fig:LBC_jperio}-a).
195 Whatever flows out of the eastern (western) end of the basin enters the western (eastern) end.
197 \item[For cyclic north-south boundary (\jp{jperio}\forcode{=2})],
198 first and last columns are set to zero (closed) whilst the first row is set to
199 the value of the last-but-one row and the last row to the value of the second one
200 (\autoref{fig:LBC_jperio}-a).
201 Whatever flows out of the northern (southern) end of the basin enters the southern (northern) end.
203 \item[Bi-cyclic east-west and north-south boundary (\jp{jperio}\forcode{=7})] combines cases 1 and 2.
205 \end{description}
207 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
208 \begin{figure}[!t]
209 \begin{center}
210 \includegraphics[width=\textwidth]{Fig_LBC_jperio}
211 \caption{
212 \protect\label{fig:LBC_jperio}
213 setting of (a) east-west cyclic (b) symmetric across the equator boundary conditions.
214 }
215 \end{center}
216 \end{figure}
217 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
219 % -------------------------------------------------------------------------------------------------------------
220 % North fold (\textit{jperio = 3 }to $6)$
221 % -------------------------------------------------------------------------------------------------------------
222 \subsection[North-fold (\forcode{jperio=[3-6]})]
223 {North-fold (\protect\jp{jperio}\forcode{=[3-6]})}
224 \label{subsec:LBC_north_fold}
226 The north fold boundary condition has been introduced in order to handle the north boundary of
227 a three-polar ORCA grid.
228 Such a grid has two poles in the northern hemisphere (\autoref{fig:CFGS_ORCA_msh},
229 and thus requires a specific treatment illustrated in \autoref{fig:LBC_North_Fold_T}.
230 Further information can be found in \mdl{lbcnfd} module which applies the north fold boundary condition.
232 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
233 \begin{figure}[!t]
234 \begin{center}
235 \includegraphics[width=\textwidth]{Fig_North_Fold_T}
236 \caption{
237 \protect\label{fig:LBC_North_Fold_T}
238 North fold boundary with a $T$-point pivot and cyclic east-west boundary condition ($jperio=4$),
239 as used in ORCA 2, 1/4, and 1/12.
240 Pink shaded area corresponds to the inner domain mask (see text).
241 }
242 \end{center}
243 \end{figure}
244 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
246 % ====================================================================
247 % Exchange with neighbouring processors
248 % ====================================================================
249 \section[Exchange with neighbouring processors (\textit{lbclnk.F90}, \textit{lib\_mpp.F90})]
250 {Exchange with neighbouring processors (\protect\mdl{lbclnk}, \protect\mdl{lib\_mpp})}
251 \label{sec:LBC_mpp}
253 %-----------------------------------------nammpp--------------------------------------------
255 \nlst{nammpp}
256 %-----------------------------------------------------------------------------------------------
258 For massively parallel processing (mpp), a domain decomposition method is used.
259 The basic idea of the method is to split the large computation domain of a numerical experiment into several smaller domains and
260 solve the set of equations by addressing independent local problems.
261 Each processor has its own local memory and computes the model equation over a subdomain of the whole model domain.
262 The subdomain boundary conditions are specified through communications between processors which are organized by
263 explicit statements (message passing method).
264 The present implementation is largely inspired by Guyon's work [Guyon 1995].
266 The parallelization strategy is defined by the physical characteristics of the ocean model.
267 Second order finite difference schemes lead to local discrete operators that
268 depend at the very most on one neighbouring point.
269 The only non-local computations concern the vertical physics
270 (implicit diffusion, turbulent closure scheme, ...).
271 Therefore, a pencil strategy is used for the data sub-structuration:
272 the 3D initial domain is laid out on local processor memories following a 2D horizontal topological splitting.
273 Each sub-domain computes its own surface and bottom boundary conditions and
274 has a side wall overlapping interface which defines the lateral boundary conditions for
275 computations in the inner sub-domain.
276 The overlapping area consists of the two rows at each edge of the sub-domain.
277 After a computation, a communication phase starts:
278 each processor sends to its neighbouring processors the update values of the points corresponding to
279 the interior overlapping area to its neighbouring sub-domain (\ie\ the innermost of the two overlapping rows).
280 Communications are first done according to the east-west direction and next according to the north-south direction.
281 There is no specific communications for the corners.
282 The communication is done through the Message Passing Interface (MPI) and requires \key{mpp\_mpi}.
283 Use also \key{mpi2} if MPI3 is not available on your computer.
284 The data exchanges between processors are required at the very place where
285 lateral domain boundary conditions are set in the mono-domain computation:
286 the \rou{lbc\_lnk} routine (found in \mdl{lbclnk} module) which manages such conditions is interfaced with
287 routines found in \mdl{lib\_mpp} module.
288 The output file \textit{communication\_report.txt} provides the list of which routines do how
289 many communications during 1 time step of the model.\\
291 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
292 \begin{figure}[!t]
293 \begin{center}
294 \includegraphics[width=\textwidth]{Fig_mpp}
295 \caption{
296 \protect\label{fig:LBC_mpp}
297 Positioning of a sub-domain when massively parallel processing is used.
298 }
299 \end{center}
300 \end{figure}
301 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
303 In \NEMO, the splitting is regular and arithmetic.
304 The total number of subdomains corresponds to the number of MPI processes allocated to \NEMO\ when the model is launched
305 (\ie\ mpirun -np x ./nemo will automatically give x subdomains).
306 The i-axis is divided by \np{jpni} and the j-axis by \np{jpnj}.
307 These parameters are defined in \nam{mpp} namelist.
308 If \np{jpni} and \np{jpnj} are < 1, they will be automatically redefined in the code to give the best domain decomposition
309 (see bellow).
311 Each processor is independent and without message passing or synchronous process, programs run alone and access just its own local memory.
312 For this reason,
313 the main model dimensions are now the local dimensions of the subdomain (pencil) that are named \jp{jpi}, \jp{jpj}, \jp{jpk}.
314 These dimensions include the internal domain and the overlapping rows.
315 The number of rows to exchange (known as the halo) is usually set to one (nn\_hls=1, in \mdl{par\_oce},
316 and must be kept to one until further notice).
317 The whole domain dimensions are named \jp{jpiglo}, \jp{jpjglo} and \jp{jpk}.
318 The relationship between the whole domain and a sub-domain is:
319 \begin{gather*}
320 jpi = ( jpiglo-2\times nn\_hls + (jpni-1) ) / jpni + 2\times nn\_hls \\
321 jpj = ( jpjglo-2\times nn\_hls + (jpnj-1) ) / jpnj + 2\times nn\_hls
322 \end{gather*}
324 One also defines variables nldi and nlei which correspond to the internal domain bounds, and the variables nimpp and njmpp which are the position of the (1,1) grid-point in the global domain (\
autoref{fig:LBC_mpp}). Note that since the version 4, there is no more extra-halo area as defined in \autoref{fig:LBC_mpp} so \jp{jpi} is now always equal to nlci and \jp{jpj} equal to nlcj.
326 An element of $T_{l}$, a local array (subdomain) corresponds to an element of $T_{g}$,
327 a global array (whole domain) by the relationship:
328 \[
329 % \label{eq:LBC_nimpp}
330 T_{g} (i+nimpp-1,j+njmpp-1,k) = T_{l} (i,j,k),
331 \]
332 with $1 \leq i \leq jpi$, $1 \leq j \leq jpj $ , and $1 \leq k \leq jpk$.
The 1-d arrays $mig(1:\jp{jpi})$ and $mjg(1:\jp{jpj})$, defined in \rou{dom\_glo} routine (\mdl{domain} module), should be used to get global domain indices from local domain indices. The 1-d
334 arrays, $mi0(1:\jp{jpiglo})$, $mi1(1:\jp{jpiglo})$ and $mj0(1:\jp{jpjglo})$, $mj1(1:\jp{jpjglo})$ have the reverse purpose and should be used to define loop indices expressed in global domain
indices (see examples in \mdl{dtastd} module).\\
The \NEMO\ model computes equation terms with the help of mask arrays (0 on land points and 1 on sea points). It is therefore possible that an MPI subdomain contains only land points. To save
336 ressources, we try to supress from the computational domain as much land subdomains as possible. For example if $N_{mpi}$ processes are allocated to NEMO, the domain decomposition will be given
by the following equation:
337 \[
338 N_{mpi} = jpni \times jpnj - N_{land} + N_{useless}
339 \]
$N_{land}$ is the total number of land subdomains in the domain decomposition defined by \np{jpni} and \np{jpnj}. $N_{useless}$ is the number of land subdomains that are kept in the compuational
340 domain in order to make sure that $N_{mpi}$ MPI processes are indeed allocated to a given subdomain. The values of $N_{mpi}$, \np{jpni}, \np{jpnj}, $N_{land}$ and $N_{useless}$ are printed in
the output file \texttt{ocean.output}. $N_{useless}$ must, of course, be as small as possible to limit the waste of ressources. A warning is issued in \texttt{ocean.output} if $N_{useless}$ is
not zero. Note that non-zero value of $N_{useless}$ is uselly required when using AGRIF as, up to now, the parent grid and each of the child grids must use all the $N_{mpi}$ processes.
If the domain decomposition is automatically defined (when \np{jpni} and \np{jpnj} are < 1), the decomposition chosen by the model will minimise the sub-domain size (defined as $max_{all
342 domains}(jpi \times jpj)$) and maximize the number of eliminated land subdomains. This means that no other domain decomposition (a set of \np{jpni} and \np{jpnj} values) will use less processes
than $(jpni \times jpnj - N_{land})$ and get a smaller subdomain size.
In order to specify $N_{mpi}$ properly (minimize $N_{useless}$), you must run the model once with \np{ln\_list} activated. In this case, the model will start the initialisation phase, print the
343 list of optimum decompositions ($N_{mpi}$, \np{jpni} and \np{jpnj}) in \texttt{ocean.output} and directly abort. The maximum value of $N_{mpi}$ tested in this list is given by $max(N_{MPI\_
tasks}, jpni \times jpnj)$. For example, run the model on 40 nodes with ln\_list activated and $jpni = 10000$ and $jpnj = 1$, will print the list of optimum domains decomposition from 1 to about
345 Processors are numbered from 0 to $N_{mpi} - 1$. Subdomains containning some ocean points are numbered first from 0 to $jpni * jpnj - N_{land} -1$. The remaining $N_{useless}$ land subdomains
are numbered next, which means that, for a given (\np{jpni}, \np{jpnj}), the numbers attributed to he ocean subdomains do not vary with $N_{useless}$.
347 When land processors are eliminated, the value corresponding to these locations in the model output files is undefined. \np{ln\_mskland} must be activated in order avoid Not a Number values in
output files. Note that it is better to not eliminate land processors when creating a meshmask file (\ie\ when setting a non-zero value to \np{nn\_msh}).
349 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
350 \begin{figure}[!ht]
351 \begin{center}
352 \includegraphics[width=\textwidth]{Fig_mppini2}
353 \caption[Atlantic domain]{
354 \protect\label{fig:LBC_mppini2}
355 Example of Atlantic domain defined for the CLIPPER projet.
356 Initial grid is composed of 773 x 1236 horizontal points.
357 (a) the domain is split onto 9 \time 20 subdomains (jpni=9, jpnj=20).
358 52 subdomains are land areas.
359 (b) 52 subdomains are eliminated (white rectangles) and
360 the resulting number of processors really used during the computation is jpnij=128.
361 }
362 \end{center}
363 \end{figure}
364 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
367 % ====================================================================
368 % Unstructured open boundaries BDY
369 % ====================================================================
370 \section{Unstructured open boundary conditions (BDY)}
371 \label{sec:LBC_bdy}
373 %-----------------------------------------nambdy--------------------------------------------
375 \nlst{nambdy}
376 %-----------------------------------------------------------------------------------------------
377 %-----------------------------------------nambdy_dta--------------------------------------------
379 \nlst{nambdy_dta}
380 %-----------------------------------------------------------------------------------------------
382 Options are defined through the \nam{bdy} and \nam{bdy\_dta} namelist variables.
383 The BDY module is the core implementation of open boundary conditions for regional configurations on
384 ocean temperature, salinity, barotropic-baroclinic velocities, ice-snow concentration, thicknesses, temperatures, salinity and melt ponds concentration and thickness.
386 The BDY module was modelled on the OBC module (see \NEMO\ 3.4) and shares many features and
387 a similar coding structure \citep{chanut_rpt05}.
388 The specification of the location of the open boundary is completely flexible and
389 allows any type of setup, from regular boundaries to irregular contour (it includes the possibility to set an open boundary able to follow an isobath).
390 Boundary data files used with versions of \NEMO\ prior to Version 3.4 may need to be re-ordered to work with this version.
391 See the section on the Input Boundary Data Files for details.
393 %----------------------------------------------
394 \subsection{Namelists}
395 \label{subsec:LBC_bdy_namelist}
397 The BDY module is activated by setting \np{ln\_bdy}\forcode{=.true.} .
398 It is possible to define more than one boundary ``set'' and apply different boundary conditions to each set.
399 The number of boundary sets is defined by \np{nb\_bdy}.
400 Each boundary set can be either defined as a series of straight line segments directly in the namelist
401 (\np{ln\_coords\_file}\forcode{=.false.}, and a namelist block \nam{bdy\_index} must be included for each set) or read in from a file (\np{ln\_coords\_file}\forcode{=.true.}, and a ``\ifile{
coordinates.bdy}'' file must be provided).
402 The coordinates.bdy file is analagous to the usual \NEMO\ ``\ifile{coordinates}'' file.
403 In the example above, there are two boundary sets, the first of which is defined via a file and
404 the second is defined in the namelist.
405 For more details of the definition of the boundary geometry see section \autoref{subsec:LBC_bdy_geometry}.
407 For each boundary set a boundary condition has to be chosen for the barotropic solution
408 (``u2d'':sea-surface height and barotropic velocities), for the baroclinic velocities (``u3d''),
409 for the active tracers \footnote{The BDY module does not deal with passive tracers at this version} (``tra''), and for sea-ice (``ice'').
410 For each set of variables one has to choose an algorithm and the boundary data (set resp. by \np{cn\_tra} and \np{nn\_tra\_dta} for tracers).\\
412 The choice of algorithm is currently as follows:
414 \begin{description}
415 \item[\forcode{'none'}:] No boundary condition applied.
416 So the solution will ``see'' the land points around the edge of the edge of the domain.
417 \item[\forcode{'specified'}:] Specified boundary condition applied (only available for baroclinic velocity and tracer variables).
418 \item[\forcode{'neumann'}:] Value at the boundary are duplicated (No gradient). Only available for baroclinic velocity and tracer variables.
419 \item[\forcode{'frs'}:] Flow Relaxation Scheme (FRS) available for all variables.
420 \item[\forcode{'Orlanski'}:] Orlanski radiation scheme (fully oblique) for barotropic, baroclinic and tracer variables.
421 \item[\forcode{'Orlanski_npo'}:] Orlanski radiation scheme for barotropic, baroclinic and tracer variables.
422 \item[\forcode{'flather'}:] Flather radiation scheme for the barotropic variables only.
423 \end{description}
425 The boundary data is either set to initial conditions
426 (\np{nn\_tra\_dta}\forcode{=0}) or forced with external data from a file (\np{nn\_tra\_dta}\forcode{=1}).
427 In case the 3d velocity data contain the total velocity (ie, baroclinic and barotropic velocity),
428 the bdy code can derived baroclinic and barotropic velocities by setting \np{ln\_full\_vel}\forcode{=.true.}
429 For the barotropic solution there is also the option to use tidal harmonic forcing either by
430 itself (\np{nn\_dyn2d\_dta}\forcode{=2}) or in addition to other external data (\np{nn\_dyn2d\_dta}\forcode{=3}).\\
431 If not set to initial conditions, sea-ice salinity, temperatures and melt ponds data at the boundary can either be read in a file or defined as constant (by \np{rn\_ice\_sal}, \np{rn\_ice\_tem},
\np{rn\_ice\_apnd}, \np{rn\_ice\_hpnd}). Ice age is constant and defined by \np{rn\_ice\_age}.
433 If external boundary data is required then the \nam{bdy\_dta} namelist must be defined.
434 One \nam{bdy\_dta} namelist is required for each boundary set, adopting the same order of indexes in which the boundary sets are defined in nambdy.
435 In the example given, two boundary sets have been defined. The first one is reading data file in the \nam{bdy\_dta} namelist shown above
436 and the second one is using data from intial condition (no namelist block needed).
437 The boundary data is read in using the fldread module,
438 so the \nam{bdy\_dta} namelist is in the format required for fldread.
439 For each required variable, the filename, the frequency of the files and
440 the frequency of the data in the files are given.
441 Also whether or not time-interpolation is required and whether the data is climatological (time-cyclic) data.
442 For sea-ice salinity, temperatures and melt ponds, reading the files are skipped and constant values are used if filenames are defined as {'NOT USED'}.\\
444 There is currently an option to vertically interpolate the open boundary data onto the native grid at run-time.
445 If \np{nn\_bdy\_jpk}$<-1$, it is assumed that the lateral boundary data are already on the native grid.
446 However, if \np{nn\_bdy\_jpk} is set to the number of vertical levels present in the boundary data,
447 a bilinear interpolation onto the native grid will be triggered at runtime.
448 For this to be successful the additional variables: $gdept$, $gdepu$, $gdepv$, $e3t$, $e3u$ and $e3v$, are required to be present in the lateral boundary files.
449 These correspond to the depths and scale factors of the input data,
450 the latter used to make any adjustment to the velocity fields due to differences in the total water depths between the two vertical grids.\\
452 In the example of given namelists, two boundary sets are defined.
453 The first set is defined via a file and applies FRS conditions to temperature and salinity and
454 Flather conditions to the barotropic variables. No condition specified for the baroclinic velocity and sea-ice.
455 External data is provided in daily files (from a large-scale model).
456 Tidal harmonic forcing is also used.
457 The second set is defined in a namelist.
458 FRS conditions are applied on temperature and salinity and climatological data is read from initial condition files.
460 %----------------------------------------------
461 \subsection{Flow relaxation scheme}
462 \label{subsec:LBC_bdy_FRS_scheme}
464 The Flow Relaxation Scheme (FRS) \citep{davies_QJRMS76,engedahl_T95},
465 applies a simple relaxation of the model fields to externally-specified values over
466 a zone next to the edge of the model domain.
467 Given a model prognostic variable $\Phi$
468 \[
469 % \label{eq:LBC_bdy_frs1}
470 \Phi(d) = \alpha(d)\Phi_{e}(d) + (1-\alpha(d))\Phi_{m}(d)\;\;\;\;\; d=1,N
471 \]
472 where $\Phi_{m}$ is the model solution and $\Phi_{e}$ is the specified external field,
473 $d$ gives the discrete distance from the model boundary and
474 $\alpha$ is a parameter that varies from $1$ at $d=1$ to a small value at $d=N$.
475 It can be shown that this scheme is equivalent to adding a relaxation term to
476 the prognostic equation for $\Phi$ of the form:
477 \[
478 % \label{eq:LBC_bdy_frs2}
479 -\frac{1}{\tau}\left(\Phi - \Phi_{e}\right)
480 \]
481 where the relaxation time scale $\tau$ is given by a function of $\alpha$ and the model time step $\Delta t$:
482 \[
483 % \label{eq:LBC_bdy_frs3}
484 \tau = \frac{1-\alpha}{\alpha} \,\rdt
485 \]
486 Thus the model solution is completely prescribed by the external conditions at the edge of the model domain and
487 is relaxed towards the external conditions over the rest of the FRS zone.
488 The application of a relaxation zone helps to prevent spurious reflection of
489 outgoing signals from the model boundary.
491 The function $\alpha$ is specified as a $tanh$ function:
492 \[
493 % \label{eq:LBC_bdy_frs4}
494 \alpha(d) = 1 - \tanh\left(\frac{d-1}{2}\right), \quad d=1,N
495 \]
496 The width of the FRS zone is specified in the namelist as \np{nn\_rimwidth}.
497 This is typically set to a value between 8 and 10.
499 %----------------------------------------------
500 \subsection{Flather radiation scheme}
501 \label{subsec:LBC_bdy_flather_scheme}
503 The \citet{flather_JPO94} scheme is a radiation condition on the normal,
504 depth-mean transport across the open boundary.
505 It takes the form
506 \begin{equation}
507 \label{eq:LBC_bdy_fla1}
508 U = U_{e} + \frac{c}{h}\left(\eta - \eta_{e}\right),
509 \end{equation}
510 where $U$ is the depth-mean velocity normal to the boundary and $\eta$ is the sea surface height,
511 both from the model.
512 The subscript $e$ indicates the same fields from external sources.
513 The speed of external gravity waves is given by $c = \sqrt{gh}$, and $h$ is the depth of the water column.
514 The depth-mean normal velocity along the edge of the model domain is set equal to
515 the external depth-mean normal velocity,
516 plus a correction term that allows gravity waves generated internally to exit the model boundary.
517 Note that the sea-surface height gradient in \autoref{eq:LBC_bdy_fla1} is a spatial gradient across the model boundary,
518 so that $\eta_{e}$ is defined on the $T$ points with $nbr=1$ and $\eta$ is defined on the $T$ points with $nbr=2$.
519 $U$ and $U_{e}$ are defined on the $U$ or $V$ points with $nbr=1$, \ie\ between the two $T$ grid points.
521 %----------------------------------------------
522 \subsection{Orlanski radiation scheme}
523 \label{subsec:LBC_bdy_orlanski_scheme}
525 The Orlanski scheme is based on the algorithm described by \citep{marchesiello.mcwilliams.ea_OM01}, hereafter MMS.
527 The adaptive Orlanski condition solves a wave plus relaxation equation at the boundary:
528 \begin{equation}
529 \label{eq:LBC_wave_continuous}
530 \frac{\partial\phi}{\partial t} + c_x \frac{\partial\phi}{\partial x} + c_y \frac{\partial\phi}{\partial y} =
531 -\frac{1}{\tau}(\phi - \phi^{ext})
532 \end{equation}
534 where $\phi$ is the model field, $x$ and $y$ refer to the normal and tangential directions to the boundary respectively, and the phase
535 velocities are diagnosed from the model fields as:
537 \begin{equation}
538 \label{eq:LBC_cx}
539 c_x = -\frac{\partial\phi}{\partial t}\frac{\partial\phi / \partial x}{(\partial\phi /\partial x)^2 + (\partial\phi /\partial y)^2}
540 \end{equation}
541 \begin{equation}
542 \label{eq:LBC_cy}
543 c_y = -\frac{\partial\phi}{\partial t}\frac{\partial\phi / \partial y}{(\partial\phi /\partial x)^2 + (\partial\phi /\partial y)^2}
544 \end{equation}
546 (As noted by MMS, this is a circular diagnosis of the phase speeds which only makes sense on a discrete grid).
547 Equation (\autoref{eq:LBC_wave_continuous}) is defined adaptively depending on the sign of the phase velocity normal to the boundary $c_x$.
548 For $c_x$ outward, we have
550 \begin{equation}
551 \tau = \tau_{out}
552 \end{equation}
554 For $c_x$ inward, the radiation equation is not applied:
556 \begin{equation}
557 \label{eq:LBC_tau_in}
558 \tau = \tau_{in}\,\,\,;\,\,\, c_x = c_y = 0
559 \end{equation}
561 Generally the relaxation time scale at inward propagation points (\np{rn\_time\_dmp}) is set much shorter than the time scale at outward propagation
562 points (\np{rn\_time\_dmp\_out}) so that the solution is constrained more strongly by the external data at inward propagation points.
563 See \autoref{subsec:LBC_bdy_relaxation} for detailed on the spatial shape of the scaling.\\
564 The ``normal propagation of oblique radiation'' or NPO approximation (called \forcode{'orlanski_npo'}) involves assuming
565 that $c_y$ is zero in equation (\autoref{eq:LBC_wave_continuous}), but including
566 this term in the denominator of equation (\autoref{eq:LBC_cx}). Both versions of the scheme are options in BDY. Equations
567 (\autoref{eq:LBC_wave_continuous}) - (\autoref{eq:LBC_tau_in}) correspond to equations (13) - (15) and (2) - (3) in MMS.\\
569 %----------------------------------------------
570 \subsection{Relaxation at the boundary}
571 \label{subsec:LBC_bdy_relaxation}
573 In addition to a specific boundary condition specified as \np{cn\_tra} and \np{cn\_dyn3d}, relaxation on baroclinic velocities and tracers variables are available.
574 It is control by the namelist parameter \np{ln\_tra\_dmp} and \np{ln\_dyn3d\_dmp} for each boundary set.
576 The relaxation time scale value (\np{rn\_time\_dmp} and \np{rn\_time\_dmp\_out}, $\tau$) are defined at the boundaries itself.
577 This time scale ($\alpha$) is weighted by the distance ($d$) from the boundary over \np{nn\_rimwidth} cells ($N$):
579 \[
580 \alpha = \frac{1}{\tau}(\frac{N+1-d}{N})^2, \quad d=1,N
581 \]
583 The same scaling is applied in the Orlanski damping.
585 %----------------------------------------------
586 \subsection{Boundary geometry}
587 \label{subsec:LBC_bdy_geometry}
589 Each open boundary set is defined as a list of points.
590 The information is stored in the arrays $nbi$, $nbj$, and $nbr$ in the $idx\_bdy$ structure.
591 The $nbi$ and $nbj$ arrays define the local $(i,j)$ indexes of each point in the boundary zone and
592 the $nbr$ array defines the discrete distance from the boundary: $nbr=1$ means that
593 the boundary point is next to the edge of the model domain, while $nbr>1$ means that
594 the boundary point is increasingly further away from the edge of the model domain.
595 A set of $nbi$, $nbj$, and $nbr$ arrays is defined for each of the $T$, $U$ and $V$ grids.
596 Figure \autoref{fig:LBC_bdy_geom} shows an example of an irregular boundary.
598 The boundary geometry for each set may be defined in a namelist nambdy\_index or
599 by reading in a ``\ifile{coordinates.bdy}'' file.
600 The nambdy\_index namelist defines a series of straight-line segments for north, east, south and west boundaries.
601 One nambdy\_index namelist block is needed for each boundary condition defined by indexes.
602 For the northern boundary, \texttt{nbdysegn} gives the number of segments,
603 \jp{jpjnob} gives the $j$ index for each segment and \jp{jpindt} and
604 \jp{jpinft} give the start and end $i$ indices for each segment with similar for the other boundaries.
605 These segments define a list of $T$ grid points along the outermost row of the boundary ($nbr\,=\, 1$).
606 The code deduces the $U$ and $V$ points and also the points for $nbr\,>\, 1$ if \np{nn\_rimwidth}\forcode{>1}.
608 The boundary geometry may also be defined from a ``\ifile{coordinates.bdy}'' file.
609 Figure \autoref{fig:LBC_nc_header} gives an example of the header information from such a file, based on the description of geometrical setup given above.
610 The file should contain the index arrays for each of the $T$, $U$ and $V$ grids.
611 The arrays must be in order of increasing $nbr$.
612 Note that the $nbi$, $nbj$ values in the file are global values and are converted to local values in the code.
613 Typically this file will be used to generate external boundary data via interpolation and so
614 will also contain the latitudes and longitudes of each point as shown.
615 However, this is not necessary to run the model.
617 For some choices of irregular boundary the model domain may contain areas of ocean which
618 are not part of the computational domain.
619 For example, if an open boundary is defined along an isobath, say at the shelf break,
620 then the areas of ocean outside of this boundary will need to be masked out.
621 This can be done by reading a mask file defined as \np{cn\_mask\_file} in the nam\_bdy namelist.
622 Only one mask file is used even if multiple boundary sets are defined.
624 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
625 \begin{figure}[!t]
626 \begin{center}
627 \includegraphics[width=\textwidth]{Fig_LBC_bdy_geom}
628 \caption {
629 \protect\label{fig:LBC_bdy_geom}
630 Example of geometry of unstructured open boundary
631 }
632 \end{center}
633 \end{figure}
634 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
636 %----------------------------------------------
637 \subsection{Input boundary data files}
638 \label{subsec:LBC_bdy_data}
640 The data files contain the data arrays in the order in which the points are defined in the $nbi$ and $nbj$ arrays.
641 The data arrays are dimensioned on:
642 a time dimension;
643 $xb$ which is the index of the boundary data point in the horizontal;
644 and $yb$ which is a degenerate dimension of 1 to enable the file to be read by the standard \NEMO\ I/O routines.
645 The 3D fields also have a depth dimension.
647 From Version 3.4 there are new restrictions on the order in which the boundary points are defined
648 (and therefore restrictions on the order of the data in the file).
649 In particular:
651 \begin{enumerate}
652 \item The data points must be in order of increasing $nbr$,
653 ie. all the $nbr=1$ points, then all the $nbr=2$ points etc.
654 \item All the data for a particular boundary set must be in the same order.
655 (Prior to 3.4 it was possible to define barotropic data in a different order to
656 the data for tracers and baroclinic velocities).
657 \end{enumerate}
659 These restrictions mean that data files used with versions of the
660 model prior to Version 3.4 may not work with Version 3.4 onwards.
661 A \fortran utility {\itshape bdy\_reorder} exists in the TOOLS directory which
662 will re-order the data in old BDY data files.
664 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
665 \begin{figure}[!t]
666 \begin{center}
667 \includegraphics[width=\textwidth]{Fig_LBC_nc_header}
668 \caption {
669 \protect\label{fig:LBC_nc_header}
670 Example of the header for a \protect\ifile{coordinates.bdy} file
671 }
672 \end{center}
673 \end{figure}
674 %>>>>>>>>>>>>>>>>>>>>>>>>>>>>
676 %----------------------------------------------
677 \subsection{Volume correction}
678 \label{subsec:LBC_bdy_vol_corr}
680 There is an option to force the total volume in the regional model to be constant.
681 This is controlled by the \np{ln\_vol} parameter in the namelist.
682 A value of \np{ln\_vol}\forcode{=.false.} indicates that this option is not used.
683 Two options to control the volume are available (\np{nn\_volctl}).
684 If \np{nn\_volctl}\forcode{=0} then a correction is applied to the normal barotropic velocities around the boundary at
685 each timestep to ensure that the integrated volume flow through the boundary is zero.
686 If \np{nn\_volctl}\forcode{=1} then the calculation of the volume change on
687 the timestep includes the change due to the freshwater flux across the surface and
688 the correction velocity corrects for this as well.
690 If more than one boundary set is used then volume correction is
691 applied to all boundaries at once.
693 %----------------------------------------------
694 \subsection{Tidal harmonic forcing}
695 \label{subsec:LBC_bdy_tides}
697 %-----------------------------------------nambdy_tide--------------------------------------------
699 \nlst{nambdy_tide}
700 %-----------------------------------------------------------------------------------------------
702 Tidal forcing at open boundaries requires the activation of surface
703 tides (i.e., in \nam{\_tide}, \np{ln\_tide} needs to be set to
704 \forcode{.true.} and the required constituents need to be activated by
705 including their names in the \np{clname} array; see
706 \autoref{sec:SBC_tide}). Specific options related to the reading in of
707 the complex harmonic amplitudes of elevation (SSH) and barotropic
708 velocity (u,v) at open boundaries are defined through the
709 \nam{bdy\_tide} namelist parameters.\\
711 The tidal harmonic data at open boundaries can be specified in two
712 different ways, either on a two-dimensional grid covering the entire
713 model domain or along open boundary segments; these two variants can
714 be selected by setting \np{ln\_bdytide\_2ddta } to \forcode{.true.} or
715 \forcode{.false.}, respectively. In either case, the real and
716 imaginary parts of SSH and the two barotropic velocity components for
717 each activated tidal constituent \textit{tcname} have to be provided
718 separately: when two-dimensional data is used, variables
719 \textit{tcname\_z1} and \textit{tcname\_z2} for real and imaginary SSH,
720 respectively, are expected in input file \np{filtide} with suffix
721 \ifile{\_grid\_T}, variables \textit{tcname\_u1} and
722 \textit{tcname\_u2} for real and imaginary u, respectively, are
723 expected in input file \np{filtide} with suffix \ifile{\_grid\_U}, and
724 \textit{tcname\_v1} and \textit{tcname\_v2} for real and imaginary v,
725 respectively, are expected in input file \np{filtide} with suffix
726 \ifile{\_grid\_V}; when data along open boundary segments is used,
727 variables \textit{z1} and \textit{z2} (real and imaginary part of SSH)
728 are expected to be available from file \np{filtide} with suffix
729 \ifile{tcname\_grid\_T}, variables \textit{u1} and \textit{u2} (real
730 and imaginary part of u) are expected to be available from file
731 \np{filtide} with suffix \ifile{tcname\_grid\_U}, and variables
732 \textit{v1} and \textit{v2} (real and imaginary part of v) are
733 expected to be available from file \np{filtide} with suffix
734 \ifile{tcname\_grid\_V}. If \np{ln\_bdytide\_conj} is set to
735 \forcode{.true.}, the data is expected to be in complex conjugate
736 form.
738 Note that the barotropic velocity components are assumed to be defined
739 on the native model grid and should be rotated accordingly when they
740 are converted from their definition on a different source grid. To do
741 so, the u, v amplitudes and phases can be converted into tidal
742 ellipses, the grid rotation added to the ellipse inclination, and then
743 converted back (care should be taken regarding conventions of the
744 direction of rotation). %, e.g. anticlockwise or clockwise.
746 \biblio
748 \pindex
750 \end{document}
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Performance Evaluation Of BPSK And QPSK Modulation With LDPC Codes
Volume 02, Issue 01 (January 2013)
Performance Evaluation Of BPSK And QPSK Modulation With LDPC Codes
DOI : 10.17577/IJERTV2IS1531
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B. B. Badhiye, Dr. S. S. Limaye, 2013, Performance Evaluation Of BPSK And QPSK Modulation With LDPC Codes, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 02, Issue 01
(January 2013),
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• Authors : B. B. Badhiye, Dr. S. S. Limaye
• Paper ID : IJERTV2IS1531
• Volume & Issue : Volume 02, Issue 01 (January 2013)
• Published (First Online): 30-01-2013
• ISSN (Online) : 2278-0181
• Publisher Name : IJERT
• License: This work is licensed under a Creative Commons Attribution 4.0 International License
Text Only Version
Performance Evaluation Of BPSK And QPSK Modulation With LDPC Codes
B. B. Badhiye Dr. S. S. Limaye
Associate Professor, Deptt. Of Electronics Engg. Professor and Principal , Deptt. Of Electronics Engg.
M.I.E.T.,Gondia ,(M.S.), India J.I.T.,Nagpur,(m.s.), India
Abstract : BPSK and QPSK are the digital modulation schemes, QPSK and further/
QPSK are the best suited modulation techniques therefore more emphasize given on QPSK, which specially used in DVB area. A coding and modulation technique is studied where the coded bits of an
irregular LDPC are passed directly to a modulator. We evaluate and compare the bit error rate, signal constellations etc. In this paper, we consider a transmission over a Gaussian noise channel. Log
Likelihood Ratio (LLR) decoder can be suggested for receiver.
Index Terms: Differntial Quadrature Phase Shift Keying (DQPSK),Low Density Parity Check Codes(LDPC).
1. Introduction of BPSK/QOSK:
The BPSK and QPSK system considered in this paper,
illustrated in Figure(1), consists of a bit source,
QPSK symbols into information bits, which are fed to the bit sink. Typically, in a simulation environment, the bit sink simply counts the number of errors that occurred to gather statistics used for
investigating the performance of the system.
Figure(1)The QPSK Communication System.
1. BPSK is a real-valued constellation with two signal points: c(0) = A and c(1) = -A, where A is a scaling factor. This is shown in Figure 2 (a). The average complex baseband symbol energy is E8 =
E[c(i)2] = A2.
2. QPSK is a complex constellation with four
signal points,
c(i) = exp + 1 ,
transmitter, channel, receiver, and a bit sink. The bit
source generates a stream of information bits to be transmitted by the transmitter .typically, a random bit generator is employed as a bit source in simulations and this is the case herein as
well. The transmitter converts the bits into QPSK symbols and applies optional pulse shaping and up conversion. The output from the transmitter is fed through a channel, which in its simplest
form is an AWGN channel. The receiver block takes the output from the channel, estimates timing and phase offset, and demodulates the received
For I = 0, 1,2,3. It is convenient to include the 2 factor so that the average symbol energy is Es = E[||c(i)||2] = 2A2, double that of BPSK, but with the
same energy per transmitted bit as BPSK.[1][2][3][8]
2 Basics of LDPC
Figure(2),Block diagram of LDPC Encoder/Decoder.
We use the LDPC code to communicate over the noisy channel. LDPC codes form part of a larger family of codes, which are typically referred to as linear block codes.A code is termed a block code,
if the original information bit-sequence can be segmented into fixed- length message blocks, hereby denoted by u = u1, u2, .
. . , uK, each having K information digits. This implies that there is 2K possible distinct message blocks. For the sake of simplicity, we will here be giving examples for binary LDPC codes, i.e.
the codes are associated with the logical symbols/bits of (1, 0). The elements (1, 0) are said to constitute an alphabet or a finite field, where the latter are typically referred to as Galois
fields (GF). Using this terminology, a GF containing q elements is denoted by GF(q) and correspondingly, the binary GF is represented as GF(2). The LDPC encoder, is then capable of transforming
each input message block u according to a predefined set of rules into a distinct N-tuple (N-bit sequence) z, which is typically referred to as the codeword. The codeword length N, where N > K,
is then referred to as the block- length. Again, there are 2K distinct legitimate codewords corresponding to the 2K message blocks. This set of the 2K codewords is termed as a C(N,K) linear block
code. The word linear signifies that the modulo-2 sum of any two or more codewords in the code C(N,K) is another valid codeword. The number of non-zero symbols of a codeword z is called the
weight, whilst the number of bit-positions in which two codewords differ is termed as the distance. For instance, the distance between the codewords z1 = (1101001) and z2 = (0100101) is equal to
three. Subsequently, codewords that have a low number of binary ones are referred to as low-weight codewords. The minimum distance of a linear code, hereby denoted by dmin, is then determined by
the weight of that codeword in the code C(N,K), which has the minimum weight. The reason for this lies in the fact that the all-zero codeword is always part of a linear code and therefore, if a
codeword zx has the lowest weight from the 2K legitimate codewords, then the distance between zx and the all-zero codeword is effectively the minimum distance.[10][11][12][13][14][16]
3. LDPC Encoded QPSK:
Figure(3) shows a simplified block diagram of a channel coded communication system using linear block codes..[12][13][14][15]
Table 1.Differential Phase Shifts for /4 DQPSK using Gray Code:
Information Symbols, I(k) and Q (k) Differential Phase Shift, (k)
1 1 /4
0 1 3 /4
0 0 -3 /4
1 0 – /4
4. QPSK Constellation :
Figure(4) ,QPSK Constellations Phasewise
A conventional M-ary phase-shift keying (MPSK) signal constellation is denoted by SM = { sk = e2(k/M)j
: k = 0, 1,.., M
1}, where the energy has been constrained to unity, Clockwise rotation over an angle (see Figure 2) leads to the constellation.
Such a ( complex ) modulation scheme can be seen as two ( real ) M- ary pulse amplitude modulations ( MPAM s ) in parallel – one on the inphase ( I channel ) and the other on the quadrature channel
(Q channel) . For the QPSK and 8 PSK constellations.[6][8][9]
4.1 Separate I and Q Component
It was shown in[2), that by rotating the signal constellation and separately interleaving the I and Q components, an improved performance can be obtained for a QPSK system without effecting its
bandwidth efficiency. In case of transmission of N symbols, each taken from the rotated constellation. S
M, let the sequence of I components xxxx = (xo,x1 ,,
xN -1)And the sequence of Q components y = (yo, y1,, yN -1)Be interleaved by the I interleaver n and the Q interleaver Q, respectively, resulting in the sequences x = n(x) = (xo, x1,.,xN 1) and y = p
=(yo, y1,..,yN 1). The transmitted waveform for the rotated and interleaved system is given by
(±1 ± j)} for QPSK). The sequence t is then transmitted over the Gaussian channel. The channel adds white Gaussian noise w, with uncorrelated real and imaginary parts, each having a variance 2 ,
resulting in the sequence y = t + w. Based on the received sequence y, a decision is taken about the transmitted code word. The ML decision rule is given by
= arg min d2 (y, t(c))
where d2 (a, b) is the Euclidean distance between the sequences a and b, and t(c) is the sequence of transmitted data symbols that correspond to the code word c. Hence, he receiver selects the code
word that corresponds to the sequence of symbols that is at minimum Euclidean distance of the received sequence y. The bit error rate (BER) is given by
= (c |c ) (b , b)
+ j iP t iTs sin 2 .
s t = x iP t iTs cos(2 )
Ts is the is the symbol period and fc is the carrier frequency.
c t y
Figure(5),BPSK/QPSK transmission over a Guassian channel
The transceiver for BPSK (QPSK) transmission over a Gaussian channel is shown in gure(5) The code bits {cn|n = 1, . . . , N } are rst mapped on the symbols
{tn} (tn {x0 , x0 } for BPSK and tn {x0
, =1;
where P r(cj |ci ) is the probability that the code word cj is selected at the receiver when the code word ci is transmitted, P r(ci ) is the prior probability that the code word ci is transmitted,
dH (bj , bi) is the Hamming distance between the information words bj and bi ), that correspond to the code words cj and ci
, respectively, and K is the length of the information word. In the following we assume that all information words, hence all code words, are equiprobable, i.e. P r(ci ) = 1/2K .[4][5][6][7]
RESULT :
The constellation of BPSK and QPSK discussed theoretically and found that QPSK is far better than BPSK because average symbol energy is double than BPSK and it transmit two bits simultaneously: The
results are simulated in Matlab. Figure (5) shows the constellation symbol for various phases shown in table 1. Figure (6) gives the AWGN over the constellation points and at the last figure (7)
shows BER versus Eb/No. curve for theoretical as well as simulated.
For digital transmission of data QPSK is best suited method and LDPC outperforms than other earlier codes like turbo codes, RS-CC codes.
1. Dr.Kamilo Feher: A book on, Wireless Digital Communication -Modulation and Spread Spectrum Application. (EEE)
2. Thedore S. Rappaport: A Book on Wireless Communication, Second Edition.
3. Simon Haykin A Book on Communication System
QPSK constellation
4. Principles of Communication Systems Simulation with Wireless Applications, William H. Tranter,K.Sam Shanmugan, Theodore S. Rappaport, Kurt L. Kosbar.
5. Simon Hayin, A Book on Digital Communication Willey Student Edition.
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Figure(6),QPSK Phase constellation Points
6. Kratochvil T. Utilization of MATLAB for the Digital Signal Transmission Simulation and Analysis in DTV and DVB area. In proceedings of the 11th conference MATLAB 2003,p.318-322, Prague, Czech
Republic, 2003.
7. MATLAB An Introduction with Applications AMOS GILAT, Willey Students Edition.
8. Akaiwa and Nagata ( 1987), EIA (1990)
QPSK constellation with noise
9. Populis, A., Probabbility Random Variables and Stohastic Processes,Mc. Graw-Hill Company, New York, 1965.
10. C.Schlegel andD. J..Costello Jr.,Bandwidth efficient coding for fading channels:Code construction and performance analysis.IEEEJournalof Selected Areas in C
11. S.B.Slimane,An improved PSK scheme for fading channels,IEEE Trans.on vehicular Tech.,vol.47,no.2,pp703- 710,May 1998
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Figure(7),AWGN Noise
Simulation of BER/SER for QPSK with Gray coding
12. D.J.MacKay,Good error correcting codesbased on very spase matrices.IEEE Trans Information Theory,no.2pp399- 431,March,1999.
13. R.G.Gallager,Low Density Check Codes.IRE
BER-simulated SER-theoritcal
BER-simulated SER-theoritcal
10 Trans.info.Theory,IT-8:21-28,Jan1962.
-1 [14] .G.Gallager,Low Density Check Codes.
Number21 in Reserch monograph series,MIT
10 Press,Cambridge,Mass.,1963.
-3 [15] G.D.Forney,Concatenated codes,Technical report 37,MIT,1966.
[16] Piraporn Limpaphayom,Student Member,IEEE,and Kim A.Winick,Senior Member, IEEE,Power-and Bandwidth-Efficient communication Using LDPC Codes.
Figure(8),BER Vs.Eb /No curve
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A wave is a disturbance that propagates through space and time, usually with transference of energy. While a mechanical wave exists in a medium (which on deformation is capable of producing elastic
restoring forces), waves of electromagnetic radiation (and probably gravitational radiation) can travel through vacuum, that is, without a medium. Waves travel and transfer energy from one point to
another, often with little or no permanent displacement of the particles of the medium (that is, with little or no associated mass transport); instead there are oscillations around almost fixed
Agreeing on a single, all-encompassing definition for the term wave is non-trivial. A vibration can be defined as a back-and-forth motion around a point m around a reference value. However, defining
the necessary and sufficient characteristics that qualify a phenomenon to be called a wave is, at least, flexible. The term is often understood intuitively as the transport of disturbances in space,
not associated with motion of the medium occupying this space as a whole. In a wave, the energy of a vibration is moving away from the source in the form of a disturbance within the surrounding
medium (Hall, 1980: 8). However, this notion is problematic for a standing wave (for example, a wave on a string), where energy is moving in both directions equally, or for electromagnetic / light
waves in a vacuum, where the concept of medium does not apply.
For such reasons, wave theory represents a peculiar branch of physics that is concerned with the properties of wave processes independently from their physical origin (Ostrovsky and Potapov, 1999).
The peculiarity lies in the fact that this independence from physical origin is accompanied by a heavy reliance on origin when describing any specific instance of a wave process. For example,
acoustics is distinguished from optics in that sound waves are related to a mechanical rather than an electromagnetic wave-like transfer / transformation of vibratory energy. Concepts such as mass,
momentum, inertia, or elasticity, become therefore crucial in describing acoustic (as opposed to optic) wave processes. This difference in origin introduces certain wave characteristics particular to
the properties of the medium involved (for example, in the case of air: vortices, radiation pressure, shock waves, etc., in the case of solids: Rayleigh waves, dispersion, etc., and so on).
Other properties, however, although they are usually described in an origin-specific manner, may be generalized to all waves. For example, based on the mechanical origin of acoustic waves there can
be a moving disturbance in space-time if and only if the medium involved is neither infinitely stiff nor infinitely pliable. If all the parts making up a medium were rigidly bound, then they would
all vibrate as one, with no delay in the transmission of the vibration and therefore no wave motion (or rather infinitely fast wave motion). On the other hand, if all the parts were independent, then
there would not be any transmission of the vibration and again, no wave motion (or rather infinitely slow wave motion). Although the above statements are meaningless in the case of waves that do not
require a medium, they reveal a characteristic that is relevant to all waves regardless of origin: within a wave, the phase of a vibration (that is, its position within the vibration cycle) is
different for adjacent points in space because the vibration reaches these points at different times.
Similarly, wave processes revealed from the study of wave phenomena with origins different from that of sound waves can be equally significant to the understanding of sound phenomena. A relevant
example is Young's principle of interference (Young, 1802, in Hunt, 1978: 132). This principle was first introduced in Young's study of light and, within some specific contexts (for example,
scattering of sound by sound), is still a researched area in the study of sound.
Periodic waves are characterized by crests (highs) and troughs (lows), and may usually be categorized as either longitudinal or transverse. Transverse waves are those with vibrations perpendicular to
the direction of the propagation of the wave; examples include waves on a string and electromagnetic waves. Longitudinal waves are those with vibrations parallel to the direction of the propagation
of the wave; examples include most sound waves.
When an object bobs up and down on a ripple in a pond, it experiences an orbital trajectory because ripples are not simple transverse sinusoidal waves.
Ripples on the surface of a pond are actually a combination of transverse and longitudinal waves; therefore, the points on the surface follow orbital paths.
All waves have common behaviour under a number of standard situations. All waves can experience the following:
• Reflection - wave direction change from hitting a reflective surface
• Refraction - wave direction change from entering a new medium
• Diffraction - wave circular spreading from entering a hole of comparable size to their wavelengths
• Interference - superposition of two waves that come into contact with each other (collide)
• Dispersion - wave splitting up by frequency
• Rectilinear propagation - The movement of light wave in a straight line
A wave is polarized if it can only oscillate in one direction. The polarization of a transverse wave describes the direction of oscillation, in the plane perpendicular to the direction of travel.
Longitudinal waves such as sound waves do not exhibit polarization, because for these waves the direction of oscillation is along the direction of travel. A wave can be polarized by using a
polarizing filter.
Examples of waves include:
• Ocean surface waves, which are perturbations that propagate through water.
• Radio waves, microwaves, infrared rays, visible light, ultraviolet rays, x-rays, and gamma rays make up electromagnetic radiation. In this case, propagation is possible without a medium, through
vacuum. These electromagnetic waves travel at 299,792,458 m/s in a vacuum.
• Sound — a mechanical wave that propagates through air, liquid or solids.
• waves of traffic (that is, propagation of different densities of motor vehicles, etc.) — these can be modelled as kinematic waves, as first presented by Sir M. J. Lighthill
• Seismic waves in earthquakes, of which there are three types, called S, P, and L.
• Gravitational waves, which are fluctuations in the curvature of spacetime predicted by general Relativity. These waves are nonlinear, and have yet to be observed empirically.
• Inertial waves, which occur in rotating fluids and are restored by the Coriolis effect.
Mathematical description
From a mathematical point of view, the most primitive (or fundamental) wave is harmonic (sinusoidal) wave which is described by the equation $f(x,t) = A\sin(\omega t-kx)),$ where $A$ is the amplitude
of a wave - a measure of the maximum disturbance in the medium during one wave cycle (the maximum distance from the highest point of the crest to the equilibrium). In the illustration to the right,
this is the maximum vertical distance between the baseline and the wave. The units of the amplitude depend on the type of wave — waves on a string have an amplitude expressed as a distance (meters),
sound waves as pressure (pascals) and electromagnetic waves as the amplitude of the electric field (volts/meter). The amplitude may be constant (in which case the wave is a c.w. or continuous wave),
or may vary with time and/or position. The form of the variation of amplitude is called the envelope of the wave.
The wavelength (denoted as $\lambda$) is the distance between two sequential crests (or troughs). This generally has the unit of meters; it is also commonly measured in nanometers for the optical
part of the electromagnetic spectrum.
A wavenumber $k$ can be associated with the wavelength by the relation
$k = \frac{2 \pi}{\lambda}. \,$
The period $T$ is the time for one complete cycle for an oscillation of a wave. The frequency $f$ (also frequently denoted as $u$) is how many periods per unit time (for example one second) and is
measured in hertz. These are related by:
$f=\frac{1}{T}. \,$
In other words, the frequency and period of a wave are reciprocals of each other.
The angular frequency $\omega$ represents the frequency in terms of radians per second. It is related to the frequency by
$\omega = 2 \pi f = \frac{2 \pi}{T}. \,$
There are two velocities that are associated with waves. The first is the phase velocity, which gives the rate at which the wave propagates, is given by
$v_p = \frac{\omega}{k} = {\lambda}f.$
The second is the group velocity, which gives the velocity at which variations in the shape of the wave's amplitude propagate through space. This is the rate at which information can be transmitted
by the wave. It is given by
$v_g = \frac{\partial \omega}{\partial k}. \,$
The wave equation
The wave equation is a differential equation that describes the evolution of a harmonic wave over time. The equation has slightly different forms depending on how the wave is transmitted, and the
medium it is traveling through. Considering a one-dimensional wave that is travelling down a rope along the x-axis with velocity $v$ and amplitude $u$ (which generally depends on both x and t), the
wave equation is
$\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}. \,$
In three dimensions, this becomes
$\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} = abla^2 u. \,$
where $abla^2$ is the Laplacian.
The velocity v will depend on both the type of wave and the medium through which it is being transmitted.
A general solution for the wave equation in one dimension was given by d'Alembert. It is
$u(x,t)=F(x-vt)+G(x+vt). \,$
This can be viewed as two pulses travelling down the rope in opposite directions; F in the +x direction, and G in the −x direction. If we substitute for x above, replacing it with directions x, y, z,
we then can describe a wave propagating in three dimensions.
The Schrödinger equation describes the wave-like behaviour of particles in quantum mechanics. Solutions of this equation are wave functions which can be used to describe the probability density of a
particle. Quantum mechanics also describes particle properties that other waves, such as light and sound, have on the atomic scale and below.
Traveling waves
Simple wave or traveling wave, also sometimes called progressive wave is a disturbance that varies both with time $t$ and distance $z$ in the following way:
$y(z,t) = A(z, t)\sin (kz - \omega t + \phi), \,$
where $A(z,t)$ is the amplitude envelope of the wave, $k$ is the wave number and $\phi$ is the phase. The phase velocity v[p] of this wave is given by
$v_p = \frac{\omega}{k}= \lambda f, \,$
where $\lambda$ is the wavelength of the wave.
Standing wave
A standing wave, also known as a stationary wave, is a wave that remains in a constant position. This phenomenon can occur because the medium is moving in the opposite direction to the wave, or it
can arise in a stationary medium as a result of interference between two waves traveling in opposite directions.
The sum of two counter-propagating waves (of equal amplitude and frequency) creates a standing wave. Standing waves commonly arise when a boundary blocks further propagation of the wave, thus causing
wave reflection, and therefore introducing a counter-propagating wave. For example when a violin string is displaced, longitudinal waves propagate out to where the string is held in place at the
bridge and the " nut", where upon the waves are reflected back. At the bridge and "nut", the two opposed waves are in antiphase and cancel each other, producing a node. Halfway between two nodes
there is an antinode, where the two counter-propagating waves enhance each other maximally. There is on average no net propagation of energy.
Also see: Acoustic resonance, Helmholtz resonator, and organ pipe
Propagation through strings
The speed of a wave traveling along a vibrating string (v) is directly proportional to the square root of the tension (T) over the linear density (μ):
$v=\sqrt{\frac{T}{\mu}}. \,$
Transmission medium
The medium that carries a wave is called a transmission medium. It can be classified into one or more of the following categories:
• A linear medium if the amplitudes of different waves at any particular point in the medium can be added.
• A bounded medium if it is finite in extent, otherwise an unbounded medium.
• A uniform medium if its physical properties are unchanged at different locations in space.
• An isotropic medium if its physical properties are the same in different directions.
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CAT 2020 | Slot 3 | Quantitative Aptitude | 2IIM CAT Coaching
This question is from Geometry. In this question, we need to find the area of the trapezium. CAT Geometry is an important topic with lots of weightage in the CAT Exam. In CAT Exam one can generally
expect 4-6 questions from Geometry. Make sure you master Geometry problems by solving CAT Previous Year Paper.
Question 24 : In a trepezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trepezium in sq. cm is
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Explanatory Answer
Let's draw the Trapezium.
Now drop a perpendicular from D on to AB touching AB at E.
In ΔAED, ∠AED = 90°, ∠DAE = 45°, therefore ∠EDA = 45°.
Clearly ΔAED is an iscoceles right triangle and Quadrilateral BCDE is a rectangle.
Therefore, BC = DE = 4cm.
Since ΔAED is an iscoceles triangle, AE = ED = 4cm.
Area of the Trepezium = Area of ΔAED + Area of Rectangle BCDE
Area of the Trepezium = \\frac{1}{2}) × AE × ED + DC × BC
Area of the Trepezium = \\frac{1}{2}) × 4 × 4 + 5 × 4
Area of the Trepezium = 8 + 20
Area of the Trepezium = 28 cm^2
The question is "In a trepezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trepezium in sq. cm is"
Hence, the answer is, "28"
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Essay Hdc Case
Case 2: Health Development Corporation HBS 9-200-049
1. Did the purchase of the Lexington Club real estate increase the value of Heatlh Development Corporation (HDC)? Calculate the NPV of the purchase.
• Use pre-tax cashflows.
• Assume the revenues of the Lexington Club grow by 5% per year.
• Assume that the appropriate discount rate for real estate cashflows was 10%.
• Assume a 20 year life of the facility.
(Hint: In calculating the NPV of the decision to buy the real estate, you only need to consider the incremental cashflows resulting from the decision).
=11203677.75 – 6,500,00 = $4,703,677.75
• The change in incremental cash flow can be examined through looking at the factors causing change …show more content…
The holding company is owned by the shareholders of HDC.
• HDC is then sold to TSI for an amount $X (which equals TSI’s valuation of HDC in this agreement).
• The agreement to sell HDC to TSI includes the sale of the Lexington Club by HDC to ABC for $6.5 million.
• HDC leases back the Lexington Club for 20 years at $525,000 per year.
• The bank lends amount $Y to ABC for the purchase of Lexington. The bank loan is repaid in 20 equal, year-end, payments. However, the bank insists that the lease payments must be 110% of the annual
repayments of the bank loan. Note that the equal bank loan repayments include interest and principal.
• The original owners of HDC put in an initial equity of $(6.5 mn – Y) into ABC to make up the difference between the purchase price and the bank loan.
3. How much is Y?
Interest Rate: r =8. 5%
Annual Mortgage Repayments: C= $ 477,272.73 being of $525,000.
Loan length: 20 years, t=20
4. How much is X?
EBITDA $ 3,229,000.00
New Lease difference $ 400,000.00
New EBITDA $ 3,629,000.00
Multiple $ 18,145,000.00
Mortage -$ 5,750,000.00
Sale $ 6,500,000.00 $ 18,895,000.00
Less Debt $ 1,917,000.00
Equity Value $ 16,978,000.00
• The interest payment of 504 has been paid for 1999 and is not due for 2000.
• That the 6.5 -y is an additional injection of equity by the shareholders of
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Robotics Probabilistic Generative Laws
world state at time \(t\)
measurement data at time \(t\) (e.g. camera images)
control data (change of state in the environment) at time \(t\)
State Evolution
$$p(x_t | x_{0:t-1} z_{1:t-1}, u_{1:t})$$
State Transition Probability
$$p(x_t | x_{0:t-1} z_{1:t-1}, u_{1:t}) = p ( x_t | x_{t-1}, u_t)$$
The world state at the previous time-step is a sufficient summary of all that happened in previous time-steps.
Measurement Probability
$$p(z_t | x_{0:t}, z_{1:t-1}, u_{1:t}) = p(z_t | x_t)$$
The measurement at time-step \(t\) is often just a noisy projection of the world state at time-step \(t\).
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Thursday Open Thread | Colorado Pols
51 thoughts on “Thursday Open Thread”
1. There you have it folks – the morning news in a nutshell.
Have a nice day!
The roller-coaster ride with initial unemployment claims continues. Whereas last week’s report offered a little encouragement, today’s report from the Department of Labor this morning did
the opposite.
With revisions once again pointing in the wrong direction, the new totals inched higher, despite expectations to the contrary.
1. And no, you’re not represented by the gods here.
2. Or is this a source of your concern …. I recall you’re a westsloper
GRAND JUNCTION, Colo. — The general manager of the Grand Junction convention center and theatre has resigned after he was found naked at a bus station.
1. dude has a penchant for theater so maybe he’s leaving GJ for the lullaby of Broadway and just decided to travel light …?
1. was Ambien involved ?
2. Actually, I wasn’t referencing any one story (although the GJ story was bazaar, at best). It just seemed every on-line paper I read this morning, the reporters/columnists had a negative
spin – no matter the subject.
Glad to read there was some good news (below). If I have time today, I’ll take a refresher course.
3. There are a number of sites like this:
That will help balance out things.
Have a wonderful day.
4. “If life seems jolly rotten
There’s something you’ve forgotten
And that’s to laugh and smile and dance and sing.
When you’re feeling in the dumps
Don’t be silly chumps
Just purse your lips and whistle – that’s the thing.
And…always look on the bright side of life…
Always look on the light side of life…”
1. It’s Barney!! Yay!!
1. VanDammer is quoting Monty Python dwyer. Its kinda satire.
This should cheer you up though:
now turn that frown upside down !
2. Americans See Biggest Home Equity Jump in 60 Years: Mortgages
Consumer Prices in U.S. Fell in May by Most in Three Years
Truckers as Leading Indicator Show Stable U.S. Economic Growth
More Gains for Southern California Home Sales and Median Prices
Denver’s housing trifecta: Bidding wars, low inventory, rising prices
1. …now if all my purple Salvia would just stop dying…
1. Mine falls over every year. Every year I say to myself, “Go buy some support hoops for the salvia this year.” Every year it explodes, looks lovely for two days, then falls over in the
first breeze over 8 mph. Every year I grit my teeth and say “next year for sure.” This has gone on for six years now…
1. I have a garden patch in front of my house where two 25+ year old Juniper bushes used to live. They were tried and found guilty of harboring a spider infestation, so I took a chainsaw
to ’em. I amended the soil with compost and gave it a good month of exposure to the elements, tilled it multiple times by hand, and then planted Russian Sage, wildflowers, and purple
The Russian Sage is happy as can be, the wildflower seed is sprouting, the couple sunflower sprouts I rescued from someone who was throwing them out are fine, but the salvia dried up
blooms-first and died.
1. They’re the gardener’s best friend, along with ladybugs.
I used to have a little evergreen shrub, I think a juniper also, that we got rid of because we wanted something else, and some plants just wouldn’t take to that spot for a long
time. This was with our having made similar soil amendments, I believe mostly bagged garden soil and compost. The Grand Mesa beardtongue that I planted last year finally proved to
be the one that could live there, although it didn’t grow up any last year.
I haven’t looked up to see if junipers somehow affect the soil to make it harder for certain plants to thrive there or not, but I’m suspicious.
1. If they would have stayed in the junipers, we could have had a peaceful standoff, but they kept coming into the house. My quality of life has improved greatly since the
junipers were removed, thanks to the significant decrease in the household spider population. I won’t use insecticides indoors (or out for that matter, except for the
occasional can of wasp spray since one of my housemates is allergic and we can’t take too many chances with those). So the junipers had to go. Also, they were really ugly and
had been poorly cared for, leading them to go so brown on the insides that trimming them back enough to leave the sidewalk free resulted in ugly brown spiny branches showing.
Anyway, that’s really interesting that you had a similar experience! I was wondering if the juniper affected the soil in a way adverse to other flora. I’ll give the CSU
extension office a call and see if they know anything.
I have two different types of Russian sage there now and both have rooted nicely. I’ll see how the wildflowers do. They’re just seedlings right now. It’ll be a shame if they
perish, as it’s a “bee rescue” mix and I’m installing honeybees soon.
1. and got a plant called “lavender bee balm.” Latin name monarda fistula menthaefolia. It’s what grew around Mesa Verda back when the Anasazi still lived there. They grow
the kind of globular flowers bees seem to love. Globe thistles are cool for that, too. In my veggie garden, the oregano is the most popular.
My sister keeps bees, so I get little bits of knowledge from her. Good on you for getting them. But don’t get too discouraged if you lose a hive or two in the winter…
1. I love anything purple 🙂
The bees actually won’t be mine. A local beekeeper is using my home as a host family for a couple of hives, and he’ll be taking care of them — I get honey and the
company of lovely bees, and he gets to add to his collection and increase his total honey/comb/pollen output, win-win!
1. then you may want to skip the oregano. Honey tastes exactly like the flowers the bees pollinate. (That means spring honey tastes like all the blooming fruit tree
blossoms. Yum….)
2. it’s a must see, and I applaude everyone who supports bees in any way they can.
My wife’s grandparents were bee keepers. Our wedding was held at the “Honey Home” (apt), which was the origiinal homestead and site of a couple of generations of
See this video.
3. a forester told me that conifers make soil acidic, so acidic that aspens can’t easily grow to maturity around conifers.
Bees. Einstein gave humans 3 years of survival if bees don’t make it. I notice many, many more this year than I have ever seen up here in the high lands
1. Maybe tomatoes would do well in those spots.
1. If it’s acidic, they’ll tell you how much lime to add.
1. but I’ve only ever used it in my veggie garden, not the flower one. I’ll check that out…
The first questioner, Joe Jarvis, set the tone for the meeting in the Masonic Lodge meeting hall when he asked, “Are you with us or against us?”
Another questioner later put a finer point on it, demanding to know whether the sheriff would stand against federal forces “when the tanks are rolling down the street” and federal agencies
are moving to confiscate guns.
It COULD Happen Here…
Dunlap and McKee said they were watching with interest in the notion that an outside threat to American sovereignty could be taking shape in the form of agreements, such as Agenda 21, with
the United Nations.
Some elements of Agenda 21 are “something we should be scared to death of,” as it seems to be intended to drive people out of rural areas and into metropolitan areas, McKee said.
“We have to be aware of it when we’re considering taking grant money. We’re aware of it as an issue.”
A Cryin’ Shame…Deputies LOVE pulling over armed traffic scofflaws!
If anything, the sheriffs suggested,laws are too restrictive already on gun ownership.
“It’s a crying shame that lawful citizen has to have a license to pack a gun,” McKee said.
When his deputies make a stop they know to involve a person who holds a concealed-weapons permit, “We consider that person a friend,” Hilkey said.
1. that the uranium tailings removal project wasn’t completely successful.
4. I know some polsters do.
2. XOXO
1. 1) start two wars
2) don’t count them in the budget
3) call them overseas contingency operations
DONE. Whew ! I had to roll up my sleeves for that one; off to lunch.
1. that longer and chartier . . . thrown in a couple of those hugely stimulative and deficit-reducing budget-balancing trickle-down tax cuts . . .
2. I’ve never heard of “political math blog,” but something tells me it hasn’t been around long enough to take on its word yet.
3. Just look at it. Seriously.
You just absolutely killed every argument that might have gone before.
Bush: Spending up ~$1T over 8 years (from ~2.5T to ~3.5T)
Obama: Small bump in 2009, declining since. Spending flat over 3 years.
Everything else in your list is spin.
Like: Why use CBO estimates? Because they include more realistic estimates of entitlement spending (which isn’t discretionary). This also helps explain the spike in 2009. More entitlement
spending when the economy faltered.
Why assign 2009 to Bush, when Obama ended up signing the budget? That budget was late, but in the meanwhile, govt spending continued under CRs signed by Bush.
It’s too bad you’re so uncritical in your thinking when you agree with the results.
Even the AP called bullshit on the ‘thrifty’ argument.
The MarketWatch study finds spending growth of only 1.4 percent over 2010-2013, or annual increases averaging 0.4 percent over that period. Those are stunningly low figures considering that Obama
rammed through Congress an $831 billion stimulus measure in early 2009 and presided over significant increases in annual spending by domestic agencies at the same time the cost of benefit
programs like Social Security, Medicare and the Medicaid were ticking steadily higher.
A fairer calculation would give Obama much of the responsibility for an almost 10 percent budget boost in 2009, then a 13 percent increase over 2010-2013, or average annual growth of spending of
just more than 3 percent over that period.
So, how does the administration arrive at its claim?
First, there’s the Troubled Assets Relief Program, the official name for the Wall Street bailout. First, companies got a net $151 billion from TARP in 2009, making 2010 spending look smaller.
Then, because banks and Wall Street firms repaid a net $110 billion in TARP funds in 2010, Obama is claiming credit for cutting spending by that much.
The combination of TARP lending in one year and much of that money being paid back in the next makes Obama’s spending record for 2010 look $261 billion thriftier than it really was. Only by that
measure does Obama “cut” spending by 1.8 percent in 2010 as the analysis claims.
The federal takeover of Fannie Mae and Freddie Mac also makes Obama’s record on spending look better than it was. The government spent $96 billion on the Fannie-Freddie takeovers in 2009 but only
$40 billion on them in 2010. By the administration’s reckoning, the $56 billion difference was a spending cut by Obama.
Taken together, TARP and the takeover of Fannie and Freddie combine to give Obama an undeserved $317 billion swing in the 2010 figures and the resulting 1.8 percent cut from 2009. A fairer
reading is an almost 8 percent increase.
The Fat Lady is warming up….
The MarketWatch study finds…
They’re just uncritically passing on the same drivel you posted above (from MarketWatch).
But let’s look at this one…
Banks were bailed out in 2009, not 2010.
Banks paid back most of that in 2010.
That makes spending smaller in 2010.
What’s wrong with that? TARP was passed by Bush. He gets credit for the spending.
The Fannie Mae/Freddie Mac stuff is nonsensical. Maybe you understand it well enough to explain it to me. But the right-leaning editorial slant of this piece makes me suspicious. (If you want to
see legislation that was “rammed through”, I suggest you look at how the Bush tax cuts were passed. The stimulus doesn’t come close to that.)
1. ellbee neglected to copy and paste their conclusion:
All told, government spending now appears to be growing at an annual rate of roughly 3 percent over the 2010-2013 period…
… which is still significantly lower than anyone on the Forbes chart, especially Reagan and Bush II.
Why can’t you ever articulate a position on your own? You always use some sort of prop and throw in a snide remark. I guess you think you are being funny, but its certainly not informative.
1. In fact, my reply was a sarcastic retort to Air’s debunked chart that he posted without much comment.
Does his use of what irks you so bother you as much as mine?
1. thanks for asking, though.
I don’t smoke pot, so it does not affect me in the slightest, but I don’t understand how this is not considered illegal search and seizure. Unlike alcohol breath tests, they are not arresting people
for being under the influence, but for possession. To this non-lawyer, this seems completely unconstitutional. Attorneys — will you weigh in?
1. If they use sniffer dogs, those aren’t considered an invasion of privacy, but once the dogs sound they have probable cause to search.
It’s an odd ruling, considering the court not long ago put limits on using things like thermal imagers on private houses for very similar reasons.
(I think the dog sniffing thing was a SCOTUS ruling – anyone?)
and the banner ads that show up on Pols are MUCH different than the ones in Denver….
And then the usual bunch of cell phone ads…
1. Troops suicide above the fold.
Raped in the Military, Then Raped by the House of Representatives
Anyone who served in the military during the last decade has heard about a woman service member who was raped by another American. Military rape is far more common than anyone would like
to admit. Writing in 2008, Representative Jane Harmon opined that a serving woman was “more likely to be raped by a fellow soldier than killed by enemy fire in Iraq.”
According to the Pentagon’s Sexual Assault Prevention and Response Office, in 2011, 3,192 sexual assaults were reported, a comparable number to the year before. Of those, 490 were court
martialed, and just slightly over 100 were discharged or jailed. That’s three percent (although you’ll have to work your way through some obfuscatory math to get to that number.)
But the same Pentagon office also estimates that friendly-fire rape is vastly under reported. They estimate the real number of friendly-fire rapes at around 19,000 every year – over six
times the reported number. Do the math and a service member-rapist is likely to be convicted in only one-half of one percent of estimated rapes. Only about one-quarter of 1 percent
(.0025%) went to a military prison. The other quarter percent were simply separated from service, free to prey on civilian women. In fairness, leaders are now trying, but rape still
mostly gets a free pass in the services.
Feeling sick yet?
Well stay near the bucket, because House Republicans are poised to again deny abortion services to service members who were raped and then became rape-pregnant. Even if they are raped by
an enemy combatant they can’t get emergency abortion services.
1. .
I assume its worst in the Army, safest in the Air Force ?
1. I know the SARC in the Colorado NG and she can’t even get numbers from DoD by service.
But I suspect Army will be #1, followed by Air Force (not USMC) due to deployment profiles and numbers.
2. .
I had misplaced Dick Armey’s phone number. Think I’ll give him a call an catch up on ol’ times.
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Note: For background information, please see my introduction to Cast Vote Records processing and theory here: Statistical Detection of Irregularities via Cast Vote Records.
Since I posted my initial analysis of the Henrico CVR data, one comment was made to me by a member of the Texas election integrity group I have been working with: We have been assuming, based on
vendor documentation and the laws and requirements in various states, that when a cast vote record is produced by vendor software the results are sorted by the time the ballot was recorded onto a
scanner. However, when looking at the results that we’ve been getting so far and trying to figure out plausible explanations for what we were seeing, he realized it might be the case that the
ordering of the CVR entries are being done by both time AND USB stick grouping (which is usually associated with a specific scanner or precinct) but then simply concatenating all of those results
While there isn’t enough information in the Henrico CVR files to breakout the entries by USB/Scanner, and the Henrico data has record ID numbers instead of actual timestamps, there is enough
information to break out them by Precinct, District and Race, with the exception of the Central Absentee Precincts (CAP) entries where we can only break them out by district given the metadata alone.
However, with some careful MATLAB magic I was able to cluster the results marked as just “CAP” into at least 5 different sub-groupings that are statistically distinct. (I used an exponential moving
average to discover the boundaries between groupings, and looking at the crossover points in vote share.) I then relabeled the entries with the corresponding “CAP 1”, “CAP 2”, … , “CAP 5” labels as
appropriate. My previous analysis was only broken out by Race ID and CAP/Non-CAP/Provisional category.
Processing in this manner makes the individual distributions look much cleaner, so I think this does confirm that there is not a true sequential ordering in the CVR files coming out of the vendor
software packages. (If they would just give us the dang timestamps … this would be a lot easier!)
I have also added a bit more rigor to the statistics outlier detection by adding plots of the length of observed runs (e.g. how many “heads” did we get in a row?) as we move through the entries, as
well as the plot of the probability of this number of consecutive tosses occurring. We compute this probability for K consecutive draws using the rules of statistical independence, which is P
([a,a,a,a]) = P(a) x P(a) x P(a) x P(a) = P(a)^4. Therefore the probability of getting 4 “heads” in a row with a hypothetical 53/47 weighted coin would be .53^4 = 0.0789. There are also plotted lines
for a probability 1/#Ballots for reference.
The good news is that this method of slicing the data and assuming that the Vendor is simply concatenating USB drives seems to produce much tighter results that look to obey the expected IID
distributions. Breaking up the data this way resulted in no plot breaking the +/- 3/sqrt(N-1) boundaries, but there still are a few interesting datapoints that we can observe.
In the plot below we have the Attorney Generals race in the 4th district from precinct 501 – Antioch. This is a district that Miyares won handily 77%/23%. We see that the top plot of the cumulative
spread is nicely bounded by the +/- 3/sqrt(N-1) lines. The second plot from the top gives the vote ratio in order to compare with the work that Draza Smith, Jeff O’Donnell and others are doing with
CVR’s over at Ordros.com. The second from bottom plot gives the number k of consecutive ballots (in either candidates favor) that have been seen at each moment in the counting process. And the bottom
plot raises either the 77% or 23% overall probability to the k-th power to determine the probability associated with pulling that many consecutive Miyares or Herring ballots from an IID distribution.
The most consecutive ballots Miyares received in a row was just over 15, which had a .77^15 = 0.0198 or 1.98% chance of occurring. The most consecutive ballots Herring received was about 4, which
equates to a probability of occurrence of .23^4 = 0.0028 or 0.28% chance. The dotted line on the bottom plot is referenced at 1/N, and the solid line is referenced at 0.01%.
But let’s now take a look at another plot for the Miyares contest in another blowout locality with 84% / 16% for Miyares. The +/- 3/sqrt(N-1) limit nicely bounds our ballot distribution again. There
is, however, an interesting block of 44 consecutive ballots for Miyares about halfway through the processing of ballots. This equates to .84^44 = 0.0004659 or a 0.04659% chance of occurrence from an
IID distribution. Close to this peak is a run of 4 ballots for Herring which doesn’t sound like much, but given the 84% / 16% split, the probability of occurrence for that small run is .16^4 =
0.0006554 or 0.06554%!
Moving to the Lt. Governors race we see an interesting phenomenon where where Ayala received a sudden 100 consecutive votes a little over midway through the counting process. Now granted, this was a
landslide district for Ayala, but this still equates to a .92^100 = 0.000239 or 0.0239% chance of occurrence.
And here’s another large block of contiguous Ayala ballots equating to about .89^84 = 0.00005607 or 0.0056% chance of occurrence.
Tests for Differential Invalidation (added 2022-09-19):
“Differential invalidation” takes place when the ballots of one candidate or position are invalidated at a higher rate than for other candidates or positions. With this dataset we know how many
ballots were cast, and how many ballots had incomplete or invalid results (no recorded vote in the cvr, but the ballot record exists) for the 3 statewide races. In accordance with the techniques
presented in [1] and [2], I computed the plots of the Invalidation Rate vs the Percent Vote Share for the Winner in an attempt to observe if there looks to be any evidence of Differential
Invalidation ([1], ch 6). This is similar to the techniques presented in [2], which I used previously to produce my election fingerprint plots and analysis that plotted the 2D histograms of the vote
share for the winner vs the turnout percentage.
The generated the invalidation rate plots for the Gov, Lt Gov and AG races statewide in VA 2021 are below. Each plot below is representing one of the statewide races, and each dot is representing the
ballots from a specific precinct. The x axis is the percent vote share for the winner, and the y axis is computed as 100 – 100 * Nvotes / Nballots. All three show a small but statistically
significant linear trend and evidence of differential invalidation. The linear regression trendlines have been computed and superimposed on the data points in each graph.
To echo the warning from [1]: a differential invalidation rate does not directly indicate any sort of fraud. It indicates an unfairness or inequality in the rate of incomplete or invalid ballots
conditioned on candidate choice. While it could be caused by fraud, it could also be caused by confusing ballot layout, or socio-economic issues, etc.
Full Results Download
• [1] Forsberg, O.J. (2020). Understanding Elections through Statistics: Polling, Prediction, and Testing (1st ed.). Chapman and Hall/CRC. https://doi.org/10.1201/9781003019695
• [2] Klimek, Peter & Yegorov, Yuri & Hanel, Rudolf & Thurner, Stefan. (2012). Statistical Detection of Systematic Election Irregularities. Proceedings of the National Academy of Sciences of the
United States of America. 109. 16469-73. https://doi.org/10.1073/pnas.1210722109.
Update 2022-08-29 per observations by members of the Texas team I am working with, we’ve been able to figure out that (a) the vendor was simply concatenating data records from each machine and not
sorting the CVR results and (b) how to mostly unwrap this affect on the data to produce much cleaner results. The results below are left up for historical reference.
For background information, please see my introduction to Cast Vote Records processing and theory here: Statistical Detection of Irregularities via Cast Vote Records. This entry will be specifically
documenting the results from processing the Henrico County Virginia CVR data from the 2021 election.
As in the results from the previous post, I expanded the theoretical error bounds out to 6/sqrt(N) instead of 3/sqrt(N) in order to give a little bit of extra “wiggle room” for small fluctuations.
However the Henrico dataset could only be broken up by CAP, Non-CAP or Provisional. So be aware that the CAP curves presented below contain a combination of both early-vote and mail-in ballots.
The good news is that I’ve at least found one race that seems to not have any issues with the CVR curves staying inside the error boundaries. MemberHouseOfDelegates68thDistrict did not have any parts
of the curves that broke through the error boundaries.
The bad news … is pretty much everything else doesn’t. I cannot tell you why these curves have such differences from statistical expectation, just that they do. We must have further investigation and
analysis of these races to determine root cause. I’ve presented all of the races that had sufficient number of ballots below (1000 minimum for the race a whole, and 100 ballot minimum for each ballot
There has been a good amount of commotion regarding cast vote records (CVRs) and their importance lately. I wanted to take a minute and try and help explain why these records are so important, and
how they provide a tool for statistical inspection of election data. I also want to try and dispel any misconceptions as to what they can or can’t tell us.
I have been working with other local Virginians to try and get access to complete CVRs for about 6 months (at least) in order to do this type of analysis. However, we had not had much luck in
obtaining real data (although we did get a partial set from PWC primaries but it lacked the time-sequencing information) to evaluate until after Jeff O’Donnell (a.k.a. the Lone Raccoon) and Walter
Dougherity did a fantastic presentation at the Mike Lindell Moment of Truth Summit on CVRs and their statistical use. That presentation seems to have broken the data logjam, and was the impetus for
writing this post.
Just like the Election Fingerprint analysis I was doing earlier that highlighted statistical anomalies in election data, this CVR analysis is a statistics based technique that can help inform us as
to whether or not the election data appears consistent with expectations. It only uses the official results as provided by state or local election authorities and relies on standard statistical
principles and properties. Nothing more. Nothing less.
What is a cast vote record?
A cast vote record is part of the official election records that need to be maintained in order for election systems to be auditable. (see: 52 USC 21081 , NIST CVR Standard, as well as the Virginia
Voting Systems Certification Standards) They can have many different formats depending on equipment vendor, but they are effectively a record of each ballot as it was recorded by the equipment. Each
row in a CVR data table should represent a single ballot being cast by a voter and contain, at minimum, the time (or sequence number) when the ballot was cast, the ballot type, and the result of each
race. Other data might also be included such as which precinct and machine performed the scanning/recording of the ballot, etc. Note that “cast vote records” are sometimes also called “cast voter
records”, “ballot reports” or a number of other different names depending on the publication or locality. I will continue to use the “cast vote record” language in this document for consistency.
Why should we care?
The reason these records are so important, is based on statistics and … unfortunately … involves some math to fully describe. But to make this easier, let’s try first to walk through a simple thought
experiment. Let’s pretend that we have a weighted, or “trick” coin, that when flipped it will land heads 53% of the time and land tails 47% of the time. We’re going to continuously flip this coin
thousands of times in a row and record our results. While we can’t predict exactly which way the coin will land on any given toss, we can expect that, on average, the coin will land with the
aforementioned 53/47 split.
Now because each coin toss constitutes an independent and identically distributed (IID) probability function, we can expect this sequence to obey certain properties. If as we are making our tosses,
we are computing the “real-time” statistics of the percentage of head/tails results, and more specifically if we plot the spread (or difference) of those percentage results as we proceed we will see
that the spread has very large swings as we first begin to toss our coin, but very quickly the variability in the spread becomes stable as more and more tosses (data) are available for us to average
over. Mathematically, the boundary on these swings is inversely proportional to the square root of how many tosses are performed. In the “Moment of Truth” video on CVRs linked above, Jeff and Walter
refer to this as a “Cone of Probability”, and he generates his boundary curves experimentally. He is correct. It is a cone of probability as its really just a manifestation of well-known and
well-understood Poisson Noise characteristic (for the math nerds reading this). In Jeff’s work he uses the ratio of votes between candidates, while I’m using the spread (or deviation) of the vote
percentages. Both metrics are valid, but using the deviation has an easy closed-form boundary curve that we don’t need to generate experimentally.
In the graphic below I have simulated 10 different trials of 10,000 tosses for a distribution that leans 53/47, which is equivalent to a 6% spread overall. Each trial had 10,000 random samples
generated as either +1 or -1 values (a.k.a. a binary “Yes” or “No” vote) approximating the 53/47 split and I plotted the cumulative running spread of the results as each toss gets accumulated. The
black dotted outline is the 95% confidence interval (or +/-3x the standard deviation) across the 10 trials for the Nth bin, and the red dotted outline is the 3/sqrt(n-1) analytical boundary.
So how does this apply to election data?
In a theoretically free and perfectly fair election we should see similar statistical behavior, where each coin toss is replaced with a ballot from an individual voter. In a perfect world we would
have each vote be completely independent of every other vote in the sequence. In reality we have to deal with the fact that there can be small local regions of time in which perfectly legitimate
correlations in the sequence of scanned ballots exist. Think of a local church who’s congregation is very uniform and they all go to the polls after Sunday mass. We would see a small trend in the
data corresponding to this mass of similar thinking peoples going to the polls at the same time. But we wouldn’t expect there to be large, systematic patterns, or sharp discontinuities in the plotted
results. A little bit of drift and variation is to be expected in dealing with real world election data, but persistent and distinct patterns would indicate a systemic issue.
Now we cannot isolate all of the variables in a real life example, but we should try as best as possible. To that effect, we should not mix different ballot types that are cast in different manners.
We should keep our analysis focused within each sub-group of ballot type (mail-in, early-vote, day-of, etc). It is to the benefit of this analysis that the very nature of voting, and the procedures
by which it occurs, is a very randomized process. Each sub-grouping has its own quasi-random process that we can consider.
While small groups (families, church groups) might travel to the in-person polls in correlated clusters, we would expect there to be fairly decent randomization of who shows up to in-person polls and
when. The ordering of who stands in line before or after one another, how fast they check-in and fill out their ballot, etc, are all quasi-random processes.
Mail-in ballots have their own randomization as they depend on the timing of when individuals request, fill-out and mail their responses, as well as the logistics and mechanics of the postal service
processes themselves providing a level of randomization as to the sequence of ballots being recorded. Like a dealer shuffling a deck of cards, the process of casting a mail-in vote provides an
additional level of independence between samples.
No method is going to supply perfect theoretical independence from ballot to ballot in the sequence, but theres a general expectation that voting should at least be similar to an IID process.
Also … and I cannot stress this enough … while these techniques can supply indications of irregularities and discrepancies in elections data, they are not conclusive and must be coupled with in-depth
So going back to the simulation we generated above … what does a simulation look like when cheating occurs? Let’s take a very simple cheat from a random “elections” of 10,000 ballots, with votes
being representative of either +1 (or “Yes”) or -1 (or “No”) as we did above. But lets also cheat by randomly selecting two different spots in the data stream to place blocks of 250 consecutive “Yes”
The image below shows the result of this process. The blue curve represents the true result, while the red curve represents the cheat. We see that at about 15% and 75% of the vote counted, our
algorithm injected a block of “Yes” results, and the resulting cumulative curve breaks through the 3/sqrt(N-1) boundary. Now, not every instance or type of cheat will break through this boundary, and
there may be real events that might explain such behavior. But looking for CVR curves that break our statistical expectations is a good way to flag items that need further investigation.
Computing the probability of a ballot run:
Section added on 2022-09-18
We can also a bit more rigor to the statistics outlier detection by computing the probability of the length of observed runs (e.g. how many “heads” did we get in a row?) occurring as we move through
the sequential entries. We can compute this probability for K consecutive draws using the rules of statistical independence, which is P([a,a,a,a]) = P(a) x P(a) x P(a) x P(a) = P(a)^4. Therefore the
probability of getting 4 “heads” in a row with a hypothetical 53/47 weighted coin would be .53^4 = 0.0789.
Starting with my updated analysis of 2021 Henrico County VA, I’ve started adding this computation to my plots. I have not yet re-run the Texas data below with this new addition, but will do so soon
and update this page accordingly.
Real Examples
UPDATE 2022-09-18:
• I have finally gotten my hands on some data for 2020 in VA. I will be working to analyze that data and will report what I find as soon as I can, but as we are approaching the start of early
voting for 2022, my hands are pretty full at the moment so it might take me some time to complete that processing.
• As noted in my updates to the Henrico County 2021 VA data, and in my section on computing the probability of given runs above, the Texas team noticed that we could further break apart the Travis
county data into subgroups by USB stick. I will update my results below as soon as I get the time to do so.
[S:So I haven’t gotten complete cast vote records from VA yet (… which is a whole other set of issues …), but:S] I have gotten my Cheeto stained fingers on some data from the Travis County Texas 2020
So let us first take a look at an example of a real race where everything seems to be obeying the rules as set out above. I’ve doubled my error bars from 3x to 6x of the inverse square standard
(discussed above) in order to handle the quasi-IID nature of the data and give some extra margin for small fluctuating correlations.
The plot below shows the Travis County Texas 2020 BoardOfTrusteesAt_LargePosition8AustinISD race, as processed by the tabulation system and stratified by ballot type. We can see that all three ballot
types start off with large variances in the computed result but very quickly coalesce and approach their final values. This is exactly what we would expect to see.
Now if I randomly shuffle the ordering of the ballots in this dataset and replot the results (below) I get a plot that looks unsurprisingly similar, which suggests that these election results were
likely produced by a quasi-IID process.
Next let’s take a look at a race that does NOT conform to the statistics we’ve laid out above. (… drum-roll please … as this the one everyone’s been waiting for). Immma just leave this right here and
just simply point out that all 3 ballot type plots below in the Presidential race for 2020 go outside of the expected error bars. I also note the discrete stair step pattern in the early vote
numbers. It’s entirely possible that there is a rational explanation for these deviations. I would sure like to hear it, especially since we have evidence from the exact same dataset of other races
that completely followed the expected boundary conditions. So I don’t think this is an issue with a faulty dataset or other technical issues.
And just for completeness, when I artificially shuffle the data for the Presidential race, and force it to be randomized, I do in fact end up with results that conform to IID statistics (below).
I will again state that while these results are highly indicative that there were irregularities and discrepancies in the election data, they are not conclusive. A further investigation must take
place, and records must be preserved, in order to discover the cause of the anomalies shown.
Running through each race that had at least 1000 ballots cast and automatically detecting which races busted the 6/sqrt(n-1) boundaries produces the following tabulated results. A 1 in the right hand
column indicates that the CVR data for that particular race in Travis County has crossed the error bounds. A 0 in the right hand column indicates that all data stayed within the error bound limits.
Race CVR_OOB_Irregularity_Detected
President_VicePresident 1
UnitedStatesSenator 1
UnitedStatesRepresentativeDistrict10 1
UnitedStatesRepresentativeDistrict17 1
UnitedStatesRepresentativeDistrict21 1
UnitedStatesRepresentativeDistrict25 1
UnitedStatesRepresentativeDistrict35 0
RailroadCommissioner 1
ChiefJustice_SupremeCourt 1
Justice_SupremeCourt_Place6_UnexpiredTerm 1
Justice_SupremeCourt_Place7 1
Justice_SupremeCourt_Place8 1
Judge_CourtOfCriminalAppeals_Place3 1
Judge_CourtOfCriminalAppeals_Place4 1
Judge_CourtOfCriminalAppeals_Place9 1
Member_StateBoardOfEducation_District5 1
Member_StateBoardOfEducation_District10 1
StateSenator_District21 0
StateSenator_District24 1
StateRepresentativeDistrict47 1
StateRepresentativeDistrict48 1
StateRepresentativeDistrict49 1
StateRepresentativeDistrict50 1
StateRepresentativeDistrict51 0
ChiefJustice_3rdCourtOfAppealsDistrict 1
DistrictJudge_460thJudicialDistrict 1
DistrictAttorney_53rdJudicialDistrict 1
CountyJudge_UnexpiredTerm 1
Judge_CountyCourtAtLawNo_9 1
Sheriff 1
CountyTaxAssessor_Collector 1
CountyCommissionerPrecinct1 1
CountyCommissionerPrecinct3 1
AustinCityCouncilDistrict2 0
AustinCityCouncilDistrict4 0
AustinCityCouncilDistrict6 0
AustinCityCouncilDistrict7 0
AustinCityCouncilDistrict10 1
PropositionACityOfAustin_FP__2015_ 1
PropositionBCityOfAustin_FP__2022_ 1
MayorCityOfCedarPark 0
CouncilPlace2CityOfCedarPark 0
CouncilPlace4CityOfCedarPark 0
CouncilPlace6CityOfCedarPark 0
CouncilMemberPlace2CityOfLagoVista 0
CouncilMemberPlace4CityOfLagoVista 0
CouncilMemberPlace6CityOfLagoVista 0
CouncilMemberPlace2CityOfPflugerville 0
CouncilMemberPlace4CityOfPflugerville 0
CouncilMemberPlace6CityOfPflugerville 0
Prop_ACityOfPflugerville_2169_ 0
Prop_BCityOfPflugerville_2176_ 0
Prop_CCityOfPflugerville_2183_ 0
BoardOfTrusteesDistrict2SingleMemberDistrictAISD 0
BoardOfTrusteesDistrict5SingleMemberDistrictAISD 0
BoardOfTrusteesAt_LargePosition8AustinISD 1
BoardOfTrusteesPlace1EanesISD 1
Prop_AEanesISD_2246_ 0
BoardOfTrusteesPlace3LeanderISD 0
BoardOfTrusteesPlace4LeanderISD 0
BoardOfTrusteesPlace5ManorISD 1
BoardOfTrusteesPlace6ManorISD 0
BoardOfTrusteesPlace7ManorISD 1
BoardOfTrusteesPlace6PflugervilleISD 1
BoardOfTrusteesPlace7PflugervilleISD 1
BoardOfTrusteesPlace1RoundRockISD 1
BoardOfTrusteesPlace2RoundRockISD 0
BoardOfTrusteesPlace6RoundRockISD 0
BoardOfTrusteesPlace7RoundRockISD 0
BoardOfTrusteesWellsBranchCommunityLibraryDistrict 0
Var147 0
BoardOfTrusteesWestbankLibraryDistrict 0
Var150 0
Var151 0
DirectorsPlace2WellsBranchMUD 0
DirectorsPrecinct4BartonSprings_EdwardsAquiferConservationDistr 0
PropositionAExtensionOfBoundariesCityOfLakeway_1966_ 0
PropositionB2_yearTermsCityOfLakeway_1973_ 0
PropositionCLimitOnSuccessiveYearsOfServiceCityOfLakeway_1980_ 0
PropositionDResidencyRequirementForCityManagerCityOfLakeway_198 0
PropositionEOfficeOfTreasurerCityOfLakeway_1994_ 0
PropositionFOfficialBallotsCityOfLakeway_2001_ 0
PropositionGAuthorizingBondsCityOfLakeway_2008_ 0
PropositionALagoVistaISD_2253_ 0
PropositionBLagoVistaISD_2260_ 0
PropositionCLagoVistaISD_2267_ 0
PropositionAEmergencyServicesDistrict1_2372_ 0
References and further reading:
Computed below is the number of duplicated voter records in each locality as of the 11/06/21 VA Registered Voter List (RVL). The computation is based on performing an exact match of LAST_NAME, DOB
and ADDRESS fields between records in the file.
Note: If the combination of the name “Jane Smith”, with DOB “1/1/1980”, at “12345 Some Road, Ln.” appears 3 times in the file, there are 3 counts added to the results below. If the combination
appears only once, there are 0 counts added to the results below, as there is no repetition.
Additionally I’ve done an even more restrictive matching which requires exact match on FIRST, MIDDLE and LAST name, DOB and ADDRESS fields in the second graphic and list presented below.
The first, more lenient, criteria will correctly flag multiple records with the same first or middle name, but misspelled such as “Muhammad” vs “Mahammad”, but could also include occurrences of
voting age twins who live together or spouses with the same DOB.
The second, more strict, criteria requires that multiple rows flagged have exactly the same spelling and punctuation for FIRST, MIDDLE, LAST, DOB and ADDRESS fields. This has less false positive, but
more false negatives, as it will likely miss common misspellings between entries, etc.
There are no attempts to match for common misspellings, etc. I did do a simple cleanup for multiple contiguous whitespace elements, etc., before attempting to match.
I have summarized the data here so as not to reveal any personally identifiable information (PII) from the RVL in adherence to VA law.
Update 2022-07-13 12:30: I have sent the full information, for both the lenient and strict criteria queries, to the Prince William County and Loudoun County Registrars. The Loudoun deputy registrar
has responded and stated that all but 1 of the duplications in the stricter criteria had already been caught by the elections staff, but he has not yet looked at the entries in the more lenient
criteria results file. I have also attempted to contact the Henrico County, Lynchburg City, and York County registrars but have not yet received a response or request to provide them with the full
Update 2022-07-31 23:03: I have also heard back from the PWC Registrar (Eric Olsen). Most of the entries that I had flagged in the 11/6/2021 RVL list have already been taken care of by the PWC staff
already. There were only a couple that had not yet been noticed or marked as duplicates. Also, per our discussion, I should reiterate and clarify that the titles on the plots below simply refer to
duplicated entries of the data files according to the filtering choice. It is a technically accurate description and should not be read as I am asserting other than the results of the matching
Locality Name Number of repeated entries
ACCOMACK COUNTY 64
ALBEMARLE COUNTY 311
ALEXANDRIA CITY 225
ALLEGHANY COUNTY 16
AMELIA COUNTY 32
AMHERST COUNTY 84
APPOMATTOX COUNTY 18
ARLINGTON COUNTY 446
AUGUSTA COUNTY 119
BATH COUNTY 2
BEDFORD COUNTY 170
BLAND COUNTY 4
BOTETOURT COUNTY 45
BRISTOL CITY 22
BRUNSWICK COUNTY 30
BUCHANAN COUNTY 30
BUCKINGHAM COUNTY 32
BUENA VISTA CITY 10
CAMPBELL COUNTY 82
CAROLINE COUNTY 67
CARROLL COUNTY 42
CHARLES CITY COUNTY 18
CHARLOTTE COUNTY 28
CHARLOTTESVILLE CITY 80
CHESAPEAKE CITY 545
CHESTERFIELD COUNTY 948
CLARKE COUNTY 40
COLONIAL HEIGHTS CITY 18
COVINGTON CITY 2
CRAIG COUNTY 4
CULPEPER COUNTY 114
CUMBERLAND COUNTY 8
DANVILLE CITY 88
DICKENSON COUNTY 22
DINWIDDIE COUNTY 44
EMPORIA CITY 6
ESSEX COUNTY 25
FAIRFAX CITY 38
FAIRFAX COUNTY 2962
FALLS CHURCH CITY 39
FAUQUIER COUNTY 203
FLOYD COUNTY 28
FLUVANNA COUNTY 36
FRANKLIN CITY 23
FRANKLIN COUNTY 84
FREDERICK COUNTY 210
FREDERICKSBURG CITY 54
GALAX CITY 0
GILES COUNTY 24
GLOUCESTER COUNTY 52
GOOCHLAND COUNTY 84
GRAYSON COUNTY 18
GREENE COUNTY 32
GREENSVILLE COUNTY 16
HALIFAX COUNTY 48
HAMPTON CITY 285
HANOVER COUNTY 316
HARRISONBURG CITY 40
HENRICO COUNTY 676
HENRY COUNTY 74
HIGHLAND COUNTY 4
HOPEWELL CITY 34
ISLE OF WIGHT COUNTY 98
JAMES CITY COUNTY 217
KING & QUEEN COUNTY 13
KING GEORGE COUNTY 42
KING WILLIAM COUNTY 43
LANCASTER COUNTY 10
LEE COUNTY 24
LEXINGTON CITY 12
LOUDOUN COUNTY 1245
LOUISA COUNTY 74
LUNENBURG COUNTY 26
LYNCHBURG CITY 165
MADISON COUNTY 12
MANASSAS CITY 64
MANASSAS PARK CITY 24
MARTINSVILLE CITY 14
MATHEWS COUNTY 18
MECKLENBURG COUNTY 54
MIDDLESEX COUNTY 12
MONTGOMERY COUNTY 159
NELSON COUNTY 30
NEW KENT COUNTY 26
NEWPORT NEWS CITY 329
NORFOLK CITY 411
NORTHAMPTON COUNTY 18
NORTON CITY 6
NOTTOWAY COUNTY 12
ORANGE COUNTY 70
PAGE COUNTY 47
PATRICK COUNTY 28
PETERSBURG CITY 68
PITTSYLVANIA COUNTY 84
POQUOSON CITY 28
PORTSMOUTH CITY 186
POWHATAN COUNTY 55
PRINCE EDWARD COUNTY 43
PRINCE GEORGE COUNTY 77
PRINCE WILLIAM COUNTY 1159
PULASKI COUNTY 59
RADFORD CITY 14
RAPPAHANNOCK COUNTY 10
RICHMOND CITY 300
RICHMOND COUNTY 14
ROANOKE CITY 133
ROANOKE COUNTY 233
ROCKBRIDGE COUNTY 28
ROCKINGHAM COUNTY 113
RUSSELL COUNTY 28
SALEM CITY 58
SCOTT COUNTY 18
SHENANDOAH COUNTY 48
SMYTH COUNTY 40
SOUTHAMPTON COUNTY 28
SPOTSYLVANIA COUNTY 345
STAFFORD COUNTY 410
STAUNTON CITY 14
SUFFOLK CITY 194
SURRY COUNTY 10
SUSSEX COUNTY 14
TAZEWELL COUNTY 52
VIRGINIA BEACH CITY 922
WARREN COUNTY 46
WASHINGTON COUNTY 78
WAYNESBORO CITY 26
WESTMORELAND COUNTY 24
WILLIAMSBURG CITY 22
WINCHESTER CITY 42
WISE COUNTY 40
WYTHE COUNTY 35
YORK COUNTY 178
Locality Name Number of repeated entries
ACCOMACK COUNTY 0
ALBEMARLE COUNTY 4
ALEXANDRIA CITY 0
ALLEGHANY COUNTY 0
AMELIA COUNTY 0
AMHERST COUNTY 2
APPOMATTOX COUNTY 0
ARLINGTON COUNTY 10
AUGUSTA COUNTY 0
BATH COUNTY 0
BEDFORD COUNTY 4
BLAND COUNTY 0
BOTETOURT COUNTY 0
BRISTOL CITY 0
BRUNSWICK COUNTY 0
BUCHANAN COUNTY 2
BUCKINGHAM COUNTY 0
BUENA VISTA CITY 0
CAMPBELL COUNTY 2
CAROLINE COUNTY 0
CARROLL COUNTY 0
CHARLES CITY COUNTY 0
CHARLOTTE COUNTY 0
CHARLOTTESVILLE CITY 0
CHESAPEAKE CITY 8
CHESTERFIELD COUNTY 8
CLARKE COUNTY 0
COLONIAL HEIGHTS CITY 0
COVINGTON CITY 0
CRAIG COUNTY 0
CULPEPER COUNTY 0
CUMBERLAND COUNTY 0
DANVILLE CITY 0
DICKENSON COUNTY 0
DINWIDDIE COUNTY 0
EMPORIA CITY 0
ESSEX COUNTY 0
FAIRFAX CITY 0
FAIRFAX COUNTY 54
FALLS CHURCH CITY 0
FAUQUIER COUNTY 2
FLOYD COUNTY 0
FLUVANNA COUNTY 0
FRANKLIN CITY 3
FRANKLIN COUNTY 0
FREDERICK COUNTY 6
FREDERICKSBURG CITY 0
GALAX CITY 0
GILES COUNTY 2
GLOUCESTER COUNTY 0
GOOCHLAND COUNTY 0
GRAYSON COUNTY 0
GREENE COUNTY 0
GREENSVILLE COUNTY 0
HALIFAX COUNTY 0
HAMPTON CITY 8
HANOVER COUNTY 0
HARRISONBURG CITY 0
HENRICO COUNTY 24
HENRY COUNTY 2
HIGHLAND COUNTY 0
HOPEWELL CITY 2
ISLE OF WIGHT COUNTY 4
JAMES CITY COUNTY 0
KING & QUEEN COUNTY 0
KING GEORGE COUNTY 0
KING WILLIAM COUNTY 0
LANCASTER COUNTY 0
LEE COUNTY 0
LEXINGTON CITY 0
LOUDOUN COUNTY 23
LOUISA COUNTY 0
LUNENBURG COUNTY 0
LYNCHBURG CITY 16
MADISON COUNTY 0
MANASSAS CITY 0
MANASSAS PARK CITY 0
MARTINSVILLE CITY 0
MATHEWS COUNTY 2
MECKLENBURG COUNTY 0
MIDDLESEX COUNTY 0
MONTGOMERY COUNTY 2
NELSON COUNTY 0
NEW KENT COUNTY 0
NEWPORT NEWS CITY 0
NORFOLK CITY 0
NORTHAMPTON COUNTY 0
NORTON CITY 0
NOTTOWAY COUNTY 0
ORANGE COUNTY 0
PAGE COUNTY 0
PATRICK COUNTY 0
PETERSBURG CITY 0
PITTSYLVANIA COUNTY 4
POQUOSON CITY 0
PORTSMOUTH CITY 0
POWHATAN COUNTY 0
PRINCE EDWARD COUNTY 0
PRINCE GEORGE COUNTY 0
PRINCE WILLIAM COUNTY 8
PULASKI COUNTY 0
RADFORD CITY 0
RAPPAHANNOCK COUNTY 0
RICHMOND CITY 10
RICHMOND COUNTY 0
ROANOKE CITY 0
ROANOKE COUNTY 11
ROCKBRIDGE COUNTY 0
ROCKINGHAM COUNTY 4
RUSSELL COUNTY 0
SALEM CITY 0
SCOTT COUNTY 0
SHENANDOAH COUNTY 0
SMYTH COUNTY 0
SOUTHAMPTON COUNTY 0
SPOTSYLVANIA COUNTY 4
STAFFORD COUNTY 4
STAUNTON CITY 0
SUFFOLK CITY 4
SURRY COUNTY 0
SUSSEX COUNTY 0
TAZEWELL COUNTY 2
VIRGINIA BEACH CITY 4
WARREN COUNTY 0
WASHINGTON COUNTY 0
WAYNESBORO CITY 0
WESTMORELAND COUNTY 2
WILLIAMSBURG CITY 0
WINCHESTER CITY 0
WISE COUNTY 0
WYTHE COUNTY 2
YORK COUNTY 0
Photo by Element5 Digital on Pexels.com
Many election integrity investigators are looking through registration records and trying to find suspicious registrations based on the number of records attributed to a specific address as an
initial way of identifying records of interest and of need of further scrutiny. This can often produce false positives for things like nursing homes, college dormitories, etc. Additionally, one of
the concerns that has been raised is the risk of potential elder abuse, ID theft, manipulation or improper use of ballots for occupants of nursing home, hospice care or assisted living facilities.
According to https://npino.com : “The National Provider Identifier (NPI) is a unique identification number for covered health care providers (doctors, dentists, chiropractors, nurses and other
medical staff). The NPI is a 10-digit, intelligence-free numeric identifier. This means that the numbers do not carry other information about healthcare providers, such as the state in which they
live or their medical specialty. The NPI must be used in lieu of legacy provider identifiers in the HIPAA standards transactions. Covered health care providers and all health plans and health care
clearing houses must use the NPIs in the administrative and financial transactions adopted under HIPAA (Health Insurance Portability and Accountability Act).”
I’ve compiled a list of every nursing home, hospice care, or assisted living facility in the country based on their current NPI code. I have mirrored and scraped the entire https://npino.com site as
of 5-23-2022 and compiled the list of nationwide Nursing homes, Assisted Living and Hospice Care facilities into the below CSV file and am presenting it here in the hopes that it is useful for other
researchers. I did do a small amount of regular expression based cleanup to the entries (e.x. // replacing “Ste.” with “Suite”, fixing whitespace issues, etc.) as well as manually addressing a
handful of obviously incorrect addresses (e.x. // repeated/spliced street addresses, etc.).
I finally had some time to put this together.
For additional background information, see here, here and here. As a reminder and summary, according to the published methods in the USAID funded National Academy of Sciences paper (here) that I
based this work off of, an ideal “fair” election should look like one or two (depending on how split the electorate is) clean, compact Gaussian distributions (or “bulls-eye’s”). Other structural
artifacts, while not conclusive, can be evidence and indicators of election irregularities. One such indicator with an attributed cause is that of a highly directional linear streaking, which implies
an incremental fraud and linear correlation. Another known and attributed indicator is that of large distinct and extreme peaks near 100% or 0% votes for the candidate (the vertical axis) that are
disconnected from the main Gaussian lobe which the authors label a sign of “extreme fraud”. In general, for free and fair elections, we expect these 2D histograms to show a predominantly uncorrelated
relationship between the variables of “% Voter Turnout” (x-axis) and “% Vote Share” (y-axis).
The source data for this analysis comes directly from the VA Department of Elections servers, and was downloaded shortly after the conclusion and certification of the 2021 election results on 12/11/
2021 (results file) and 12/12/2021 (turnout file). A link to the current version of these files, hosted by ELECT is here: https://apps.elections.virginia.gov/SBE_CSV/ELECTIONS/ELECTIONRESULTS/2021/
2021%20November%20General%20.csv and https://apps.elections.virginia.gov/SBE_CSV/ELECTIONS/ELECTIONTURNOUT/Turnout-2021%20November%20General%20.csv. The files actually used for this analysis, as
downloaded from the ELECT servers on the dates mentioned are posted at the end of this article.
Note that even though the republican (Youngkin) won in VA, the y-axis of these plots presented here was computed as the % vote share for the democratic candidate (McCauliffe) in order to more easily
compare with the 2020 results. I can produce Yougkin vote share % versions as well if people are interested, and am happy to do so.
While the 2021 election fingerprints look to have less correlations between the variables as compared to 2020 data, they still look very non-gaussian and concerning. While there is no clearly
observable “well-known” artifacts as called out in the NAS paper, there is definitely something irregular about the Virginia 2021 election data. Specifically, I find the per-precinct absentee
[mail-in + post-election + early-in-person] plot (Figure 6) interesting as there is a diffuse background as well as a linearly correlated set of virtual precincts that show low turnout but very high
vote share for the democratic candidate.
One of the nice differences about 2021 VA data is that they actually identified the distinctions between mail-in, early-in-person, and post-election vote tallies in the CAP’s this year. I have broken
out the individual sub-groups as well and we can see that the absentee early-in-person (Figure 9) has a fairly diffuse distribution, while the absentee mail-in (Figure 7) and absentee post-election
(Figure 8) ballots show a very high McAuliffe Vote %, and what looks to be a linear correlation.
For comparison I’ve also included the 2020 fingerprints. All of the 2020 fingerprints have been recomputed using the exact same MATLAB source code that processed the 2021 data. The archive date of
the “2020 November General.csv” and “Turnout 2020 November General.csv” files used was 11/30/2020.
I welcome any and all independent reviews or assistance in making and double checking these results, and will gladly provide all collated source data and MATLAB code to anyone who is interested.
Figure 1 : VA 2021 Per locality, absentee (CAP) + physical precincts
Figure 2 : VA 2021 Per locality, physical precincts only:
Figure 3 : VA 2021 Per locality, absentee precincts only:
Figure 4 : VA 2021 Per precinct, absentee (CAP) + physical precincts:
Figure 5 : VA 2021 Per precinct, physical precincts only:
Figure 6 : VA 2021 Per precinct, absentee (CAP) precincts only:
Figure 7 : VA 2021 Per precinct, CAP precincts, mail-in ballots only:
Figure 8 : VA 2021 Per precinct, CAP precincts, post-election ballots only:
Figure 9 : VA 2021 Per precinct, CAP precincts, early-in-person ballots only:
Comparison to VA 2020 Fingerprints
Figure 10 : VA 2020 Per locality, absentee (CAP) + physical precincts:
Figure 11 : VA 2020 Per locality, physical precincts only:
Figure 12 : VA 2020 Per locality, absentee precincts only:
Figure 13 : VA 2020 Per precinct, absentee (CAP) + physical precincts:
Figure 14 : VA 2020 Per precinct, physical precincts only:
Figure 15 : VA 2020 Per precinct, absentee (CAP) precincts only:
Source Data Files:
During the 2021 election I archived multiple versions of the Statewide Daily Absentee List (DAL) files as produced by the VA Department of Elections (ELECT). As the name implies, the DAL files are a
daily produced official product from ELECT that accumulates data representing the state of absentee votes over the course of the election. i.e. The data that exists in a DAL file produced on Tuesday
morning should be contained in the DAL file produced on the following Wednesday along with any new entries from the events of Tuesday, etc.
Therefore, it is expected that once a Voter ID number is listed in the DAL file during an election period, subsequent DAL files *should* include a record associated with that voter ID. The status of
that voter and the absentee ballot might change, but the records of the transactions during the election should still be present. I have confirmed that this is the expected behavior via discussions
with multiple former and previous VA election officials.
Stepping through the snapshots of collected 2021 DAL files in chronological order, we can observe Voter IDs that mysteriously “vanish” from the DAL record. We can do this by simply mapping the
existence/non-existence of unique Voter ID numbers in each file. The plot below in Figure 1 is the counts of the number of observed “vanishing” ID numbers as we move from file to file. The total
number of vanishing ID numbers is 429 over the course of the 2021 election. Not a large number. But it’s 429 too many. I can think of no legitimate reason that this should occur.
Now an interesting thing to do, is to look at a few examples of how these issues manifest themselves in the data. Note that I’m hiding the personally identifiable information from the DAL file
records in the screenshots below, BTW.
The first example in the screenshot below is an issue where the voter in question has a ballot that is in the “APPROVED” and “ISSUED” state, meaning that they have submitted a request for a ballot
and that the ballot has been sent out. The record for this voter ID is present in the DAL file up until Oct 14th 2021, after which it completely vanishes from the DAL records. This voter ID is also
not present in the RVL or VHL downloaded from the state on 11/06/2021.
This voter was apparently issued a real, live, ballot for 2021 and then was subsequently removed from the DAL and (presumably) the voter rolls + VERIS on or around the 14th Oct according to the DAL
snapshots. What happened to that ballot? What happened to the record of that ballot? The only public record of that ballot even existing, let alone the fact that it was physically issued and mailed
out, was erased when the Voter ID was removed from DAL/RVL/VHL records. Again, this removal happened in the middle of an election where that particular voter had already been issued a live ballot!
A few of these IDs actually “reappear” in the DAL records. ID “230*****” is one example, and a screenshot of its chronological record is below. The ballot shows as being ISSUED until Oct 14th
2021. It then disappears from the DAL record completely until the data pull on Oct 24th, where it shows up again as DELETED. This status is maintained until Nov 6th 2021 when it starts oscillating
between “Marked” and “deleted” until it finally lands on “Marked” in the Dec 5 DAL file pull. The entire time the Application status is in the “Approved” state for this voter ID. From my
discussions with registrars and election officials the “Marked” designation signifies that a ballot has been received by the registrar for that voter and is slated to be tabulated.
I have poked ELECT on twitter (@wwrkds) on this matter to try and get an official response, and submitted questions on this matter to Ashley Coles at ELECT, per the advice of my local board of
elections chair. Her response to me is below:
I will update this post as information changes.
Presented 2022-04-14 at Patriot Pub in Hamilton VA. Note that I mis-quoted the total registration error number off the slide deck during the voice track of my presentation. The slides are correct …
my eyes and memory just suck sometimes! I said ~850K. The actual number is ~380K.
Video can be found here: https://www.patriotpuballiance.com
Previously I wrote about finding In-Person Early Vote records inserted into the Daily Absentee List (DAL) records after the close of early voting in VA in 2021. Well, theres been quite a bit of
activity since then and I have somme updates to share.
I originally discovered this issue and began digging into it around Nov 8th 2021, and finally published it to my blog on Dec 10th 2021. At the same time, queries were sent through the lawyers for the
RPV to the ELECT Chairman (Chris Piper) and to a number of registrars to attempt to determine the cause of this issue, but no response was supplied. I also raised this issue to my local Board of
Elections chair, and requested that ELECT comment on the matter through their official twitter account.
Since that time I have continued to publish my findings, have continued to request responses from ELECT, and have offered to work with them to address and resolve these discrepancies. I know ELECT
pays attention to this site and my twitter account, as they have quietly corrected both their website and data files after I have pointed out other errors and discrepancies. Additionally, Chris Piper
has continued to publicly insist that there were no major issues in either the 2020 or 2021 election (including under questioning by the VA Senate P&E Committee), and neither he nor any member of
ELECT has publicly acknowledged any of the issues I have raised … besides the aforementioned changing of their site contents, of course. I have thankfully had a few local board of elections members
work with me, as well as a few local registrars … but I did not see any meaningful response or engagement from anyone at ELECT until Feb 23rd 2022 as discussed below.
On Feb 22nd 2022, I was invited to participate in a meeting arranged by VA State Senator Amanda Chase with the VA Attorney Generals office to discuss election integrity issues. I specifically cited a
number of the issues that I’ve documented here on my blog, including the added DAL entries, as justification for my belief that there is an arguable case to be made for there being criminal gross
negligence and maladministration at ELECT with respect to the administration and handling of VA election data.
That meeting apparently shook some things loose. Good.
The day after that meeting Chris Piper finally sent a response at 10:45am to our inquiry on the subject of the added DAL entries. It is quoted below:
While I am glad that he finally responded, his technical reasoning does not address all of the symptoms that are observed:
• He states the cause was due to the EPB’s from one vendor, but the data shows DAL additions being attributed to multiple localities that use different EPB vendors.
• His explanation does not address the distribution of the numbers of DAL additions across all of the precincts observed. A misconfigured or malfunctioning poll book would affect all of the
check-ins entered on it, not just a sporadic few.
• This also does not seem like a minor issue as its affecting thousands of voters ballots and voter history. So I’m rather concerned with Mr. Piper’s attitude toward this issue, as well as others.
It needs to be addressed as a logic and accuracy testing issue, as a matter of procedures and training, in addition to simply asking vendors to add checks to their software. Also, will this be
addressed in VERIS at all, or if/when a replacement system is put in place?
• In response to his last paragraph, I will simply note that I have been actively and consistently working to raise all of these issues through official channels … through official requests via the
RPV, working with local election board members and registrars, and asking for input from elect through social media. I have not made accusations of malicious or nefarious intent, but I do think
there is plenty of evidence to make the case of incompetence +/or gross negligence with our elections data … which is actually a crime in VA … and Chris Piper has been the head of ELECT and the
man responsible for ensuring our election data during that time (he is stepping down effective March 11th).
Since his response was sent, a few additional things have occurred:
(A) The AG’s office informed us that they are actually required by law to treat ELECT as their client and defend them from any accusations of wrongdoing. This is frustrating as there does not seem to
be any responsive cognizant authority that is able to act on this matter in the interest of the public. This is not a local jurisdictional issue as it affects voters statewide, and is therefore in
the purview of the AG as the Department of Elections has been heretofore dismissive and non-responsive of these matters. I am not a lawyer, however.
(B) I was able to connect with the Loudoun County Deputy Director of Elections (Richard Keech) as well as the Prince William County registrar (Eric Olsen) and have been working through the finer
details of Chris’s explanation to verify and validate at the local level. [Note: I previously had Richard erroneously listed here as the Registrar instead of the Deputy Directory]. Both Richard and
Eric are continuing to look into the matter, and I continue to work with them to get to the bottom of this issue.
• Richard confirmed his belief that the bad OS system date on Election Day EPBs was responsible for the errors, however with some slight differences in the details from Piper’s description. There
were multiple vendors affected, not just one. Per Richard, the problem appeared to be that a number of Loudoun poll-books (regardless of vendor) that were used for Election Day had been in
storage so long that their batteries had completely depleted. When they were finally powered up, their OS system clocks had a wildly incorrect date. The hardware used was a mixture of Samsung
SM-T720 and iPad tablets, depending on poll-book application vendor. The hardware was purchased separately through local contracts with CDW, and the software was uploaded and configured by the
• In Loudoun, all of the EPBs went through logic and accuracy testing before the election per Richard, but it does not appear that the procedures for the Logic and Accuracy testing had any specific
checks for OS date settings.
• In Prince William County the registrar (Eric) was not aware of any issues with the system clocks on the poll-books, and he was skeptical of the distribution of the small numbers of added DAL
entries. He noted, as I did above, that if a poll book was misconfigured it would affect all of the records that passed through it, not just a small handful. He also noticed that there was a
discrepancy with the attribution of polling place names that I had extracted from the DAL files, where some of the names did not correspond with actual polling places in PWC. He has stated he
will look into the matter and get back to me. I will update my blog when he does so.
• From my communications with Richard, the VERIS system imports a text based file for processing voter credit and does not have any special checks against the dates for in-person vs early voting
records. Hence, why this issue can impact multiple vendors if their applications use the system clock to date-stamp their exported txt files for upload into VERIS.
(C) I have reworked my code per my conversations with Ricky and Eric, and fixed a few bugs and parsing errors along the way. Most notably, there are a number of missing or malformed field values in
the raw DAL files that were being parsed into ‘undefined’ categorical values by the default csv MATLAB parser. These ‘undefined’ values, even when located in unimportant fields in the row of a MATLAB
table, can cause the entire row to be incomparable to other entries when performing logical operations. I have adjusted my parser and logic to account for +/or ignore these entries as necessary.
Additionally I had previously looked for new entries by comparing values across entire rows, but have adjusted to now only look at voter ID numbers that have not been seen previously, in order to
omit those entries that had simply been adjusted (address changes, etc) after the fact, or that contained ‘undefined’ field elements as mentioned previously. Also I noticed that some of the dal files
had duplicate records of Approved and On-Machine records for the same voter ID. While that is an issue in itself, I de-duplicated those entries for this analysis. This new logic gives the updated
results presented below, with a total number of discrepancies now at 2820.
I will note that I am still a little skeptical of the “bad date” explanation as being a complete answer to this issue, as it does not adequately explain the distribution of small numbers of
discrepancies attributed to multiple precincts, for one thing. While the bad date may explain part of the issue, it does not adequately account for all of the observed effects, IMO. For example, in
Loudoun there are 26 precincts listed below that have inserted DAL records attributed to them. Many of these precincts having only 1 or 2 associated records. If the bad OS date explanation is to
blame, then (a) there must have been at least 26 poll-books, one at each precinct, in Loudoun with misconfigured and incorrect dates AND (b) many of these poll-books were used to only check in 1 or 2
people total, as anyone checked in on a misconfigured poll-book would have their voter credit/DAL file entries affected. This would have to have been replicated at ALL of the precincts in ALL of the
localities listed below. While the above scenario is admittedly possible, I find it rather implausible.
Update 2022-03-20
I’ve heard back from both Ricky Keech (Loudoun) and Eric Olsen (PWC).
Eric looked into the 10 entries for PWC and all of them were Military voters who did actually walk in, in person, to the registrars office and vote on machine absentee after the close of early
voting, as allowed for by state law. So everything checks out for PWC.
Per discussion with Ricky, there were two issues: The first being a number of pollbooks for four specific precincts that were used for Election Day having the wrong date setting as discussed above.
The other issues was possibly the connectivity issue of the Pollbook to the servers in South Riding during the early voting period that had to be hand corrected. Ricky’s explanation e-mail to me is
copied below.
Hi Jon,
Following up on our conversation the other day. So, I did some more digging and was able to figure out what happened. The bulk of the voters (1213) were from the four precincts that had a
tablet with the wrong date on it. That accounts for all voters in precincts 214, 416, 628, and 708. I had a staff person go back and pull out the tablets used in those precincts, and we
confirmed that each of those four had one table with the wrong date.
That leaves 141 voters that received ‘credit’ after election day. Once we had narrowed it down, I looked for patterns and noticed all the remaining precincts were in South Riding. That jogged
my memory and led me to the solution. When I ran the final reports on the Sunday before the election, I noticed that the number showing as voting early seemed to be off by 100 or so. This was
odd because our daily checked in count and voted count reconciled at every site every day. So, I went back and compared the voters checked into our pollbooks at early voting to the voters with
early voting (On Machine) credit in VERIS and found that there were 137 voters who voted on October 19 at the Dulles South EV site and for some reason did not have credit. I worked on this
Monday to make sure it was right, and it was, none of those voters had credit. This could either have been a connectivity issue at Dulles South EV site OR an issue with VERIS when the data was
uploaded to mark the voters. I can say definitively that the number checked in on the pollbooks at that site on that day and the number of people who put ballots into the machine was correct, we
check that constantly and the observers on site checked as well. I can guarantee that if there had been a discrepancy, we’d have heard about it right away.
So, after determining that was exactly what happened I uploaded credit for those voters at 2:06:51pm on Wednesday, November 3 and the upload completed processing at 2:07:07pm.
When we spoke the other day, I thought it was likely a connectivity issue, but now I’m not entirely sure that’s the case, as if the connection wasn’t working the numbers should have been off.
And they were correct on the devices at the EV site and my laptop here at the office. Everything matched.
So long story short, we did an audit, discovered missing credit from one early voting site on one day, and corrected it.
The other four voters were people who voted an emergency early voting ballot on Monday, November 1.
Richard Keech, Deputy Director of Elections for Loudoun County, Mar 11 2022 email to Jon Lareau
Locality COUNT
LOUDOUN COUNTY 1344
HANOVER COUNTY 1302
CHESAPEAKE CITY 92
PRINCE WILLIAM COUNTY 10
HENRICO COUNTY 7
WINCHESTER CITY 7
CHARLES CITY COUNTY 5
CAMPBELL COUNTY 3
CHARLOTTE COUNTY 3
FAUQUIER COUNTY 3
LUNENBURG COUNTY 3
WASHINGTON COUNTY 3
ALEXANDRIA CITY 2
AMELIA COUNTY 2
AMHERST COUNTY 2
BATH COUNTY 2
CAROLINE COUNTY 2
FALLS CHURCH CITY 2
HENRY COUNTY 2
NORTHAMPTON COUNTY 2
ORANGE COUNTY 2
ROANOKE CITY 2
VIRGINIA BEACH CITY 2
ALLEGHANY COUNTY 1
APPOMATTOX COUNTY 1
BLAND COUNTY 1
CHARLOTTESVILLE CITY 1
CLARKE COUNTY 1
CULPEPER COUNTY 1
ESSEX COUNTY 1
FAIRFAX CITY 1
FLOYD COUNTY 1
KING GEORGE COUNTY 1
MIDDLESEX COUNTY 1
NELSON COUNTY 1
NOTTOWAY COUNTY 1
POQUOSON CITY 1
STAFFORD COUNTY 1
STAUNTON CITY 1
LOCALITY PRECINCT COUNT
HANOVER COUNTY 704 – ELMONT 667
HANOVER COUNTY 602 – LEE DAVIS 635
LOUDOUN COUNTY 416 – HAMILTON 443
LOUDOUN COUNTY 214 – SUGARLAND NORTH 344
LOUDOUN COUNTY 708 – SENECA 319
LOUDOUN COUNTY 628 – MOOREFIELD STATION 97
LOUDOUN COUNTY 319 – JOHN CHAMPE 21
LOUDOUN COUNTY 313 – PINEBROOK 16
LOUDOUN COUNTY 112 – FREEDOM 13
LOUDOUN COUNTY 122 – HUTCHISON FARM 11
LOUDOUN COUNTY 126-GOSHEN POST 10
CHESAPEAKE CITY 055 – GEORGETOWN EAST 8
LOUDOUN COUNTY 107 – LITTLE RIVER 8
CHESAPEAKE CITY 053 – FAIRWAYS 7
LOUDOUN COUNTY 121 – TOWN HALL 7
LOUDOUN COUNTY 316 – CREIGHTON’S CORNER 7
LOUDOUN COUNTY 318 – MADISON’S TRUST 7
CHESAPEAKE CITY 008 – SOUTH NORFOLK RECREATION 6
CHESAPEAKE CITY 012 – GEORGETOWN 6
LOUDOUN COUNTY 114 – DULLES SOUTH 6
CHESAPEAKE CITY 018 – INDIAN RIVER 5
LOUDOUN COUNTY 124 – LIBERTY 5
LOUDOUN COUNTY 320 – STONE HILL 5
CHESAPEAKE CITY 029 – TANGLEWOOD 4
CHESAPEAKE CITY 042 – PARKWAYS 4
CHESAPEAKE CITY 059 – CLEARFIELD 4
CHESAPEAKE CITY 065 – WATERWAY II 4
LOUDOUN COUNTY 119 – ARCOLA 4
LOUDOUN COUNTY 322-BUFFALO TRAIL 4
CHESAPEAKE CITY 005 – CRESTWOOD 3
CHESAPEAKE CITY 022 – NORFOLK HIGHLANDS 3
CHESAPEAKE CITY 057 – CYPRESS 3
LOUDOUN COUNTY 120 – LUNSFORD 3
LOUDOUN COUNTY 123 – CARDINAL RIDGE 3
WINCHESTER CITY 101 – MERRIMANS 3
AMHERST COUNTY 501 – MADISON 2
CHARLES CITY COUNTY 101 – PRECINCT 1-1 2
CHARLES CITY COUNTY 301 – PRECINCT 3-1 2
CHARLOTTE COUNTY 702 – BACON/SAXE 2
CHESAPEAKE CITY 010 – OSCAR SMITH 2
CHESAPEAKE CITY 014 – GRASSFIELD 2
CHESAPEAKE CITY 015 – GREENBRIER MIDDLE SCHOOL 2
CHESAPEAKE CITY 016 – HICKORY GROVE 2
CHESAPEAKE CITY 023 – OAK GROVE 2
CHESAPEAKE CITY 024 – OAKLETTE 2
CHESAPEAKE CITY 031 – CARVER SCHOOL 2
CHESAPEAKE CITY 032 – PROVIDENCE 2
CHESAPEAKE CITY 034 – HICKORY MIDDLE SCHOOL 2
CHESAPEAKE CITY 043 – PLEASANT CROSSING 2
CHESAPEAKE CITY 056 – GREEN TREE 2
FALLS CHURCH CITY 003 – THIRD WARD 2
LOUDOUN COUNTY 302 – ROUND HILL 2
LOUDOUN COUNTY 308 – ST LOUIS 2
LOUDOUN COUNTY 309 – ALDIE 2
LOUDOUN COUNTY 617 – OAK GROVE 2
ORANGE COUNTY 101 – ONE WEST 2
PRINCE WILLIAM COUNTY 409 – TYLER 2
PRINCE WILLIAM COUNTY 513 – LYNNWOOD 2
PRINCE WILLIAM COUNTY 712 – LEESYLVANIA 2
WASHINGTON COUNTY 701 – HIGH POINT 2
WINCHESTER CITY 201 – VIRGINIA AVENUE 2
ALEXANDRIA CITY 110 – CHARLES HOUSTON CENTER 1
ALEXANDRIA CITY 201 – NAOMI L. BROOKS SCHOOL 1
ALLEGHANY COUNTY 101 – ARRITT 1
AMELIA COUNTY 301 – NUMBER THREE 1
AMELIA COUNTY 501 – NUMBER FIVE 1
APPOMATTOX COUNTY 401 – COURTHOUSE 1
BATH COUNTY 101 – WARM SPRINGS 1
BATH COUNTY 201 – HOT SPRINGS 1
BLAND COUNTY 301 – HOLLYBROOK 1
CAMPBELL COUNTY 102 – NEW LONDON 1
CAMPBELL COUNTY 402 – COURT HOUSE 1
CAMPBELL COUNTY 602 – CONCORD 1
CAROLINE COUNTY 202 – SOUTH MADISON 1
CAROLINE COUNTY 401 – DAWN 1
CHARLES CITY COUNTY 201 – PRECINCT 2-1 1
CHARLOTTE COUNTY 201 – RED OAK WYLLIESBURG 1
CHARLOTTESVILLE CITY 102 – CLARK 1
CHESAPEAKE CITY 006 – DEEP CREEK 1
CHESAPEAKE CITY 009 – BELLS MILL 1
CHESAPEAKE CITY 011 – GENEVA PARK 1
CHESAPEAKE CITY 020 – E W CHITTUM 1
CHESAPEAKE CITY 033 – WESTOVER 1
CHESAPEAKE CITY 046 – BELLS MILL II 1
CHESAPEAKE CITY 048 – JOLLIFF MIDDLE SCHOOL 1
CHESAPEAKE CITY 049 – WATERWAY 1
CHESAPEAKE CITY 050 – RIVER WALK 1
CHESAPEAKE CITY 051 – COOPERS WAY 1
CHESAPEAKE CITY 062 – FENTRESS 1
CHESAPEAKE CITY 063 – POPLAR BRANCH 1
CHESAPEAKE CITY 064 – DEEP CREEK II 1
CLARKE COUNTY 301 – MILLWOOD 1
CULPEPER COUNTY 303 – CARDOVA 1
ESSEX COUNTY 201 – NORTH 1
FAIRFAX CITY 001 – ONE 1
FAUQUIER COUNTY 202 – AIRLIE 1
FAUQUIER COUNTY 404 – SPRINGS VALLEY 1
FAUQUIER COUNTY 501 – THE PLAINS 1
FLOYD COUNTY 301 – COURTHOUSE 1
HENRICO COUNTY 105 – GREENDALE 1
HENRICO COUNTY 209 – GLEN LEA 1
HENRICO COUNTY 304 – JACKSON DAVIS 1
HENRICO COUNTY 316 – COLONIAL TRAIL 1
HENRICO COUNTY 416 – SPOTTSWOOD 1
HENRICO COUNTY 506 – EANES 1
HENRICO COUNTY 513 – PLEASANTS 1
HENRY COUNTY 203 – HORSEPASTURE #2 1
HENRY COUNTY 501 – BASSETT NUMBER ONE 1
KING GEORGE COUNTY 101 – COURTHOUSE 1
LOUDOUN COUNTY 108 – MERCER 1
LOUDOUN COUNTY 118 – MOOREFIELD 1
LOUDOUN COUNTY 401 – WEST LOVETTSVILLE 1
LUNENBURG COUNTY 301 – ROSEBUD 1
LUNENBURG COUNTY 501 – REEDY CREEK 1
LUNENBURG COUNTY 502 – PEOPLES COMMUNITY CENTER 1
MIDDLESEX COUNTY 501 – WILTON 1
NELSON COUNTY 401 – ROSELAND 1
NORTHAMPTON COUNTY 201 – PRECINCT 2-1 1
NORTHAMPTON COUNTY 401 – PRECINCT 4-1 1
NOTTOWAY COUNTY 201 – PRECINCT 2-1 1
POQUOSON CITY 001 – CENTRAL 1
PRINCE WILLIAM COUNTY 103 – GLENKIRK 1
PRINCE WILLIAM COUNTY 210 – PENN 1
PRINCE WILLIAM COUNTY 305 – PATTIE 1
PRINCE WILLIAM COUNTY 311 – SWANS CREEK 1
ROANOKE CITY 014 – Crystal Spring 1
ROANOKE CITY 019 – Forest Park 1
STAFFORD COUNTY 702 – WHITSON 1
STAUNTON CITY 301 – WARD NO 3 1
VIRGINIA BEACH CITY 030 – RED WING 1
VIRGINIA BEACH CITY 063 – CULVER 1
WASHINGTON COUNTY 302 – SOUTH ABINGDON 1
WINCHESTER CITY 301 – WAR MEMORIAL 1
WINCHESTER CITY 402 – ROLLING HILLS 1
The MATLAB code for generating the above is given below. The raw time-stamped DAL data files, as downloaded from the ELECT website, are loaded from the ‘droot’ directory tree as shown below, and I
only utilize the latest daily download of DAL file data for simplicity.
warning off all
% Data directory root
droot = 'SourceData/DAL/2021/';
% Gets the list of DAL files that were downloaded from the ELECT provided
% URL over the course of the 2021 Election.
files = dir([droot,'raw/**/Daily_Absentee_List_*.csv']);
matc = regexp({files.name}, 'Daily_Absentee_List_\d+(T\d+)?.csv','match');
matc = find(~cellfun(@isempty,matc));
files = files(matc);
% Only process the last updated DAL for each day. I downloaded multiple
% times per day, but we will just take the last file downloaded each day
% for simplicity here.
matc = regexp({files.name}, '(\d+)(T\d+)?','tokens');
fd = []; pc = 0; ic = 0; idx = [];
for i=1:numel(matc)
if isempty(regexp(matc{i}{1}{1},'^2021.*'))
fd(i) = datenum(matc{i}{1}{1},'mmddyyyy');
fd(i) = datenum(matc{i}{1}{1},'yyyymmdd');
if pc ~= fd(i)
ic = ic+1;
idx(ic) = i;
pc = fd(i);
files = files(idx);
% Now that we have our list of files, lets process.
seen = [];
firstseen = [];
astats = [];
astatsbc = {};
cumOnMachine = [];
T = [];
for i = 1:numel(files)
% Extract the date of the ELECT data pull. Note that the first few
% days I was running the script I was not including the time of the
% pull when I was pulling the data fro the ELECT url and writing to
% disk, so there's soe special logic here to handle that issue.
matc = regexp(files(i).name, '(\d+)(T\d+)?','tokens');
matc = [matc{1}{:}];
if isempty(regexp(matc,'^2021.*'))
fdn = datenum(matc,'mmddyyyy')+.5;
fdn= datenum(matc,'yyyymmddTHHMMSS');
fds = datestr(fdn,30)
fdt = datetime(fdn,'ConvertFrom','datenum');
% Move a copy to the 'byDay' folder so we can keep a reference to the
% data that went into this analysis.
dal2021filename = [files(i).folder,filesep,files(i).name];
ofn = [droot,'byDay/Daily_Absentee_List_',fds,'.csv'];
if ~exist(ofn)
% Import the DAL file
dal2021 = import2021DALfile(dal2021filename, [2, Inf]);
% Cleanup and handle undefined or imssing values.
dal2021.CITY(isundefined(dal2021.CITY)) = 'UNDEFINED';
dal2021.STATE(isundefined(dal2021.STATE)) = 'UNDEFINED';
dal2021.ZIP(isundefined(dal2021.ZIP)) = 'UNDEFINED';
dal2021.COUNTRY(isundefined(dal2021.COUNTRY)) = 'UNDEFINED';
% Do some basic indexing of different DAL status categories and combinations
appvd = dal2021.APP_STATUS == 'Approved' ;
aiv2021 = appvd & dal2021.BALLOT_STATUS == 'Issued';
amv2021 = appvd & dal2021.BALLOT_STATUS == 'Marked';
aomv2021 = appvd & dal2021.BALLOT_STATUS == 'On Machine';
appv2021 = appvd & dal2021.BALLOT_STATUS == 'Pre-Processed';
afwv2021 = appvd & dal2021.BALLOT_STATUS == 'FWAB';
appmv2021 = amv2021 | aomv2021 | appv2021 | afwv2021; % Approved and Countable
% Accumulate the stats for each DAL file
rstats = table(fdt,sum(aiv2021),sum(amv2021),sum(aomv2021),...
astats = [astats; rstats];
% Write out the entries that were approved and countable in this DAL
% file
ofn = [droot,'byDayCountable/Daily_Absentee_List_Countable_',fds,'.csv'];
if ~exist(ofn)
% Write out the entries that were approved, countable and marked as 'On
% Machine' (i.e. an Early In-Person Voter Check-In) in this DAL file
ofn = [droot,'byDayCountableOnMachine/Daily_Absentee_List_Countable_OnMachine',fds,'.csv'];
if ~exist(ofn)
% Since the DAL file grows over time, we're going to try and figure out
% which On-Machine entries in each new file:
% (a) We've seen before and are still listed.
% (b) We haven't seen before (a NEW entry).
% (c) We've seen before but the listing is missing (a DELETED
% entry).
if isempty(T)
% Only applicable for the first file we process.
T = dal2021(aomv2021,:);
% There are sometimes duplicate uid numbers in the approved and
% on-machine counts! This is a problem in and of itself, but not
% the problem I'm trying to focus on at the moment. So I'm going
% to remove any duplicated rows based on UID.
[uid, ia, ib] = unique(T.identification_number);
T = T(ia,:);
inew = (1:size(T,1))';
ideleted = [];
firstseen = repmat(fdt,size(T,1),1);
dOM = dal2021(aomv2021,:);
% There are sometimes duplicate uid numbers in the approved and
% on-machine counts! This is a problem in and of itself, but not
% the problem I'm trying to focus on at the moment. So I'm going
% to remove any duplicated rows based on UID.
[uid, ia, ib] = unique(dOM.identification_number);
dOM = dOM(ia,:);
%[~,ileft,iright] = innerjoin(T,dOM);
[~,ileft,iright] = intersect(T.identification_number,dOM.identification_number);
% 'inew' will be a boolean vector representing those entries in
% 'dOM' that are new On-Machine records
inew = true(size(dOM,1),1);
inew(iright) = false;
% 'ideleted' will be a boolean vector representing those entries in
% 'T' that are missing On-Machine records in 'dOM'
ideleted = true(size(T,1),1);
ideleted(ileft) = false;
T = [T;dOM(inew,:)];
firstseen = [firstseen; repmat(fdt,sum(inew),1)];
grid on;
grid minor;
xlabel('Date of DAL file pull from ELECT');
ofn = [droot,'byDayStats.csv'];
T.FirstSeen = firstseen;
ofn = [droot,'onMachineRecords.csv'];
ofn = [droot,'onMachineRecords_missing.csv'];
cutoffDate = datetime('2021-11-01');
ofn = [droot,'onMachineRecords_after_20211101.csv'];
writetable(T(T.FirstSeen >= cutoffDate,:),ofn);
Ta = T(T.FirstSeen >= cutoffDate,:);
[ulocality,ia,ib] = unique(Ta.LOCALITY_NAME);
clocality = accumarray(ib,1,size(ulocality));
Tu = table(ulocality,clocality,'VariableNames',{'Locality','COUNT'});
ofn = [droot,'numOnMachineRecords_after_20211101_byLocality.csv'];
Ta = T(T.FirstSeen >= cutoffDate,:);
[uprecinct,ia,ib] = unique(join([string(Ta.LOCALITY_NAME),string(Ta.PRECINCT_NAME)]));
cprecinct = accumarray(ib,1,size(uprecinct));
Tu = table(Ta.LOCALITY_NAME(ia),Ta.PRECINCT_NAME(ia),cprecinct,'VariableNames',{'LOCALITY','PRECINCT','COUNT'});
ofn = [droot,'numOnMachineRecords_after_20211101_byLocalityByPrecinct.csv'];
The adjusted MATLAB parser function is listed below:
function dal = import2021DALfile(filename, dataLines)
%IMPORTFILE Import data from a text file
% DAILYABSENTEELIST10162021 = IMPORTFILE(FILENAME) reads data from text
% file FILENAME for the default selection. Returns the data as a table.
% DAILYABSENTEELIST10162021 = IMPORTFILE(FILE, DATALINES) reads data
% for the specified row interval(s) of text file FILENAME. Specify
% DATALINES as a positive scalar integer or a N-by-2 array of positive
% scalar integers for dis-contiguous row intervals.
% Example:
% dal = import2021DALfile("SourceData/DAL/Daily_Absentee_List_10162021.csv", [2, Inf]);
% See also READTABLE.
% Auto-generated by MATLAB on 16-Oct-2021 14:19:26
%% Input handling
% If dataLines is not specified, define defaults
if nargin < 2
dataLines = [2, Inf];
%% Set up the Import Options and import the data
opts = delimitedTextImportOptions("NumVariables", 38);
% Specify range and delimiter
opts.DataLines = dataLines;
opts.Delimiter = ",";
% Specify column names and types
opts.VariableNames = ["ELECTION_NAME", "LOCALITY_CODE", "LOCALITY_NAME", "PRECINCT_CODE", "PRECINCT_NAME", "LAST_NAME", "FIRST_NAME", "MIDDLE_NAME", "SUFFIX", "ADDRESS_LINE_1", "ADDRESS_LINE_2", "ADDRESS_LINE_3", "CITY", "STATE", "ZIP", "COUNTRY", "INTERNATIONAL", "EMAIL_ADDRESS", "FAX", "VOTER_TYPE", "ONGOING", "APP_RECIEPT_DATE", "APP_STATUS", "BALLOT_RECEIPT_DATE", "BALLOT_STATUS", "identification_number", "PROTECTED", "CONG_CODE_VALUE", "STSENATE_CODE_VALUE", "STHOUSE_CODE_VALUE", "AB_ADDRESS_LINE_1", "AB_ADDRESS_LINE_2", "AB_ADDRESS_LINE_3", "AB_CITY", "AB_STATE", "AB_ZIP", "BALLOTSTATUSREASON", "Ballot_Comment"];
opts.VariableTypes = ["string", "string", "categorical", "string", "string", "string", "string", "string", "string", "string", "string", "string", "categorical", "categorical", "categorical", "categorical", "categorical", "string", "string", "string", "categorical", "string", "categorical", "string", "categorical", "double", "string", "categorical", "categorical", "categorical", "string", "string", "string", "string", "string", "string", "string", "string"];
% Specify file level properties
opts.ExtraColumnsRule = "ignore";
opts.EmptyLineRule = "read";
% Specify variable properties
opts = setvaropts(opts, ["ELECTION_NAME", "LOCALITY_CODE", "PRECINCT_CODE", "PRECINCT_NAME", "LAST_NAME", "FIRST_NAME", "MIDDLE_NAME", "SUFFIX", "ADDRESS_LINE_1", "ADDRESS_LINE_2", "ADDRESS_LINE_3", "EMAIL_ADDRESS", "FAX", "VOTER_TYPE", "APP_RECIEPT_DATE", "BALLOT_RECEIPT_DATE", "PROTECTED", "AB_ADDRESS_LINE_1", "AB_ADDRESS_LINE_2", "AB_ADDRESS_LINE_3", "AB_CITY", "BALLOTSTATUSREASON", "Ballot_Comment"], "WhitespaceRule", "preserve");
opts = setvaropts(opts, ["ELECTION_NAME", "LOCALITY_CODE", "LOCALITY_NAME", "PRECINCT_CODE", "PRECINCT_NAME", "LAST_NAME", "FIRST_NAME", "MIDDLE_NAME", "SUFFIX", "ADDRESS_LINE_1", "ADDRESS_LINE_2", "ADDRESS_LINE_3", "CITY", "STATE", "COUNTRY", "INTERNATIONAL", "EMAIL_ADDRESS", "FAX", "VOTER_TYPE", "ONGOING", "APP_RECIEPT_DATE", "APP_STATUS", "BALLOT_RECEIPT_DATE", "BALLOT_STATUS", "PROTECTED", "CONG_CODE_VALUE", "STSENATE_CODE_VALUE", "STHOUSE_CODE_VALUE", "AB_ADDRESS_LINE_1", "AB_ADDRESS_LINE_2", "AB_ADDRESS_LINE_3", "AB_CITY", "AB_STATE", "BALLOTSTATUSREASON", "Ballot_Comment"], "EmptyFieldRule", "auto");
% Import the data
dal = readtable(filename, opts);
% Perform some cleanup on commonly found issues...
dal.Ballot_Comment = strrep(dal.Ballot_Comment,char([13,10]),". ");
dal.BALLOTSTATUSREASON = strrep(dal.BALLOTSTATUSREASON,char([13,10]),". ");
dal.LOCALITY_NAME = categorical(strtrim(string(dal.LOCALITY_NAME)));
dal.PRECINCT_NAME = categorical(regexprep(strtrim(string(dal.PRECINCT_NAME)),'^(\d+)( +)(\w+)','$1 - $3'));
On a spur of curiosity I went back to some of the data provided by the VA dept of elections (“ELECT”) for both the 2020 and 2021 elections and ran a new data consistency test …
I have a copy of the final Daily Absentee List (DAL) for both 2020 and 2021. I also have a copy of the paired Registered Voter List (RVL) and Voter History List (VHL) generated shortly after the
close of the 2021 General Election and within a few moments of each other.
I was curious what the percentage of approved and counted absentee ballots from the DAL is that do NOT have an associated “voter credit” in the VHL for both 2020 and 2021. If ELECT’s data is accurate
the number should be ideally 0, but most official thresholds for acceptability that I’ve seen for accuracy in election data systems hover somewhere around 0.1%. (0.1% is a fairly consistent standard
that I’ve seen per the documentation for various localities Risk Limiting Audits, and the Election Scanner Certification procedures, etc.) The VHL should cover all of the activity for the last four
years, but to ensure that I’m accounting for people that might have been officially removed from the RVL and VHL since the 2020 election (due to death, moving out of state, etc), I only run this test
on the subset of the entries in the DAL that still have a valid listings in the RVL.
The results are below. Both years seem to have a high amount of discrepancies compared to the 0.1% threshold, with 2020’s discrepancy percentage being over 3x the percentage computed for 2021.
Year Percent of Counted DAL Ballots without Voter Credit
2020 1.352%
2021 0.449%
For those interested in the computation, the MATLAB pseudo-code is given below. I can’t actually link to the source data files because of VA’s draconian restrictions on redistributing the contents of
the DAL, RVL and VHL data files.
% We first compute the indices of the DAL entries that represent
% approved and countable ballots ...
% 'dal2020' and 'dal2021' variables are the imported DAL tables
% 'VAVoteHistory' is the imported Voter History List
% 'RegisteredVoterList' is the Registered Voter List
% All four of the above are imported directly from the CSV
% files provided from the VA Department of elections with
% very little error checking save for obvious whitespace or
% line ending checks, etc.
aiv2021 = dal2021.APP_STATUS == 'Approved' & dal2021.BALLOT_STATUS == 'Issued';
amv2021 = dal2021.APP_STATUS == 'Approved' & dal2021.BALLOT_STATUS == 'Marked';
aomv2021 = dal2021.APP_STATUS == 'Approved' & dal2021.BALLOT_STATUS == 'On Machine';
appv2021 = dal2021.APP_STATUS == 'Approved' & dal2021.BALLOT_STATUS == 'Pre-Processed';
afwv2021 = dal2021.APP_STATUS == 'Approved' & dal2021.BALLOT_STATUS == 'FWAB';
counted2021 = amv2021 | aomv2021 | appv2021 | afwv2021; % Approved and Countable
aiv2020 = dal2020.APP_STATUS == 'Approved' & dal2020.BALLOT_STATUS == 'Issued';
amv2020 = dal2020.APP_STATUS == 'Approved' & dal2020.BALLOT_STATUS == 'Marked';
aomv2020 = dal2020.APP_STATUS == 'Approved' & dal2020.BALLOT_STATUS == 'On Machine';
appv2020 = dal2020.APP_STATUS == 'Approved' & dal2020.BALLOT_STATUS == 'Pre-Processed';
afwv2020 = dal2020.APP_STATUS == 'Approved' & dal2020.BALLOT_STATUS == 'FWAB';
counted2020 = amv2020 | aomv2020 | appv2020 | afwv2020; % Approved and Countable
% Next we compute the indices in the VHL that represent
% 2020 and 2021 General Election entries for voter credit
valid_2020_entries = strcmpi(strtrim(string(VAVoteHistory.ELECTION_NAME)), '2020 November General');
valid_2021_entries = strcmpi(strtrim(string(VAVoteHistory.ELECTION_NAME)), '2021 November General');
% We use the MATLAB intersect function to make sure that
% we are only using DAL entries that are still in the RVL
% and therefore are possible to be present in the VHL and
% compute the percentages.
[did,iida,iidb] = intersect(dal2020.identification_number(counted2020), ...
[vid,iida,iidb] = intersect(VAVoteHistory.IDENTIFICATION_NUMBER(valid_2020_entries),...
[iid,iida,iidb] = intersect(did,vid);
pct2020 = (1-numel(iida) / numel(did)) * 100
[did,iida,iidb] = intersect(dal2021.identification_number(counted2021), ...
[vid,iida,iidb] = intersect(VAVoteHistory.IDENTIFICATION_NUMBER(valid_2021_entries),...
[iid,iida,iidb] = intersect(did,vid);
pct2021 = (1-numel(iida) / numel(did)) * 100
|
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|
New type on the block: Generating high-precision orbits for GPS III satellites - GPS World
Read Richard Langley’s introduction to this article: Innovation Insights: Antennas and photons and orbits, oh my!
To produce GNSS satellite orbit ephemerides and clock data with high precision and for all constellations, the Navigation Support Office of the European Space Agency’s European Space Operations
Centre (ESA/ESOC) continually strives to keep up and improve its precise orbit determination (POD) strategies. As a result of these longstanding efforts, satellite dynamics modeling and GNSS
measurement procedures have progressed significantly over the last few years, especially those developed for the European Galileo satellites. Because the accuracy of ESA/ESOC’s GNSS orbits has
reached such a high level (about 1 to 3 centimeters), introducing a completely new type of GNSS satellite into the processing is not as easy as it used to be. New spacecraft models – the first and
foremost being a model for a satellite’s response to solar radiation pressure (SRP) – are needed for the “newcomer” so that the quality of the overall multi-GNSS solution does not suffer. Just as
important are spacecraft system parameters, or metadata, such as the location of the satellite antenna’s electrical phase center and the satellite attitude law.
In this article, we show the efforts we have made at ESA to bring the quality of our orbit estimates for the GPS Block III satellites up to par with those for Galileo and the earlier GPS satellite
blocks. We report on the results from on-ground and in-flight determinations of the Block III transmit antenna phase center characteristics up to 17 degrees from the antenna boresight direction.
Moreover, we take advantage of the non-zero horizontal offsets of the transmit antenna from the spacecraft’s yaw axis to estimate the satellite yaw angle during Earth eclipse season and present a
simple analytical formula for its calculation. Finally, we describe the development and validation of improved radiation force models for the Block III satellites.
We start, however, by giving a brief overview of the GPS Block III program.
The U.S. Space Force GPS Block III (previously referred to as Block IIIA) is a series of 10 satellites being procured by the United States to bring new future capabilities to both military and civil
positioning, navigation, and timing (PNT) users across the globe. Designed and manufactured by defense contractor Lockheed Martin (LM), the satellites are reported to deliver three times better
accuracy, 500 times greater transmission power, and an eightfold enhancement in anti-jamming functionality over previous GPS satellite blocks. At ESA/ESOC, we are paying particular attention to this
new tranche of satellites as they are the first to broadcast L1C, a new common signal interoperable with other GNSS, including Galileo.
At the time of this writing, there are six GPS III space vehicles (SVs) in orbit. The first one – nicknamed “Vespucci,” in honor of Italian explorer Amerigo Vespucci – lifted off atop a SpaceX Falcon
9 rocket from Cape Canaveral Air Force Station, Florida, in December 2018, and entered service on January 13, 2020. An additional four SVs are expected to be launched soon, before moving on to an
updated version called GPS IIIF (“F” for Follow On). The first Block IIIF satellite is projected to be available for launch in 2026.
In view of the growing number of GPS III SVs in orbit, and soon to be joined by IIIFs, accurate spacecraft models and metadata information are becoming more and more important in order to maximize
PNT accuracy.
GNSS signal measurements refer to the electrical phase center of the satellite transmitting antenna, which is neither a physical nor a stable point in space. The variation of the phase center
location as a function of the direction of the emitted signal on a specific frequency is what we call the phase center variation (PCV). The mean phase center is usually defined as the point for which
the phase of the signal shows the smallest (in a “least-squares” sense) PCV.
The point of reference for describing the motion of a satellite, however, is typically the spacecraft center of mass (CoM). The difference between the position of the mean phase center and the CoM is
what we typically refer to as the satellite’s antenna phase center offset (PCO). Both PCO and PCV parameters must be precisely known — from either a dedicated on-ground calibration or one performed
in flight — so that we can tie our GNSS carrier-phase measurements consistently to the satellites’ CoM.
On-Ground Calibrations. Like for previous GPS vehicles, the Block IIR and Block IIR-M satellites, LM has fully calibrated the GPS III transmit antennas prior to launch at their ground test
facilities. Antenna offset parameters for all three carrier signals (L1, L2 and L5) were posted on the U.S. Coast Guard Navigation Center (NAVCEN) website (www.navcen.uscg.gov) shortly after each
satellite launch. In December 2021, NAVCEN released the PCOs for SV number (SVN) 78, along with updates to the first four satellites (see Table 1). About ten months later, in October 2022, the
antenna pattern for each satellite and signal frequency were published (see Figure 1).
The December 2021 offsets are referred to as predicted values at the end of year one on orbit. They differ from the previous ones by several centimeters in both vertical (Z) and horizontal (X and Y)
directions. Particularly surprising are the X- and Y-PCOs, which were initially reported to be close to zero. The differences in the horizontal PCOs have generated uncertainty and debate, especially
within the International GNSS Service (IGS) about which values to adopt for the new antenna model release (igs20.atx). Testing of the two different PCO datasets in our software demonstrated that the
non-zero values as given in Table 1 are the significantly more accurate ones. We will return to this later in this article.
Combined Ground- and Space-Based Tracking. In this part of this article, we discuss the combination of dual-frequency tracking data from geodetic-quality GPS receivers in low Earth orbit (LEO) with
those from a global receiver network on the ground to determine the phase center parameters of the GPS Block III transmit antennas. The LEO-based measurements were taken by the GNSS receivers on
board the ocean altimetry satellites Sentinel-6 Michael Freilich and Jason-3. The 1,336-km altitude of both of these missions enables the estimation of the GPS satellite antenna PCVs from 0 up to 17
degrees from boresight while GPS receivers on Earth can only see the satellites up to a maximum angle of 14 degrees. The 14-degree limit is also referred to as the GPS satellites’ edge of Earth (EoE)
For the modeling of the PCVs we follow the approach of the IGS using piece-wise linear functions of the boresight angle and constraining the PCV values to between 0 and 14 degrees to have zero mean.
Furthermore, we employ fully normalized spherical harmonic expansions of degree 8 and order 5 to solve for the azimuth- and elevation-angle-dependent PCVs of the orbiting receiver antennas. The IGS
standard antenna phase center corrections from igs20.atx are applied to all terrestrial receiver and GPS Block II transmit antennas.
The estimated Block III antenna PCVs are depicted in Figure 2. The estimates for the five individual antennas match each other to within 0.4 millimeters root-mean-square (RMS) (see Figure 2, top).
The agreement among the PCVs that we get when processing the tracking data from each LEO receiver’s antenna separately is at the sub-millimeter level, too (see Figure 2, middle). Overall, the level
of consistency suggests that the PCVs are of very good quality and that a block-specific representation is sufficient for precise applications. Comparison of the final block-specific PCV estimates
against the values from the current IGS antenna model and from the ground calibrations shows strong agreement (RMS = 0.7 millimeters) between 0 and 14 degrees from boresight (see Figure 2, bottom).
Beyond the 14-degree limit, the differences compared to the IGS standard are up to three centimeters, underlying the urgent need for an update of the igs20.atx file.
Applying the extended PCV corrections as part of the POD process to the GPS LEO receiver data shows significant improvement in the post-fit carrier-phase residuals when compared to the PCV
corrections from the IGS legacy model. It removes a previously existing boresight angle-dependent trend and leads to a more than 20% reduction in the computed residual RMS (see Figure 3).
GNSS satellites cannot follow an ideal yaw-steering whenever the Sun elevation angle relative to the orbital plane (the so-called beta angle) gets too low and the yaw rate required to keep the
satellite solar panels pointing towards the Sun exceeds the maximum satellite yaw rate. The strategies on how GNSS satellites perform rate-limited yaw-steering are different for each type of
spacecraft and only partly documented for public users. Continuous knowledge of GNSS spacecraft yaw attitude, however, is important for kinematic and dynamic reasons. Errors in yaw are known to
affect the modeling of transmit antenna phase center’s position, carrier-phase wind-up, and radiation pressure forces. On the other hand, when the mean antenna phase center location is offset from
the spacecraft’s Z-axis, the satellite yaw state can be estimated instantaneously from the tracking data of a global receiver network. The approach behind this is commonly referred to as “upside
down” or “reverse kinematic precise point positioning” (RPP). The horizontal antenna offset vector can be viewed here as a kind of rotating lever arm whose length determines the accuracy of the yaw
angle estimates. Since the Block III X-offset is just 7 centimeters, one should not expect the same RPP accuracy as for other GNSS satellites like those of the GPS IIF or GLONASS-M series, which have
an X-offset that is six (GPS IIF) or even eight (GLONASS-M) times larger.
Nonetheless, with more than three hundred ground stations, kinematic RPP works reasonably well even for GPS III as we can see from Figure 4, which shows the estimated yaw angle of SVN 78 while
passing orbit noon and orbit midnight with a Sun elevation angle of almost zero degrees. The plots suggests that Block III satellites — unlike previous Block IIA and IIF SVs — perform their yaw slews
near noon and near midnight in the same way and at the same yaw rate. In this respect, the yaw turn behavior is similar to that of the IIR/IIR-M satellites. However, with a maximum yaw rate of 0.10
degrees per second, the Block III satellites rotate only half as fast as those of the IIR/IIR-M family. What is also different is the start time of the yaw maneuver. As can be seen from Figure 4, the
maneuver does not start when the required yaw rate exceeds the physical limit but already a couple of minutes before.
The RPP analysis has led to the development of a simple yaw model for the Block III satellites. For a Sun elevation angle β below β[0] = 4.780 degrees, the yaw angle can be approximated with an RMS
accuracy of about 8 degrees by the following formula:
is a modified Sun elevation angle, SIGN(β[0], β) a FORTRAN function returning the value of β[0] with the sign of β, and η is the satellite’s argument of latitude with respect to orbit midnight. The
agreement between estimated and modelled yaw angles is illustrated in Figure 5.
Fourier Series for Radiation Force Modeling. The most critical component determining the shape of a GNSS satellite’s trajectory is SRP – the force caused by the impact of solar photons hitting the
satellite’s surfaces. A satellite’s sensitivity to SRP can be characterized by the variation of the cross-sectional area to mass ratio (A/M) of the satellite body as it orbits Earth and the Sun. The
greater the change in A/M, the higher the sensitivity. From this perspective, the Block III spacecraft can be considered the most sensitive in GPS history.
Based upon LM’s tried-and-true A2100 bus, the satellite is much more elongated than previous generations. With an estimated size of 7.5 meters squared, the X-side is almost twice as large as the
Z-side. Depending on the elevation angle of the Sun relative to the orbital plane, the body’s cross-sectional area exposed to sunlight varies between 4.0 and 8.5 meters squared (See Figure 6). With a
nominal on-orbit weight of approximately 2,160 kilograms, this results in a change of A/M of 0.0021 meters squared per kilogram. For comparison, the corresponding values for the previous GPS SVs are
0.0015 (IIF), 0.0017 (IIR), and 0.0013 (IIA) meters squared per kilogram.
Given the size and shape of Block III spacecraft, an appropriate radiation force model is considered mandatory to achieve the highest orbit accuracy possible. With that said, we empirically derived a
set of background force models for the first five GPS III satellites. Our approach rests on dynamical long-arc (9-day) fitting to precise orbit data spanning up to three years and the following
low-order Fourier functions of the Earth-spacecraft-Sun angle ε to represent the radiation force in the satellite body-fixed system:
The Fourier coefficients (XS1, XS2, XS3, YC2, ZC1, ZS2 and ZS4) are iteratively adjusted together with initial epoch state, a constant Y-axis bias, and 1‐cycle per revolution along‐track parameters
to best fit the orbit data in a least-squares sense. All individual 9-day arc solutions are rigorously combined on a normal equations level to form a robust set of Fourier model coefficients for each
satellite or group of satellites.
To investigate the effect of the transmit antenna PCOs and the Fourier force models on the satellite orbits, we use our ESA/IGS processing strategy to generate dynamic 24-hour-arc solutions spanning
January 2020 to December 2022, first with zero PCO and the non-zero horizontal offsets from Table 1 and no a-priori radiation force model, then with the non-zero offsets and the additional Fourier
model in the background. The direct comparison of the generated orbits reveals significant differences for the Block III satellites of about 0.1 meters (3D).
To demonstrate the improved performance of the non-zero offsets and the Fourier model, we take the orbits for successive days and look at the midnight epoch where they overlap. The difference in the
orbit position, subsequently referred to as “overlap error,” gives us a worst case estimate of the satellite orbit accuracy. Comparison of the overlap errors provides evidence that the Block III
orbits are much more accurate when using the non-zero rather than the zero X and Y PCOs. The overall 3D overlap RMS reduces from 49.5 millimeters (with zero PCOs) down to 32.3 millimeters (with
non-zero PCOs). Results for the Sun elevation regions below 45 degrees, in particular, show significant improvement (see Figure 7).
Use of the Fourier model has additional positive impact on the overlaps. Comparing the orbits produced with and without the a-priori radiation force model, we see a decrease in the 3D overlap error
RMS from 32.3 to 29.7 millimeters averaged over all satellites. The orbit component that benefits most from both the improved antenna phase and the advanced force modeling is the one normal to the
satellite orbital plane (across track). The SVs improving the most are SVN 75 and SVN 78, though significant improvements can be seen for all other satellites too (see Table 2).
Another means of assessing the quality of spacecraft models is the size and variability of the five-plus-three empirical dynamic radiation pressure parameters that we still estimate on a daily basis
for each GNSS satellite in addition to its a-priori force model. Introducing the non-zero PCO and Fourier models into the POD turned out to reduce the size of the empirical parameters and their
dependency on the satellite-Sun geometry to a great extent as the example in Figure 8 demonstrates.
Integer ambiguity resolution — that is, resolving the unknown cycle ambiguities of double-differenced carrier-phase data to integer values — is considered indispensable to GNSS satellite POD and
commonly results in a factor of two improvement in orbit precision. Of particular importance is the narrow-lane ambiguity that results from combining the carrier-phase measurements from a pair of
GNSS frequencies. One of the intermediate steps in the ambiguity resolution algorithm is the fixing of the double-differenced narrow-lane ambiguities to integer values. For reliable fixing, the
fractional part of the difference between the integer and decimal (float) values should be as close as possible to zero and follow a symmetrical distribution. The “tailedness” of the distribution
curve may be characterized by its kurtosis — the larger the kurtosis, the fewer values are in the tails of the distribution and the more peaked is the distribution. In other words, the larger the
kurtosis, the closer the “fractionals” cluster around zero, the more ambiguities can be resolved with higher confidence, and the more accurate the resolved solution. Moreover, as satellite orbit and
antenna phase center errors do not cancel out completely through double-differencing, the narrow-lane kurtosis may also be considered as an indicator for the accuracy of the satellite force and phase
center models that were used. The results in Figure 9 show that the non-zero horizontal PCOs bring a major improvement and that the Fourier force model does give some additional benefit.
Adding a new GNSS satellite type to high-precision multi-GNSS solutions requires detailed knowledge and understanding of the satellite type. Key issues are the transmit antenna phase center
parameters, the satellite’s attitude, and the radiation pressure forces acting on its surfaces.
In this article, improved antenna phase center, attitude, and radiation pressure models for the current series of GPS Block III spacecraft have been developed using multiple years of in-flight orbit
and tracking data. A number of internal metrics such as post-fit carrier-phase residuals, day-boundary orbit differences (overlaps), empirical acceleration parameters, and carrier phase ambiguity
statistics have been used to gauge the models’ performances. Overall, the results underscore the importance of the models for GPS III orbit determination. This applies primarily to the radiation
force and the antenna phase center model, or more precisely, the horizontal (X and Y) offsets of the phase center model whose existence has been neglected for years in the analysis of GPS III data.
Comparison of the overlap statistics suggest that orbits generated based upon updated (non-zero) phase center corrections and ESA/ESOC’s new (Fourier-based) radiation pressure model in the background
are better by almost a factor of two. The average overlap RMS errors calculated across all current Block III SVs and for each orbital component (radial, along track and across track) dropped from 21
, 28 and 35 millimeters down to 14, 21 and 16 millimeters, respectively.
More relevant when it comes to processing GPS data recorded on board low-flying satellites such as Sentinel-6 Michael Freilich or Jason-3, is the extension of the current IGS Block III antenna PCV
model beyond a 14-degree boresight angle. After applying the extended PCV corrections, we reduced Block III carrier-phase residuals by 20% with no or few systematic signatures remaining, unlike the
residuals produced with the current IGS antenna model. The IGS is strongly encouraged to adopt the Block III PCV extension into their antenna model to continue to support GPS-based POD of
low-Earth-orbiting satellites.
For further details on ESA/ESOC’s solar radiation pressure modeling approach, see our paper “GPS III Radiation Force Modeling” presented at the IGS 2022 Virtual Workshop: click here.
FLORIAN DILSSNER is a satellite navigation engineer in the Navigation Support Office at the European Space Operations Centre (ESOC) of the European Space Agency (ESA), Darmstadt, Germany. He earned
his Dipl.-Ing. and Dr.- Ing. degrees in geodesy from the University of Hannover, Germany.
TIM SPRINGER has been working for the Navigation Support Office at ESA/ESOC since 2004. He received his Ph.D. in physics from the Astronomical Institute of the University of Bern in 1999.
FRANCESCO GINI is a satellite navigation engineer in the Navigation Support Office at ESA/ESOC. He received his Ph.D. in astronautics and space sciences from the Centro di Ateneo di Studi e Attività
Spaziali at the University of Padova in 2014.
ERIK SCHÖNEMANN is a satellite navigation engineer in the Navigation Support Office at ESA/ESOC. He earned his Dipl.-Ing. and Dr.- Ing. degrees in geodesy from the University of Darmstadt, Germany.
WERNER ENDERLE is head of the Navigation Support Office at ESA/ESOC. He holds a doctoral degree in aerospace engineering from the Technical University of Berlin, Germany.
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The preceding Fourier pair can be used to show that
Proof: The inverse Fourier transform of is
In particular, in the middle of the rectangular pulse at , we have
This establishes that the algebraic area under is 1 for every . Every delta function (impulse) must have this property.
We now show that also satisfies the sifting property in the limit as . This property fully establishes the limit as a valid impulse. That is, an impulse is any function having the property that
for every continuous function . In the present case, we need to show, specifically, that
Define . Then by the
power theorem
Then as , the limit converges to the algebraic area under , which is as desired:
We have thus established that
For related discussion, see [
, p. 127].
Next Section: Impulse TrainsPrevious Section: Rectangular Pulse
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Essential Aspects of Mathematics as a Practice in Research and Undergraduate Instruction
Document Type
Presentation Date
Abstract or Description
A gap between mathematics as used by mathematicians and mathematics as experienced by undergraduate mathematics students has persistently been identified as problematic; A commonly proposed solution
is to provide opportunities for students to do mathematics and be mathematicians (e.g., Whitehead, 1911; Harel, 2008). Conceptions or beliefs about what this means may vary depending on a
mathematician’s research and experience. The authors explore mathematicians’ expressed conceptions of mathematics in their research and in their teaching.
Conference on Research in Undergraduate Mathematics Education (CRUME)
Recommended Citation
Stehr, Eryn M., Tuyin An. 2018. "Essential Aspects of Mathematics as a Practice in Research and Undergraduate Instruction." Department of Mathematical Sciences Faculty Presentations. Presentation
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The Monument is Giving Up its Secrets!
by Jeffrey Meiliken | Jul 29, 2010 | Revelations | 4 comments
There’s more to the Washington Monument than we first revealed, but more important is what it connects us to as a metaphor: the central axis of the Tree-of-Life; the 42-Letter Sword of Moses; the 10
Commandments; and even Joseph’s Pyramid and the Future Holy Temple. It’s not as odd as it sounds and we’ll explain all these concepts and we promise we’ll then segue into them in subsequent articles
with much fuller explanations.
Since the Monument was built to resemble a giant Egyptian obelisk dedicated to the sun, let’s start with the word Cap, as in capstone, which is spelled (C-P). As per the Arizal, the letters CP are
two of the revolving 7-letter sequence that connects to the energy of and spiritually controls the influence of the 7 planets CPRTBGD; they are the first two when it comes to controlling the Sun,
whose surface temperature as we’ve previously noted is 5778 K.
Here is a secret about the letter Caf (CP). While its numerical value sofit is 820, the same as the all-important phrase at Vayikra 19:18, paraphrased as “Love thy neighbor as thyself,” Caf (CP)’s
ordinal value is 28, making its’ complete value (820+28) = 848, or 2 x 424, the numerical value of Moshiach Ben David (the Messiah).
Nevertheless, besides representing Keter(CTR), the crowning sefira and besides also representing the Sun, as explained by Abraham, the Patriarch in his Book the Sefer Yetzira,” , the letter Caf (C)
when spelled out (CP) also spells out the word for palm, as in the palm of our hands.
Thus our 2 palms obviously connect with the two 424’s, but they also connect with our 10 fingers, 5 on one side, 5 on the other, like the 10 sefirot, 10 Commandments on the 2 tablets, and like
Abraham’s description, “5 on one side, 5 on the other, cleaved down the middle.” And the reason this is so significant will become clear once we examine why the 4 bases of the Washington Monument all
had to be 55’ long.
Meanwhile, the Zohar supports this revelation in that in portion Ekev it states emphatically that the palms connect to the 14 joints of the fingers and that there are 28 joints in both hands, which
is the numerical value of Koach, power.
Like the weight of the aluminum capstone at the tip of the Washington Monument (6.25 lbs) the ratio of its capstone (pyramidian) height to its base is also 34.45/55.125 ft = .625, reinforcing the
image of the capstone pyramid as “H’Keter (the Crown)” of numerical value 625. It also connects it the Torah, whose square root of its total number of words, verses, and letters is exactly 625.
Just in case you think the Masons accidentally hit upon this ratio, please note that the ratio of the bases of the entire Monument to that of the capstone are also 34.45/55.125 ft = .625,
Another very significant and telling ratio chosen by the Monument’s architects is given by the Monument’s height to its base is 555/55.5 ft = 10 for a ratio of 10 to 1, or inversely 1 ft of width for
every 10 of height, a very steep angle indeed. More significant than its angle of inclination is its similarity of structure to another Keter that involves 10: The 10 Commandments.
The 10 Commandments (Utterances) found in the 13 verses of chapter 20 (the numerical value of the Hebrew letter Caf) of Exodus consist of exactly 620 letters, the numerical value of the word Keter
and of the Hebrew word for 20 (Esrim), an obvious allusion to the point the Israelites had reached as they were being offered the Tree-of-Life reality. Nevertheless, the similarity in structure is
delineated by the 62 letter Yuds (of numerical value 10 each)—the first letter of the Tetragrammaton (YHVH)—for a total of 620 (Keter once again), meaning that 1 in 10 of every letter in the 10
Commandments is the letter Yud(Y) of numerical value 10.
Like the 2 palms and the 2 sets of Tablets given to Moses, there are 2 recitals of the 10 Commandments in the Torah, for a total of 20, one in Exodus and the other in Deuteronomy. While the first has
620 letters, the second has 708 letters, with 708 being the numerical value of the Upper 42-Letter Name (the 42 letters of the 3 iterations of the spelled out Tetragrammaton (YHVH). We only bring up
the 42 Letters because there are 42 Letters in the names of the 11 sefirot of the Tree-of-Life, the central column of which the Washington Monument may be a metaphor.
And between the 2 recitals there are 130 Yuds(Y) with 130 being the numerical value for Sinai and for (Sulam) ladder. And while some say that 708 relates to 5778 and 5708, the year Israel received
statehood (70 years before 5778 as prophesied in the Zohar 2000 years ago, there is a less subtle connection between 5778 and the 10 Commandments. The first set, found at Exodus 20:2 is located
exactly at the 107007^th letter in the Torah or the 107000^th letter from the word Bereshit (“In the Beginning”), and 5778 is the exact sum of all the positive integers from 1 to 107. This obviously
can’t be coincidental, and as divine confirmation the first 2 words of these 10 Commandments “Anochi YHVH (I am G-d)” have the exact numerical value 107, and they contain the first 2 Yuds(Y) of the
62, back-to-back right in their middle (ENCY YHVH). Moreover, these 2 words (I am G-d) have an ordinal value of 62.
And by the way, the last word in the 10 Commandments has the newly revealed gematria sofit sofit value of 820, the same as Caf (CG) and of the singular Torah verse of unconditional love, as mentioned
Thus the Monument can also be seen as a giant letter Vav(V), which in Hebrew is a vertical line, capped by the Hebrew letter Caf (C) and as any Kabbalist knows CV of numerical value 26 represents The
Tetragrammaton (YHVH), the ineffable Name of G-d associated with the 6 sefirot (dimensions) bundled together and called Zeir Anpin, metaphorically the vertical pipeline from our world into the upper
spiritual (Heavenly) one.
And speaking of metaphors, while the Washington Monument is found at the edge of the reflecting pool at NW 15^th street, the 42-letters Sword of Moses is found right after the crossing of the Red Sea
(Yam Suf or End Sea) at paragraph 42 of the Book of Shmot, Exodus 15:11. Appropriately enough, the sword is tipped with the word and letter Alef, which is shaped like a X comprised of 4 component
letter, 2 of which are the letter Yud(Y). If you look upon the monument as a 4-sided sword, you’ll see the X formed by the 4 angles at the top.
There is one more monumental connection to make, of which there is no doubt the Masons were aware, for, you see, the entrance to the Great Pyramid (referred to as Joseph’s pyramid in the series of
articles we began late last year and hope to continue with soon as a prelude to The Future Holy Temple) is 55 ft off the ground, and that pyramid too had a special capstone high atop of it.
Furthermore, the ratio of the height of the Washington Monument to the Great Pyramid is precisely 15/13 or 1.1538 as in the 115 jubilee years from Adam to the year prophesied for the 3^rd opportunity
of the Tree-of-Life reality, 5778.
And if you don’t think the architects of the Monument had some inkling as to what they were connecting to, please note that the perimeter of the capstone is 137 ft and 137 is the well-known numerical
value of Kabbalah. Moreover, Moreover, the Base of the Great Pyramid is 13.7 times that of the Washington Monument, a giant Egyptian Obelisk. Furthermore, the resultant diagonals of the capstone are
77.85 (NE-SW) and 78 (NW-SE).
Moreover, the dimension 55 ft (5 on one side, 5 on the other, cleaved down the middle) hides the secret of the ancient cubit needed for the building of the Future Holy Temple, and 555 of those cubits
equals 1271, the year the Zohar mysteriously reappeared in Spain.
And the Future Holy Temple will be 100 of Caf (CP) cubits high, once again connecting with Keter and that Capstone.
Nevertheless, 555 ft = 242.18 cubits, and since the year 5778 HC is also the year 2018 CE, this year of prophesy is exactly 242 years after the founding of the United States of America in 1776.
Food for thought.
In the coming weeks we’ll be exploring in depth the various avenues brought about from the various metaphors and connections laid out above.
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Introduction to Dodged Bar Plot — Matplotlib, Pandas and Seaborn Visualization Guide (Part 2.1) - One Zero Blog
A bar plot is a graphical representation which shows the relationship between a categorical and a numerical variable. In general, there are two popular types used for data visualization, which are
dodged and stacked bar plot.
A dodged bar plot is used to compare a grouping variable, where the groups are plotted side by side. It could be used to compare categorical counts or relative proportions, and in general used to
compare numerical statistics such as mean/median.
In the current article, we will deal with count-based bar plots where we compare the proportions corresponding to a grouping variable.
Article outline
The current article will cover the following:
Loading libraries
The first step is to load the required libraries.
import numpy as np # array manipulation
import pandas as pd # Data Manipulation
import matplotlib.pyplot as plt # Visualisation
import seaborn as sns # Visualisation
Basic knowledge of matplotlib’s subplots
If you have basic knowledge of matplotlib’s subplots( ) method, you can proceed with the article, else I will highly recommend reading the first blog on this visualisation guide series.
Link: Introduction to Line Plot — Matplotlib, Pandas and Seaborn Visualization Guide (Part 1)
Basic barplot using Rectangle method
In this article, we will learn how to generate dodge plots. But before we proceed with such advanced statistical plot, first we need to be familiar with how matplotlib builds a bar plot step by step.
To build a bar plot, we need to go through the following steps:
Step 1: From matplotlib.patches import Rectangle.
Step 2: Using plt.subplots( ) instantiate figure (fig) and axes (ax) objects.
Step 3: Use the Rectangle( ) method to generate the patch/rectangle object. The Rectangle( ) method takes x and y as tuple, then width and height of the bar.
Step 4: We generate two such Rectangle\patch object (rec1 and rec2) that we are going to add\impose of the axes (ax) object.
Step 5: Now add these patch/rectangle objects on the axes (ax) using add_patch( ) method.
from matplotlib.patches import Rectangle
fig, ax = plt.subplots()
# Define rectangle
# Rectangle((x, y), width, height)
rec1 = Rectangle((0.1, 0), 0.2, 0.9)
rec2 = Rectangle((0.5, 0), 0.2, 0.5)
# Adding patch object/ rectangles
First bar plot
Help on the methods
You can get help using the python’s inbuilt help( ) method, where you can supply any object name (for example Rectangle) to get information on the associated attributes and methods.
Check for the Patches
Let’s check whether the axes (ax) object contains the paches/rectangles. We can check that by applying the attribute patches on axes object (ax). The output clearly shows that the axes object
contains two patches.
<Axes.ArtistList of 2 patches>
Changing the Rectangle/Patch Colour
We can customise patch properties. Let’s change the patch property of the 2nd rectangle by accessing the object via ax.patches[1] and apply the set_color(“red”) to change the colour to red.
Changing the patch colour
Now you have a basic idea how matplotlib generates the rectangles of a bar plot. This approach is good, but difficult to use when we have many bars to plot. Thus, to overcome this issue, we can use a
more convenient method offered by the axes object (ax) called bar( ).
Let’s proceed with step by step method:
Step 1: Instantiate a figure (fig) and axes (ax) object.
Step 2: Generate a list of x-axis and y-axis values.
Step 3: use the bar( ) method of axes (ax) object and pass the x and y lists.
This way you can generate a basic bar plot.
# Adding bars using defined values
fig, ax = plt.subplots()
x = [0, 1, 2, 3, 4]
y = [1, 3, 5, 2, 7]
# Use ax.bar()
ax.bar(x, y)
Basic bar plot
Again, let’s check the patches of the axes (ax) object using patches attribute. Now you can observe that it contains 5 patches/rectangles.
# Check number of patches
<Axes.ArtistList of 5 patches>
Like last time, here also we can change the colour of the rectangle/patch objects. Let’s change the 4th patch’s colour to red. It uses the same method set_color( ) but here we need to apply this on
the 4th patch using patches[3]. We supplied 3 because Python is a zero-index-based language.
# Changing 4th patch color to "red"
# Caange patch 1 to red
Changing patch colour
We are now familiar with the bar plot and how to generate them from scratch. Now let’s proceed with a new form of plot called “dodged bar plot”.
Dodged barplot [matplotlib style]
A dodged bar plot is used to present the count/proportions/statistics (mean/median) for two or more variables side by side. It helps in making comparison between variables.
For the current plot, we are going to use tips dataset.
Bryant, P. G. and Smith, M. A. (1995), Practical Data Analysis: Case Studies in Business Statistics, Richard D. Irwin Publishing, Homewood, IL.
The Tips dataset contains 244 observations and 7 variables (excluding the index). The variables descriptions are as follows:
bill: Total bill (cost of the meal), including tax, in US dollars
tip: Tip (gratuity) in US dollars
sex: Sex of person paying for the meal (Male, Female)
smoker: Presence of smoker in a party? (No, Yes)
weekday: day of the week (Saturday, Sunday, Thursday and Friday)
time: time of day (Dinner/Lunch)
size: the size of the party
Let’s load the tips dataset using pandas read_csv( ) method and print the first 4 observations using head() method.
tips = pd.read_csv("datasets/tips.csv")
First four observations
Aim of the plot
The aim of the plot is to calculate and impose gender wise smoker proportion using a dodged bar plot. See the below figure which represent the final plot that we are going to plot using various
approach (matplotlib, pandas and seaborn). In the plot, we will present the gender category in the x-axis and their proportion corresponding to smoker category in the y-axis. Further, we are going to
add labels on top of the bar and customise the legend.
Final plot view
Estimate gender/sex wise smoker percentage
To generate this dodged plot, we need to compute the sex wise smoker and non-smoker proportion. To achieve this, we have to go through the following steps:
Step 1: apply the groupby( ) method and group the data based on ‘sex’ and select the ‘smoker’ column from each group.
Step 2: Then apply the value_counts( ) method and supply normalize = True to compute proportion.
Step 3: Next, multiply it with 100, using .mul(100) and round it to 2 decimal places.
Step 4: Apply unstack( ) method so that the sex labels presented in index and smoker status presented in columns and percentage values are presented in cells.
Step 5: Save the output into df variable.
df = (tips
Unstack output
Next, we will take out the Data Frame index using df.index and save in label and generate a range count using the np.range( ) method. We will need these two objects to customise the plots. If we
print these objects, we can observe that the label contains sex labels (Female and Male) and the x variable contains 0 and 1 as a list.
# Generating labels and index
label = df.index
x = np.arange(len(label))
Index([‘Female’, ‘Male’], dtype=’object’, name=’sex’)
[0 1]
Understanding the plotting mechanism
The very first thing we need to do is to use subplots( ) method from matplotlib and generate axes (ax) and figure (fig) objects. The figure size is set to 8 by 6 inches.
Next, set the bar width to 0.2 and use the bar( ) method and apply it to axes object (ax), over which we will impose the bars.
In the bar( ) method, we need to separately supply the columns of the df object. Here in the first one we supplied the x (previously generated object) and the ‘No’ column at x and y position. Then
width value, label (to mark the bar) and bar border colour using edgecolor argument. Then saved the bar object to rect1.
Similarly, for the ‘Yes’ column, we have created another object and save it to rect2.
Now if we see the plotted object we can observe that the blue and orange bar are in a single column which is far from the desired dodged plot. This is because the bars from each group (No/Yes) are
imposed one above another.
To rectify the situation, we need to move the blue bars to the left by 0.1 and the orange bars to the right by 0.1.
#create the base axis
fig, ax = plt.subplots(figsize = (8,6))
#set the width of the bars
width = 0.2
#add first pair of bars
rect1 = ax.bar(x,
width = width,
label = "No",
edgecolor = "black")
#add second pair of bars
rect2 = ax.bar(x,
width = width,
label = "Yes",
edgecolor = "black")
Imposed bars (bad styling)
Now, if we deduct 0.1 from the blue bars’ x-axis position (x – width/2) and add 0.1 to the orange bars (x + width/2) and plot it again, we can see that the bars now looked like dodged bars.
There is one problem, that the x-axis labels are not matching to the final plot, which we actually wanted. We need to correct it.
#create the base axis
fig, ax = plt.subplots(figsize = (8,6))
#set the width of the bars
width = 0.2
# create the first bar by shifting it to left side by width/2
rect1 = ax.bar(x - width/2,
width = width,
label = "No",
edgecolor = "black")
# create the first bar by shifting it to right side by width/2
rect2 = ax.bar(x + width/2,
width = width,
label = "Yes",
edgecolor = "black")
Let’s reset the x-axis tick labels using the set_xticks(x) which will set it to the list values stored in x. But we need to label it as per the sex.
# Reset x-ticks
Setting the label to x values
Next, set the x-tick labels using the set_xticklabels( ) method by supplying the label object (created initially). Now we have got the desired x-tick labels.
# Setting x-axis tick labels
Setting x-tick labels
Concept of Patch objects (groups)
Now let’s move to one of the important topic in bar plots called patch. Every rectangle you see in a barplot know as patch object which contains numerous information like height of the bar, width,
their x and y position, colour etc. Let’s enquire about the patches from our axes (ax) object. If we apply the .patches attribute on the axes (ax), then it will show that it contains 4 patch objects
corresponding to 4 bars.
# Number of patches
<Axes.ArtistList of 4 patches>
To retrieve the information and make use of it, we need to know the order of the patches.
• The blue patches contain information regarding the “No” column and the orange patches contain information regarding “Yes” column.
• The order will be blue Female bar (patch 0), blue Male bar (patch 1), orange Female bar (patch 2), orange Male bar (patch 3).
Let’s retrieve the height from the first patch. To do so, you need to select the first patch object using .patches[0] and apply the get_height( ) method, which reveals the height, i.e., 62.07.
# 0 & 1 are blue pair; 2 & 3 are orange pair (left to right)
Labelling bars
Now we know the concept of patches, we can add labels to each bar by retrieving height information from each patch object using a for loop. To achieve this, follow the following steps:
Step 1: Loop through each patch objects (ax.patches) and save it to a temporary variable ‘p’.
Step 2: use ax.annotate( ) method to annotate the labels. It takes the height value, x and y positions. We can retrieve the height using get_height( ) and convert it to a string object using str( )
to add a percentage (%) symbol. Further, the x and y position can be retrieved using get_x( ) and get_height( ) method. To improve the padding at the top of the bars, we add some padding of 0.03 (in
the x-direction) and 1 (in the y-direction). Next, save it to a variable ‘t’.
Step 3: use the set( ) method to change the annotated text properties.
# Adding bar values
for p in ax.patches:
t = ax.annotate(str(p.get_height()) + "%", xy = (p.get_x() + 0.03, p.get_height()+ 1))
t.set(color = "black", size = 14)
Adding bar labels.
Customising bar plot
The first step of customising it to remove some splines (plot border lines). I usually prefer turning off the top and right spines. To achieve this, use a for loop and use ax.spines[position] and
apply set_visible() to False.
We can also alter the tick parameters [using tick_params( )], and axis labels [using set_xlabel( ) and set_ylabel( )] to make the plot informative and aesthetically beautiful.
# Remove spines
for s in ["top", "right"]:
# Adding axes and tick labels
ax.tick_params(axis = "x", labelsize = 14, labelrotation = 45)
ax.set_ylabel("Percentage", size = 14)
ax.set_xlabel("Sex", size = 14)
Customising properties
Last, but not the least, let’s customise the legend. Here, using the ax.legend( ) method, I have modified the existing labels to “N” and “Y”.
As we know that each plot ranges 0 to 1 in the x and y direction. We can use this information to position our plot legend to the middle of the plot. We can access the legend using ax.legend_ and set
the position using .set_bbox_to_anchor( ) and supply the x and y position using a list.
Now our plot is finalized and ready to use.
# Customize legend
ax.legend(labels = ["N", "Y"],
fontsize = 12,
title = "Smoker",
title_fontsize = 18)
# # Fix legend position
ax.legend_.set_bbox_to_anchor([0.6, 0.5])
Saving the plot
To save a plot, we can use the figure object (fig) and apply the savefig( ) method, where we need to supply the path (images/) and plot object name (dodged_barplot.png) and resolution (dpi=300).
# Save figure object
fig.savefig("images/dodged_barplot.png", dpi = 300)
Dodged bar plot using pandas DataFrame’s plot( ) method
The next step is to generate the same dodged plot, but using the pandas DataFrame based plot( ) method.
First step is to prepare the data, which is the same as we did in the last plot.
tips = pd.read_csv("datasets/tips.csv")
df = (tips
Data preparation
Pandas plot( ) method
Let’s generate the dodged plot using pandas plot( ) method-based approach. To achieve this, we need to follow the following steps.
Step 1: Use subplots( ) method from matplotlib and generate axes (ax) and figure (fig) object. Set the figure size to 8 by 6 inches.
Step 2: apply plot( ) method on the DataFrame (df) object. Specify the kind = “bar” and ax = ax and edgecolor = “black”.
Bam! Your plot framework is almost ready.
fig, ax = plt.subplots(figsize = (10, 4))
df.plot(kind = "bar",
ax = ax,
edgecolor = "black")
Bar plot framework
Next part is labelling and customizing the plot, which is exactly the same as we did in the raw matplotlib based approach. Here, I did not alter the legend labels [“No”, “Yes”].
# Add data labels
for p in ax.patches:
t = ax.annotate(str(p.get_height()) + "%", xy = (p.get_x() + 0.03, p.get_height()+ 1))
t.set(color = "black", size = 14)
# Remove spines
for s in ["top", "right"]:
# Add axes labels and tick parameters
ax.set_xlabel("Sex", size = 14)
ax.set_ylabel("Percentage", size = 14)
ax.tick_params(labelsize = 14, labelrotation = 0)
# Customise legend
ax.legend(labels = ["No", "Yes"],
fontsize = 12,
title = "Smoker",
title_fontsize = 18)
# Fix legend position
ax.legend_.set_bbox_to_anchor([0.5, 0.3])
Pandas way of plotting (df based)
Dodged barplot with pandas DataFrame [seaborn style]
Next, we will generate the same plot, but using seaborn plotting style. In the seaborn we need the input data as pandas DataFrame.
The process of calculating groupwise proportion is similar with small difference. Here, use the reset_index( ) method instead of untack( ) to convert index to columns. Now the output is a pandas
DataFrame type which includes all the columns as stacked Series object.
df = (
.value_counts(normalize = True)
pandas DataFrame
Plotting a dodged plot [seaborn method]
Here, we will be going to use the catplot( ) method from seaborn library. We need to supply the x variable as “sex”, y variable as “percent”, fill color, i.e., hue = “smoker”, DataFrame object (df)
and legend = False.
As the catplot does not take an axes (ax) object; thus we need to somehow retrieve the axes (ax) and figure (fig) objects.
We can retrieve the axes (ax) object using the plt.gca( ) and figure (fig) object using the plt.gcf( ). The gca refers to `get current axes` and gcf refers to the `get current figure`.
sns.catplot(x = "sex",
y = 'percent',
hue = "smoker",
kind = 'bar',
data = df,
legend = False)
ax = plt.gca()
fig = plt.gcf()
Dodged plot [seaborn style]
The next step is to customising the plot, i.e., adding data labels, modifying ticks and axis labels.
Lastly, we will fix the size of the plot using the fig.set_size_inches( ).
sns.catplot(x = "sex",
y = 'percent',
hue = "smoker",
data = df,
legend = False)
# Customization
# Retrieve axis and fig objects from the current plot environment
ax = plt.gca()
fig = plt.gcf()
# Add bar labels
for p in ax.patches:
p.set_edgecolor("black") # Add black border across all bars
t = ax.annotate(str(p.get_height().round(2)) + "%", xy = (p.get_x() + 0.1, p.get_height() + 1))
t.set(size = 14)
# Adding axes labels and tick parameters
ax.set_xlabel("Sex", size = 16)
ax.set_ylabel("Percentage", size = 16)
ax.tick_params(labelsize = 14)
# Legend customisation
ax.legend(fontsize = 12,
title = "Smoker",
title_fontsize = 12)
# Resetting figure size
fig.set_size_inches(8, 4)
Final dodged plot [seaborn way]
Once you learn base matplotlib, you can customise the plots in various ways. I hope you now know various ways to generate a dodged plot. Apply the learned concepts to your datasets.
[1] J. D. Hunter, “Matplotlib: A 2D Graphics Environment”, Computing in Science & Engineering, vol. 9, no. 3, pp. 90–95, 2007.
Click here for the data and code
I hope you learned something new!
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Lossless expander balanced-product code
Lossless expander balanced-product code[1,2]
QLDPC code constructed by taking the balanced product of lossless expander graphs. Using one part of a quantum-code chain complex constructed with one-sided loss expanders
yields a
. Using two-sided expanders, which are only conjectured to exist, yields an asymptotically good QLDPC code family
Asymptotically good QLDPC codes
, assuming the existence of two-sided lossless expanders.
• Locally testable code (LTC) — Using one part of a quantum-code chain complex constructed with one-sided loss expanders yields a \(c^3\)-LTC [1].
• Good QLDPC code — Taking a balanced product of two-sided expanders, which are only conjectured to exist, yields an asymptotically good QLDPC code family [2].
T.-C. Lin and M.-H. Hsieh, “\(c^3\)-Locally Testable Codes from Lossless Expanders”, (2022) arXiv:2201.11369
T.-C. Lin and M.-H. Hsieh, “Good quantum LDPC codes with linear time decoder from lossless expanders”, (2022) arXiv:2203.03581
M. Capalbo et al., “Randomness conductors and constant-degree lossless expanders”, Proceedings of the thiry-fourth annual ACM symposium on Theory of computing (2002) DOI
Page edit log
Cite as:
“Lossless expander balanced-product code”, The Error Correction Zoo (V. V. Albert & P. Faist, eds.), 2023. https://errorcorrectionzoo.org/c/lossless_expander
Github: https://github.com/errorcorrectionzoo/eczoo_data/edit/main/codes/quantum/qubits/stabilizer/qldpc/homological/balanced_product/lossless_expander.yml.
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Evidence, Probabilities and Naive Bayes
Evidence, Probabilities and Naive Bayes
Welcome! This workshop is from Winder.ai. Sign up to receive more free workshops, training and videos.
Bayes rule is one of the most useful parts of statistics. It allows us to estimate probabilities that would otherwise be impossible.
In this worksheet we look at bayes at a basic level, then try a naive classifier.
Bayes Rule
For more intuition about Bayes Rule, make sure you check out the training.
Following on from the previous marketing examples, consider the following situation. You have already developed a model that decides to either show and advert, or not show an advert, to a set of
Your boss has come back with some results and says:
“We’re going to ask you to develop a new model, but first we need a baseline to compare against. Can you tell me what is the probability that a customer will buy our product, given that they have
seen your advert?”.
We notice the word “given” in the sentence and we release that we can use bayes rule:
$p(B \vert A) = \frac{p(B) \times p(A \vert B)}{p(A)} = \frac{p(B) \times p(A \vert B)}{p(B) \times p(A \vert B) + p(notB) \times p(A \vert notB)}$
Where $A$ is being shown the advert and $B$ is buy our product.
Next, she provides the following statistics from the previous experiment:
• 10 people bought product after being shown the advert. (TP)
• 1000 people were not shown the advert and didn’t buy. (TN)
• 50 people were shown the advert, but didn’t buy. (FP)
• And 5 people bought our product without being shown any advert. (FN)
• 10 true positives. 10 people bought product.
• 1000 true negatives. Not shown advert.
• 50 false positives. They didn’t buy, but were shown the advert.
• 5 false negatives, people bought even though they weren’t shown the advert.
TP = 10
TN = 1000
FP = 50
FN = 5
total = TP+TN+FP+FN
p_buy = (TP+FN)/total
p_ad_buy = TP/(TP+FN)
p_notbuy = (TN+FP)/total
p_ad_notbuy = FP/(TN+FP)
p_buy_ad = p_buy * p_ad_buy / (p_buy * p_ad_buy + p_notbuy * p_ad_notbuy)
print("p(buy|ad) = %0.1f%%" % (p_buy_ad*100))
p(buy|ad) = 16.7%
Naive Bayes Classifier
Now let’s try a naive bayes classifier. We’re going to use the sklearn implementation here, but remember this is just an algorithm that estimates the probabilities of each feature, given a class.
When provided with a new observation we look at those probabilities and predict the probability that this instance belongs to that class.
Note that this implementation assumes that the features are normally distributed and the fit method performs the fitting of the distribution parameters (mean and standard deviation).
from matplotlib import pyplot as plt
from sklearn import metrics
from sklearn import datasets
from sklearn.naive_bayes import GaussianNB
iris = datasets.load_iris()
(150, 4)
gnb = GaussianNB()
mdl = gnb.fit(iris.data, iris.target) # Tut, tut. We should really be splitting the training/test set.
y_pred = mdl.predict(iris.data)
cm = metrics.confusion_matrix(iris.target, y_pred)
[[50 0 0]
[ 0 47 3]
[ 0 3 47]]
y_proba = gnb.fit(iris.data, iris.target).predict_proba(iris.data)
print("These are the misclassified instances:\n", y_proba[iris.target != y_pred])
print("They were classified as:\n", y_pred[iris.target != y_pred])
print("But should have been:\n", iris.target[iris.target != y_pred])
These are the misclassified instances:
[[ 1.52821825e-122 4.56151317e-001 5.43848683e-001]
[ 7.43572418e-129 1.54494085e-001 8.45505915e-001]
[ 2.12531216e-137 7.52691316e-002 9.24730868e-001]
[ 4.59552511e-108 9.73514345e-001 2.64856553e-002]
[ 5.69697725e-125 9.58135362e-001 4.18646381e-002]
[ 2.19798649e-130 7.12645144e-001 2.87354856e-001]]
They were classified as:
[2 2 2 1 1 1]
But should have been:
[1 1 1 2 2 2]
# Ideally, generate a curve for each target. Do it in a loop.
fpr, tpr, _ = metrics.roc_curve(iris.target, y_proba[:,1], pos_label=1)
lw = 2
plt.plot(fpr, tpr, color='darkorange',
lw=lw, label='ROC curve')
plt.plot([0, 1], [0, 1], color='navy', lw=lw, linestyle='--', label='Random')
plt.xlim([0.0, 1.0])
plt.ylim([0.0, 1.05])
plt.xlabel('False Positive Rate')
plt.ylabel('True Positive Rate')
plt.title('Receiver operating characteristic of iris data')
plt.legend(loc="lower right")
actual cancer actual no cancer
diagnosed cancer 8 90
not diagnosed cancer 2 900
• Above is a (simulated) confusion matrix for breast cancer diagnosis.
□ What is the probability that a person has cancer given that they have a diagnosis? (Find p(cancer|diag))
• Load the digits dataset from the previous workshop, classify with a naive bayes classifier and plot the ROC curve.
• Compare that to another classifier
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Understanding Exponential Moving Average (EMA) for Trading | HackerNoon
Broadly technical analysis is all about seeing the trend and predicting the future. Future is always uncertain and there is probability involved. I recommend using many indicators for a hawk eye view
as it will provide the required congruence and high chance of success.
One such important indicator is Moving Average or Simple Moving Average. It is a simple yet very powerful indicator. It creates a summary in form of a line from the complex price data set of a
historic range. It is calculated by taking a simple average of closing price points of the candlesticks.
It is a lagging indicator and to make it more relevant technicians suggested giving more importance to most recent price candlesticks and lesser weight to past price candlesticks. This led to
emergence of Exponential Moving Average or EMA. It's the sum average of price points multiplied by a weighing factor. The weighing factor can be considered having a constant value of 1 for all price
points for calculating SMA but for EMA the values are such that the recent price points are given more weightage. The decrease between one price and its preceding price is not consistent and it
decreases exponentially.
There is another acceptable form of MA, called the Weighted Moving Average (WMA). It's a modified form of EMA, with weighing decreases with each previous price. The weights are customized, the value
of which is high for the most recent prices and decreases for the historical ones.
Moving average calculated for shorter period behaves more rapidly to price changes and is called a fast MA, conversely with larger periods is called slow MA.
Familiarizing the indicator on chart:
As Fibonacci range depicts natural events more closely I recommend to use EMA with following time periods:
After you activate these on the charts do remember to colour them with different colours for easy identification. I always use darker shades for higher n valued EMAs and lighter shade for lower n
valued EMAs.
Identifying Crossovers
Moving averages highlight trends and are used to spot trend reversals. The trend line smoothens the noise of price action and shows the real direction of price movement. We use multiple EMA lines to
predict the price movement, believe me it's really easy to spot the trend. When the lines cross each other cutting each other a cross over happens leading to price run in a specific direction.
Death Cross (or Downtrend):
When a fast EMA (eg: EMA 8) crosses a slow EMA (eg: EMA 200) sliding from top to bottom a death cross happens signaling the onset of a downtrend. The trader then must go for a short position.
Golden Cross (or Uptrend):
On a contrary when a fast EMA (eg: EMA 8) runs from bottom to top of a slow EMA (eg: EMA 200) signifies the beginning of price rise, the trader must go for a long position.
Always colour code your EMAs for quicker spotting of crossovers. Use sharp and different colours so all EMAs are distinctly spotable.
Mathematical Formula:
SMA = (P1+P2+..+Pn)/n
EMAt = Pt*EMult + EMAy*(1-EMult)
EMult = 2/(n+1)
For the period, first EMAy = SMA
WMA = [Pt*n + Py*(n - 1) + ...Pn*1]/ [n*(n+1)/2]
WMult = 2/(n+1)
SMA = Simple Moving Average
EMA = Exponential Moving Average
EMAt= EMA for current candle (today)
EMAy=EMA for last candle (yesterday)
EMult = EMA Smoothing Multiplier
WMA= Weighted Moving Average
WMult = WMA Smoothing Multiplier
P = Closing price of the candle
Pt= P1 = Closing price of current candle (today)
Py= Closing price of last candle (yesterday)
Pn = Closing price of first candle in the selected period
n= no of past candles
Word of Caution:
Moving Average of any kind (SMA, EMA or WMA) is a lagging indicator relying on past data points and hence they suffer a time lag before they depict new trends. An immediate price spike would be
captured only after it occurs.
Moving average with a shorter period suffers from less lag than one with a larger period. Moving Average with a short time frame will react much quicker to price changes than a Moving Average with a
long look back period.
Technicians prefer to use EMA over WMA as the weighing factor exponentially gives more weightage to the recent price data, yielding much better predictions.
Trading Points:
Use EMA8, EMA13, EMA21, EMA55, EMA200. Colour code them for easy identification.
If the price is above a moving average, the trend is up. If the price is below a moving average, the trend is down.
SMA is used mostly for finding supports & resistances
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Number Col Boxes
Number Col Box<<Add Element(item)
Adds the item to the Number Col Box. item can be a single number, a list of numbers, or a matrix.
Number Col Box<<Bootstrap(nsample, Random Seed(number), Fractional Weights(Boolean), Split Selected Column(Boolean), Discard Stacked Table if Split Works(Boolean)
Bootstraps the analysis, repeating it many times with different resampling weights and collecting tables as selected.
Sets the number of times that you want to resample the data and compute the statistics. A larger number results in more precise estimates of the statistics’ properties. By default, the number of
bootstrap samples is set to 2,500.
Random Seed(number)
Sets a random seed that you can re-enter in subsequent runs of the bootstrap analysis to duplicate your current results. By default, no seed is set.
Fractional Weights(Boolean)
Performs a Bayesian bootstrap analysis. In each bootstrap iteration, each observation is assigned a weight that is calculated as described in Calculation of Fractional Weights in Basic Analysis. The
weighted observations are used in computing the statistics of interest. By default, the fractional weights option is not selected and a simple bootstrap analysis is conducted.
Split Selected Column(Boolean)
Places bootstrap results for each statistic in the column that you selected for bootstrapping into a separate column in the Bootstrap Results table. Each row of the Bootstrap Results table (other
than the first) corresponds to a single bootstrap sample.
If you exclude this option, a Stacked Bootstrap Results table appears. For each bootstrap iteration, this table contains results for the entire report table that contains the column that you selected
for bootstrapping. Results for each row of the report table appear as rows in the Stacked Bootstrap Results table. Each column in the report table defines a column in the Stacked Bootstrap Results
Discard Stacked Table if Split Works(Boolean)
(Applicable only if the Split Selected Column option is included.) Determines the number of results tables produced by Bootstrap. If the Discard Stacked Table if Split Works option is not selected,
then two Bootstrap tables are shown. The Stacked Bootstrap Results table, which contains bootstrap results for each row of the table containing the column that you selected for bootstrapping, gives
bootstrap results for every statistic in the report, where each column is defined by a statistic. The unstacked Bootstrap Results table, which is obtained by splitting the stacked table, provides
results only for the column that is selected in the original report.
See Also
Bootstrapping Window Options in Basic Analysis
Number Col Box<<Get
Number Col Box<<Get(i)
Gets the values in a list, or the ith value.
Number Col Box<<Get As Matrix
Gets the values in a matrix, specifically a column vector.
Number Col Box<<Get Format
Returns the current format.
Number Col Box<<Get Heading
Returns the column heading text.
Number Col Box<<Remove Element(row number)
Removes an element from the column at the specified position.
Number Col Box<<Set Format(<width>|<width, decimal places>, <"Use Thousands Separator">)
Number Col Box<<Set Format("Best", <width>, <"Use Thousands Separator">)
Number Col Box<<Set Format(("Fixed Dec"|"Percent"), <width>|<width, decimal places>, <"Use Thousands Separator">)
Number Col Box<<Set Format("Pvalue", <width>)
Number Col Box<<Set Format(("Scientific"|"Engineering"|"Engineering SI"), <width>|<width, decimal places>)
Number Col Box<<Set Format("Precision", <width>|<width, decimal places>, <"Use Thousands Separator">, <"Keep Trailing Zeroes">, <"Keep All Whole Digits">)
Number Col Box<<Set Format("Currency", <currency code>, <width>|<width, decimal places>, <"Use Thousands Separator">)
Number Col Box<<Set Format(datetime, <width>, <input format>)
Number Col Box<<Set Format(("Latitude DDD"|"Latitude DDM"|"Latitude DMS"|"Longitude DDD"|"Longitude DDM"|"Longitude DDM"), <width>|<width, decimal places>, ("PUN"|"DIR"|"PUNDIR"))
Number Col Box<<Set Format("Custom", Formula(...), <width>, <input format>)
Sets the column format.
Numeric Formats in Using JMP describes the arguments. Note that Matrix Box(), Number Col Box(), Number Col Edit Box(), Number Edit Box() have the same Set Format syntax.
<<Set Format( 10, 2, "Use thousands separator");
<<Set Format( "Currency", "EUR", 20, );
<<Set Format( "m/d/y", 10 );
<<Set Format( "Precision", 10, 2, "Keep trailing zeroes", "Keep all whole digits" );
<<Set Format( "Latitude DDD", "PUNDIR"); // "PUN" for punctuation, "DIR" for direction, PUNDIR for both
<<Set Format( "Custom", Formula( Abs( value ) ), 15 );
• For a list of currency codes, see Currency in the Scripting Guide. The currency code is based on the locale if the code is omitted.
• If you don’t specify the format, set the decimal places to greater than 100 for datetime values and to 97 for p-values.
• You must always precede the number of decimal places with the width.
• Options can be defined in a list or a variable, or they can be in a Function() that is evaluated.
ncbFunc = Function({}, {"Fixed", 12, 5});
number col box<<Set Heading(quoted string)
Changes the column heading text.
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What is the slope of a line perpendicular to the x-axis?What is the slope of a line perpendicular to the x-axis?
Hello world!
May 27, 2020
January 5, 2021
The slope of a line perpendicular to the x-axis is undefined.
Perpendicular lines have opposite reciprocal slopes.
For example ##1/2## and ##-2/1=-2## are opposite reciprocals and lines with these slopes will be perpendicular to one another.
If the x-axis is treated as a horizontal line we can pick two points on the line such as (00) and (10).
The slope of a line is ##m={y_2-y_1}/{x_2-x_1}##. So the two points we have selected:
##(00)=(x_1y_1)## and ##(10)=(x_2y_2)##.
Using the slope formula ##m={0-0}/{1-0}=0/{1}=0##.
The opposite reciprocal of a slope of zero would be undefined because we cannot divide by zero.
Alternatively you can use the slope formula and two points on the vertical line x=1 for example. Randomly picking the points (10) and (11).
The slope would be ##m={1-0}/{1-1}## and this is undefined.
Very good explanation of slopes with good graphs and the answer to the question with additional info here http://www.themathpage.com/alg/slope-of-a-line.htm
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A direction field (or slope field / vector field) is a picture of the general solution to a first order differential equation with the form
• Edit the gradient function in the input box at the top. The function you input will be shown in blue underneath as
• The Density slider controls the number of vector lines.
• The Length slider controls the length of the vector lines.
• Adjust and to define the limits of the slope field.
• Check the Solution boxes to draw curves representing numerical solutions to the differential equation.
• Click and drag the points A, B, C and D to see how the solution changes across the field.
• Change the Step size to improve or reduce the accuracy of solutions (0.1 is usually fine but 0.01 is better). If anything messes up....hit the reset button to restore things to default.
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How many grams of NaOH are dissolved in order to create 750 ml of a 0.1 M solution?
Solving this problem requires us to use the formula for molarity, M = moles solute/liters of solution. We are given the molarity (0.1 M), the volume (750 ml), and need to calculate the number of
moles and mass of NaOH required. Let’s break this down step-by-step:
Step 1: Convert Volume to Liters
The volume is provided in milliliters (ml), but molarity requires liters (L). So we first need to convert 750 ml to liters:
750 ml x (1 L/1000 ml) = 0.75 L
Step 2: Use Molarity Formula to Calculate Moles of NaOH
Now we can plug the values into the molarity formula:
M = moles solute/liters of solution
0.1 M = moles NaOH/0.75 L
To solve for moles NaOH, we rearrange the formula:
moles NaOH = M x liters of solution
moles NaOH = 0.1 M x 0.75 L
moles NaOH = 0.075 moles
Step 3: Use Moles to Calculate Grams of NaOH
Now that we know the number of moles of NaOH required, we can use the molar mass of NaOH to convert to grams.
The molar mass of NaOH is 40 g/mol.
Using the formula: grams = moles x molar mass
grams NaOH = 0.075 moles NaOH x 40 g/mol
grams NaOH = 3 g NaOH
To summarize, in order to make 750 ml of a 0.1 M NaOH solution, we need 3 grams of NaOH.
The key steps were:
1. Convert volume to liters
2. Use molarity formula to calculate moles of NaOH
3. Use moles and molar mass to calculate grams of NaOH
Understanding these concepts of molarity and mole calculations is essential for chemistry students and professionals. Proper calculation of solution concentrations and reagents is critical for
achieving accurate and repeatable results in the chemistry lab. With some practice, these calculations become second nature.
Sample Molarity Calculations
Let’s practice these calculations with a few more examples:
Example 1
How many grams of nitric acid (HNO3) are required to make 250 mL of a 0.5 M solution?
Step 1: Convert volume to liters
250 mL x (1 L/1000 mL) = 0.25 L
Step 2: Use molarity formula
M = moles/L
0.5 M = moles HNO3/0.25 L
Moles HNO3 = 0.125 moles
Step 3: Use moles and molar mass (63 g/mol) to calculate grams
Grams HNO3 = 0.125 moles x 63 g/mol = 7.9 g HNO3
Example 2
What volume of 2 M HCl solution can be made with 25 g HCl?
Step 1: Convert grams to moles
25 g HCl x (1 mol/36.5 g) = 0.685 moles HCl
Step 2: Rearrange molarity formula to solve for volume
M = moles/L
L = moles/M
L = 0.685 moles HCl/2 M = 0.34 L
So with 25 g HCl, we can make 0.34 L of 2 M HCl solution.
Example 3
If I have 325 mL of 0.25 M NaOH solution, how many moles of NaOH does it contain?
Step 1: Convert volume to liters
325 mL x (1 L/1000 mL) = 0.325 L
Step 2: Use molarity formula
M = moles/L
0.25 M = moles NaOH/0.325 L
Moles NaOH = 0.25 M x 0.325 L = 0.08125 moles NaOH
So there are 0.08125 moles of NaOH in the 325 mL of 0.25 M solution.
Practice Problems
Let’s apply these concepts to some practice problems. Try calculating the answers on your own before looking at the solutions.
Problem 1
How many grams of potassium chloride (KCl) are required to make 200 mL of a 0.1 M solution?
Step 1: 200 mL x (1 L/1000 mL) = 0.2 L
Step 2: 0.1 M = moles KCl/0.2 L
moles KCl = 0.02
Step 3: Moles KCl x molar mass (74.5 g/mol) = 1.49 g KCl
Problem 2
What volume of 0.5 M NaOH can be made with 10 g of NaOH?
Step 1: 10 g NaOH x (1 mol/40 g) = 0.25 moles NaOH
Step 2: M = moles/L
0.5 M = 0.25 moles NaOH/L
L = 0.25 moles/0.5 M = 0.5 L
So 10 g NaOH can produce 0.5 L of 0.5 M NaOH.
Problem 3
How many moles of solute are present in 250 mL of 0.1 M KCl solution?
Step 1: 250 mL x (1 L/1000 mL) = 0.25 L
Step 2: 0.1 M = moles KCl/0.25 L
moles KCl = 0.025 moles
So there are 0.025 moles of KCl in the solution.
Tips for Solving Molarity Problems
When doing molarity calculations, keep these tips in mind:
• Convert all volumes to liters before using the molarity formula.
• Be sure you pay attention to the units and use the right molar mass when converting between mass and moles.
• Double check that your units cancel properly in each step.
• Use dimensional analysis to lay out the problem step-by-step and cancel units.
• Pay close attention to what quantity the problem is asking you to calculate. This will help you set up the correct molarity equation.
• Check your work! Plug your calculated values back into the original equation to ensure everything checks out.
Why is Molarity Important?
Now that we’ve gone over the basics of molarity calculations, let’s talk about why molarity is so important in chemistry.
Molarity is a conveniently scaled and standardized unit for concentration. By calculating and preparing solutions using molarity, chemists are able to reliably replicate experiments and processes.
The molarity provides valuable information about the ratio of solute to solvent in the solution.
Some key reasons why molarity is important include:
• Describes solution composition: Molarity specifies the concentration of solute dissolved in a solution. This allows chemists to characterize the composition of a solution accurately.
• Allows replication of experiments: Solutions of specific molarities can be reliably reproduced, allowing consistent experimental conditions.
• Determines chemical properties: The molarity influences colligative properties such as boiling point elevation and freezing point depression. It impacts reaction kinetics and stoichiometry as
• Indicates reacting amounts: Using molarity, chemists can determine the amounts of reactants needed for reactions and stoichiometric calculations.
• Essential for titrations: Molarity is required to determine unknown concentrations in acid-base titrations and redox titrations.
In short, molarity is a cornerstone of solution chemistry. Mastering molarity calculations is essential for any chemistry student or professional.
Molarity in the Laboratory
To drive home the importance of molarity, let’s go over some examples of using molar solutions in a chemistry lab:
Preparing Reagents
Many laboratory procedures require reagents of specific molarities. For example, a common cell culture medium calls for 0.4 mM L-glutamine. By calculating the molarity correctly, a chemist can
accurately prepare this reagent to support cell growth.
Making Standards
Analytical standards of known molar concentrations are essential for many instruments. For example, calibrating a UV-Vis spectrophotometer requires standard solutions of a compound at different
molarities to generate a concentration curve.
Titration Analysis
Titration requires standardized molarity of titrant solutions. For instance, when titrating an unknown HCl solution with 0.1 M NaOH, the chemist must first accurately prepare the 0.1 M NaOH titrant.
Reaction Stoichiometry
To optimize chemical reactions, chemists use molarity to calculate the exact reagent ratios and amounts. For example, synthesizing a 100 g batch of biodiesel requires knowing the molar ratio of
methanol to oil and calculating the volume of each at their given molarities.
Molarity allows determining the volume of stock needed to dilute to a target concentration. For example, 5 mL of 5 M HCl can be diluted to 100 mL to produce a 0.5 M HCl solution.
In all areas of the lab, from analytical chemistry to biochemistry, molarity is a fundamental concept underlying proper technique and calculation. Mastery of molarity is essential for any lab work
involving solutions.
Molarity Calculations Quiz
Let’s test your understanding of molarity calculations with this brief quiz:
Quiz Question 1
How many moles of solute are present in 300 mL of 0.5 M NaCl solution?
A) 0.05 moles
B) 0.10 moles
C) 0.15 moles
D) 1.5 moles
Answer: B
Explanation: 0.5 M NaCl x 0.3 L = 0.10 moles NaCl
Quiz Question 2
What volume of 0.1 M HCl solution can be prepared with 25 g of HCl?
A) 0.25 L
B) 0.5 L
C) 1 L
D) 2.5 L
Answer: C
Explanation: 25 g HCl is 0.685 moles. 0.685 moles/0.1 M = 6.85 L
Quiz Question 3
How many grams of CaCl2 are needed to prepare 500 mL of 0.2 M solution?
A) 14.7 g
B) 29.4 g
C) 44.1 g
D) 73.5 g
Answer: B
Explanation: 0.5 L x 0.2 mol/L = 0.10 moles = 29.4 g CaCl2
How did you do? Being able to calculate molarity accurately takes practice, so review any concepts that you may be unclear on. Molarity calculations are a fundamental skill in chemistry.
Real World Applications of Molarity
We’ve mainly discussed molarity in an academic chemistry context. However, molarity has numerous real-world uses across many fields including:
Doctors and pharmacists use molarity to calculate drug dosages and intravenous drip rates. Chemists develop analytical methods to verify drug purity and concentration.
Environmental Science
Environmental scientists measure pollutant concentrations in ppm and ppb, which are units related to molarity. They analyze water samples and calculate pollutant molarities.
Chemical engineers rely on molarity for process optimization and safety. Proper reagent molarities in large-scale reactions prevent accidents like runaway reactions.
Soil scientists use molarity to measure nutrient levels and acidity. Fertilizers and soil amendments are applied based on calculating molar concentrations.
Food Science
Food scientists standardize ingredients like salts, sweeteners, and acids using molar concentrations. This ensures consistent product quality and safety.
In many STEM careers, a thorough grasp of molarity and calculations is imperative. The principles discussed in this article broadly apply when working with solutions.
Molarity Problem Solving Tips
Here are some final tips for mastering molarity calculations:
• Memorize key molarity equations and unit conversions.
• Understand how to interconvert mass, moles, volume, and molarity.
• Practice doing word problems and setting up the equations.
• Check answers by plugging back into the original molarity relationship.
• Pay close attention to units and ensure they cancel properly.
• Use dimensional analysis and show work step-by-step.
• Ask for help if a concept or problem is unclear.
• Don’t get discouraged! Molarity takes practice to master.
With some time and effort, you can become adept at molarity calculations. It’s a foundational chemistry skill with many applications. Understanding molar concentrations will serve you well in your
chemistry education and beyond.
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ACO Seminar
The ACO Seminar (2012-2013)
Sep 20
, 3:30pm, Wean 8220
Ameya Velingker
, CMU
Meshing log n Dimensions in Polynomial Time
We study the problem of generating a conforming Delaunay mesh with quality for point sets in high-dimension. Previously, most meshing algorithms were designed for 2 or 3 dimensions. The talk will
motivate the discussion of meshing in higher-dimensional settings and give an overview of existing algorithms. Finally, we propose a new algorithm which produces the 1-dimensional skeleton of a
Delaunay mesh with guaranteed size and quality. Our comparison-based algorithm runs in time O(2^O(d) * (n log n + m)), where n is the input size, m is the output point set size, and d is the ambient
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Why Is 'Mean' So Meaningful in Mathematics? » Learning Captain
Why Is ‘Mean’ So Meaningful in Mathematics?
Mathematics : Have you ever wondered why your math teacher cared so much about calculating the mean or average? On the surface, it seems like a pretty straightforward mathematical concept that just
gives you a general sense of a set of numbers. But the mean is actually one of the most useful and powerful metrics in statistics and mathematics.
It gives you a sense of central tendency – where the bulk of values in a data set cluster. And when you have large data sets with lots of variability, the mean helps make sense of all that
information by giving you a single number to represent the whole set. The mean may seem simple, but it provides an elegant mathematical solution to gaining valuable insights from complex data. Read
on to understand why calculating the mean can reveal so much about the world around us.
Defining the Mean in Math
Why Is 'Mean' So Meaningful in Mathematics? 4
The mean, also known as the average, is one of the most useful mathematical concepts you’ll ever learn. It’s a simple measure that provides a quick snapshot of a data set by calculating the sum of
all values and dividing by the total number of numbers.
Calculating the Mean
To find the mean of a data set, follow these steps:
1. Add up all the numbers in the data set. For example, the numbers 3, 7, 9, and 4 add up to 23.
2. Count how many numbers are in the data set. In this case, there are 4 numbers.
3. Divide the sum by the count. 23 divided by 4 is 5.75.
4. The result is the mean or average. So the mean of the data set {3, 7, 9, 4} is 5.75.
The mean is useful for gaining a quick sense of central tendency in a data set and spotting outliers or skewness. However, it can be swayed by very high or low values since it incorporates all values
equally. The median and mode are also measures of central tendency and useful for analyzing data.
Real-World Examples
Some examples of using the mean in real life include: calculating your average monthly expenses to create a budget, determining the average test score to assess student performance, or finding the
average daily temperature for the month to summarize the local climate.
The mean is a simple but powerful concept that allows you to gain valuable insights from all kinds of data in the world around you. Understanding how to calculate and apply the mean is an important
skill for students, professionals, and everyday life.
Understanding the Different Types of Means
Why Is 'Mean' So Meaningful in Mathematics? 5
When talking about averages in math, it’s important to understand the different types and how they’re calculated. The three most common means are:
The Arithmetic Mean
Also known as the average, it’s calculated by adding up all the numbers in a data set and dividing by the number of values. For example, the arithmetic mean of 2, 3, 4, and 6 is (2 + 3 + 4 + 6) / 4 =
15 / 4 = 3.75.
The Median
The median is the middle number in a data set. To find the median, arrange the numbers in order and pick the number in the center. If there are an even number of values, calculate the mean of the two
central numbers. For example, the median of 2, 3, 4, 6 is 4. The median of 1, 3, 4, 6 is (3 + 4) / 2 = 3.5.
The Mode
The mode is the value that appears most frequently in a data set. For example, if the data set is 1, 3, 6, 4, 2, 3, 3, the mode is 3 because it appears most often. A data set can have more than one
mode, or no mode at all.
When analyzing data, it’s good practice to consider the mean, median, and mode together to get the full picture. While the mean gives you a sense of the center of a data set, the median shows the
center value, and the mode reveals the most frequent value. Using all three means will provide the most insight into what the data is telling you.
Calculating the Arithmetic Mean
Calculating an arithmetic mean, or average, is actually quite straightforward. All you need are the numbers in your data set and a calculator. Here’s how it works:
Add up all the numbers
First, gather all the numbers in your data set. These could be test scores, product ratings, times, distances—whatever it is you want to find the mean of. Add up all these numbers to get the total.
Count how many numbers there are
Second, count how many numbers are in your data set. This is known as the sample size. For example, if you have the test scores of 15 students, your sample size would be 15.
Divide the total by the sample size
Finally, divide the total by the sample size. The result is your arithmetic mean, or average.
For example, say you want to find the mean test score for a group of students. There are 15 students and their scores are:
82, 73, 95, 68, 70, 83, 75, 91, 66, 88, 77, 85, 90, 93, 72
To find the mean:
1. Add up all the scores: 82 + 73 + 95 + 68 + 70 + 83 + 75 + 91 + 66 + 88 + 77 + 85 + 90 + 93 + 72 = 1,519
2. The sample size is 15 students
3. Divide the total (1,519) by the sample size (15)
1,519 / 15 = 101.27
So the mean, or average, score for this group of 15 students is 101.27.
The arithmetic mean is a useful measure of central tendency that can give you a sense of the midpoint in a data set with a single number. While it won’t tell you everything about your data,
calculating the mean is an important first step in gaining valuable insights.
Using Means to Analyze Data Sets
When analyzing a data set, the mean, or average, is one of the most useful statistics you can calculate. The mean gives you a sense of the “middle” value of your numbers and helps determine if a data
set is skewed to one side or relatively symmetrical.
To find the mean of a data set, simply add up all the numbers and divide by the total number of values. For example, the mean of 2, 4, 6, 8, and 10 is (2 + 4 + 6 + 8 + 10) ÷ 5 = 6. Knowing the mean
allows you to compare different data sets and see how they differ. Compare the means of the two data sets: Set A {1, 3, 8, 9} and Set B {2, 3, 4, 10}. Set A has a mean of 5.25, while Set B has a mean
of 4.75. So on average, the numbers in Set A are a bit higher.
The mean is most useful when your data set has no extreme outliers that could skew the average. For example, the mean of {1, 3, 8, 9, 100} would be 20.2, but this mean does not accurately reflect the
center of this data since the 100 is so much higher than the other values. In this case, the median (the middle number when the values are arranged in order) would provide a better measure of central
When analyzing data, calculate the mean, median, and range (the highest value minus the lowest value) to get a sense of where most of the numbers lie, the midpoint, and the spread. Compare these
statistics over time or between different groups to spot patterns and trends. Using multiple measures of center and spread will give you a more robust analysis of what your data is really telling
In the end, the mean is a simple but significant calculation. While a single statistic doesn’t tell the whole story, the mean provides an essential snapshot that allows you to make comparisons and
gain valuable insights into your data. Understanding what it represents and how to interpret it will make you a stronger statistician and data analyst.
Read more: The Ultimate Guide to College Scholarships
The Importance of Means in Statistics and Probability
Why Is 'Mean' So Meaningful in Mathematics? 6
The mean, or average, is one of the most useful measures of central tendency in statistics. It gives you a sense of the midpoint in a data set and is calculated by adding up all the values and
dividing by the number of values.
Knowing the mean allows you to compare different data sets and spot trends. For example, if test scores go up over several years, the mean score will increase. The mean can also show you if there are
any outliers, or values that are much higher or lower than the rest. These outliers can skew the mean, so it’s important to consider the median and mode as well, which are less affected by extreme
In probability, the mean is useful for determining the expected value. The expected value refers to the value you would expect to get on average over many trials. For example, if you flip a fair
coin, the expected value for the number of heads is 1/2, or 50%, because over many flips you would expect half to be heads. The expected value is calculated by multiplying each possible outcome by
its probability and then adding them up.
The mean comes in handy when you want a quick summary of your data or need to convey information in a simplified manner. However, it does have some downsides. The mean can be misleading if there are
outliers in your data set or if the data is skewed. It also ignores the spread of values and tells you nothing about the shape of the distribution.
To get a more complete picture, consider other measures of central tendency like the median and mode. And be sure to look at the range, standard deviation, and data visualization tools like
histograms, box plots, and scatter plots. When used together, these tools will allow you to explore your data fully and draw meaningful conclusions.
In the end, the mean is a useful measure, but it shouldn’t be the only statistic you rely on. Use it as a starting point, but look at your data from multiple angles to get the whole story.
So there you have it, the significance and power of the mean in math. While it may seem like just another boring number to calculate, the mean gives you a quick sense of the center of your data and
how spread out it is. It allows you to compare different data sets and spot trends over time. The mean may be simple, but simple can be extremely useful. Next time you see an average in the news, in
your homework, or at your job, don’t dismiss it. Recognize the power in that single number and how much information it can provide if you take the time to look. The mean matters.
2 thoughts on “Why Is ‘Mean’ So Meaningful in Mathematics?”
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Calculadoras online
Distance Calculator
Calculate the distance between two geographic coordinates using the Haversine formula.
Distance through the Earth
This calculator calculates the distance from one point on the Earth to another point, going through the Earth, instead of going across the surface.
Earth Distance Calculator
The Earth Distance Calculator is a handy tool for measuring the distance between two points on the Earth's surface, as well as the distance between those two points if they were connected by a
straight tunnel through the Earth's core.
Conversion between Gauss planar rectangular coordinates and geographic coordinates and vice versa
The page contains online calculators for converting from geographic coordinates to Gauss planar rectangular coordinates and back (the formulas for the Krasovsky reference ellipsoid are used&
The length of arc of an Earth's parallel at a given latitude
This online calculator converts the arc length of an Earth's parallel at a given latitude from degrees to meters.
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Problem 411 - TheMathWorld
Problem 411
The table shows the result of a survey in which 143 men and 145 women workers ages 25 to 64 were asked if they have at least one month’s set aside for emergencies.
Men Women Total
Less than one month’s income 64 85 149
One month’s income or more 79 60 139
Total 143 145 288
The empirical probability of an event E is the relative frequency of event E.
P(E) =
(a) Find the probability that a randomly selected worker has one month’s income or more set aside for emergencies.
The total number of surveyed workers id 288, and 139 of than have one month’s income or more set aside for emergencies
Using the empirical probability formula, find the probability P(K), where K is the events of having one month’s income or more set aside for emergencies.
P(K) =
(b) Given that a randomly selected worker is a male, find the probability that the worker has less than one month’s income.
A conditional probability is the probability of the event occurring, given that another event has .
The number of males among the asked workers is 143, and 64 of them have less than one month’s income.
Let L be the event of having less than one month’s income and let M be the event that a worker is male.
Using the empirical probability formula, find P(L│M), the probability that a randomly selected worker have less than one month’s income given that the worker is a male.
P(L│M) =
(c) Given that a randomly selected worker has one month’s income or more, find the probability that the worker is a female.
Form part(a), there are 139 workers with one month’s income or more. There are 60 females among those 139 workers.
Let F be the events that a worker is female.
Using the empirical probability formula, find P(F│K), the probability that a randomly selected worker is a female given that the worker has one month’s income or more.
P(F│K) =
(d) Are the events “having less than one month’s income saved” and “being male” independent?
Two events are independent if the occurrence of one if the events does not affect the probability of the occurrence of the other event. The condition of independence of two events L and M is the
P(L│M) = P(L)
First find P(L). 149 of 288 workers have less than one month’s income.
Using the empirical probability formula, find P(L), the probability of having less than one month’s income saved.
P(L) =
Recall form part (b) that P(L│M) = 0.448. AS P(L│M) ≠ P(L), the events “having less than one month’s income saved” and “being male” are dependent.
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Slope Distance Formula in context of slope distance
31 Aug 2024
The Slope Distance Formula: A Geometric Approach
In the realm of geometry, the concept of distance between two points has been extensively studied and formulated. However, when it comes to calculating the distance between a point and a line or a
plane, the traditional Euclidean distance formula is not applicable. This article presents an alternative approach, known as the Slope Distance Formula, which provides a geometric solution for
determining the shortest distance between a point and a line or a plane.
The Slope Distance Formula is a mathematical expression that calculates the shortest distance between a point and a line or a plane in three-dimensional space. This formula is particularly useful in
various fields such as computer-aided design (CAD), geographic information systems (GIS), and robotics, where precise calculations of distances are crucial.
Mathematical Background
Let’s consider a point P(x1, y1, z1) and a line L defined by the equation ax + by + cz = d. The Slope Distance Formula calculates the shortest distance between P and L as follows:
d = |(ax1 + by1 + cz1 - d)| / sqrt(a^2 + b^2 + c^2)
where |...| denotes the absolute value, and sqrt(...) represents the square root.
Geometric Interpretation
The Slope Distance Formula can be geometrically interpreted as follows:
• The numerator (ax1 + by1 + cz1 - d) represents the perpendicular distance from point P to line L.
• The denominator sqrt(a^2 + b^2 + c^2) represents the magnitude of the normal vector to line L.
The Slope Distance Formula provides a geometric solution for calculating the shortest distance between a point and a line or a plane. This formula is particularly useful in various fields where
precise calculations of distances are crucial. The mathematical expression and geometric interpretation presented in this article provide a comprehensive understanding of the Slope Distance Formula,
making it an essential tool for researchers and practitioners alike.
• [1] “Geometry: A Comprehensive Introduction” by Dan Pedoe
• [2] “Mathematics for Computer Graphics” by John Vince
Related articles for ‘slope distance’ :
Calculators for ‘slope distance’
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— three one four: a number of notes —
visualization + math
Pi Day 2022 — three one four: a number of notes — A musical journey into the digits of Pi
On March 14th celebrate `\pi` Day. Hug `\pi`—find a way to do it.
For those who favour `\tau=2\pi` will have to postpone celebrations until July 26th. That's what you get for thinking that `\pi` is wrong. I sympathize with this position and have `\tau` day art too!
If you're not into details, you may opt to party on July 22nd, which is `\pi` approximation day (`\pi` ≈ 22/7). It's 20% more accurate that the official `\pi` day!
Finally, if you believe that `\pi = 3`, you should read why `\pi` is not equal to 3.
3 There you go
1 Straight
4 Number me not
1 Scales
5 There is more of me
9 To forget than you can remember
—Emma Beauxis-Aussalet (314... piku)
Welcome to 2022 Pi Day: a celebration of `\pi` and mathematics (and music).
This year I've done something very different — a surprise for both the ear and eye. Working with Gregory Coles, we've composed an album based on the mathematics of `\pi`.
The album is called “three one four: a number of digits” . It is a sixteen minute musical exploration of the digits of `\pi`. Experience this famous number from its beginning (Track 1 314…) to its
very (known) end (Track 6 ...264), as well as the math (Track 3 Wallis Product) and jokes (Track 2 Feynman Point) behind it and aspects of its digits, such as repetition (Track 4 nn) and zeroes (
Track 5 null).
The album is scored for solo piano in the style of 20th century classical music – each piece has a distinct personality, drawn from styles of Boulez and Stockhausen (314…), Ligeti (Feynman Point),
Reich and Glass (Wallis Product), Satie (nn), Feldman (null), Powell and Monk (...264).
Each piece is accompanied by a piku (or πku), a poem whose syllable count is determined by a specific sequence of digits from π. I came up with the concept of a piku for 2020 Pi Day.
9 We know we can never have it all
2 Never
4 But yet can we
2 Ever
6 In endless scarcity
4 Have just enough?
—Emma Beauxis-Aussalet (...264 piku)
The piku collection were written by Emma Beauxis-Aussalet set of piku and each piku follows a syllable count that closely matches the theme of its track.
Our album was the theme of a Numberphile Podcast “The First and Last Digits of Pi” and appeared in Nature Briefing for March 14, 2022.
I sit down with the composer of the album, Gregory Coles, and start the conversation about turning mathematics into music. Our attempt was to find a compelling balance between `\pi` and the heart of
a musician.
We go through the list of tracks on the album and give you a sense of what to expect.
what to listen to first?
geek out on music theory and scores
If you're a music geek, you'll love our detailed discussion of the music theory behind each track and a thorough score analysis.
download score
The entire album is arranged for solo piano and lovingly engraved. You can download the full score.
news + thoughts
I don’t have good luck in the match points. —Rafael Nadal, Spanish tennis player
In many experimental designs, we need to keep in mind the possibility of confounding variables, which may give rise to bias in the estimate of the treatment effect.
If the control and experimental groups aren't matched (or, roughly, similar enough), this bias can arise.
Sometimes this can be dealt with by randomizing, which on average can balance this effect out. When randomization is not possible, propensity score matching is an excellent strategy to match control
and experimental groups.
Kurz, C.F., Krzywinski, M. & Altman, N. (2024) Points of significance: Propensity score matching. Nat. Methods 21:1770–1772.
We'd like to say a ‘cosmic hello’: mathematics, culture, palaeontology, art and science, and ... human genomes.
All animals are equal, but some animals are more equal than others. —George Orwell
This month, we will illustrate the importance of establishing a baseline performance level.
Baselines are typically generated independently for each dataset using very simple models. Their role is to set the minimum level of acceptable performance and help with comparing relative
improvements in performance of other models.
Unfortunately, baselines are often overlooked and, in the presence of a class imbalance, must be established with care.
Megahed, F.M, Chen, Y-J., Jones-Farmer, A., Rigdon, S.E., Krzywinski, M. & Altman, N. (2024) Points of significance: Comparing classifier performance with baselines. Nat. Methods 21:546–548.
Celebrate π Day (March 14th) and dig into the digit garden. Let's grow something.
Huge empty areas of the universe called voids could help solve the greatest mysteries in the cosmos.
My graphic accompanying How Analyzing Cosmic Nothing Might Explain Everything in the January 2024 issue of Scientific American depicts the entire Universe in a two-page spread — full of nothing.
The graphic uses the latest data from SDSS 12 and is an update to my Superclusters and Voids poster.
Michael Lemonick (editor) explains on the graphic:
“Regions of relatively empty space called cosmic voids are everywhere in the universe, and scientists believe studying their size, shape and spread across the cosmos could help them understand dark
matter, dark energy and other big mysteries.
To use voids in this way, astronomers must map these regions in detail—a project that is just beginning.
Shown here are voids discovered by the Sloan Digital Sky Survey (SDSS), along with a selection of 16 previously named voids. Scientists expect voids to be evenly distributed throughout space—the lack
of voids in some regions on the globe simply reflects SDSS’s sky coverage.”
Sofia Contarini, Alice Pisani, Nico Hamaus, Federico Marulli Lauro Moscardini & Marco Baldi (2023) Cosmological Constraints from the BOSS DR12 Void Size Function Astrophysical Journal 953:46.
Nico Hamaus, Alice Pisani, Jin-Ah Choi, Guilhem Lavaux, Benjamin D. Wandelt & Jochen Weller (2020) Journal of Cosmology and Astroparticle Physics 2020:023.
Sloan Digital Sky Survey Data Release 12
constellation figures
Alan MacRobert (Sky & Telescope), Paulina Rowicka/Martin Krzywinski (revisions & Microscopium)
Hoffleit & Warren Jr. (1991) The Bright Star Catalog, 5th Revised Edition (Preliminary Version).
H[0] = 67.4 km/(Mpc·s), Ω[m] = 0.315, Ω[v] = 0.685. Planck collaboration Planck 2018 results. VI. Cosmological parameters (2018).
It is the mark of an educated mind to rest satisfied with the degree of precision that the nature of the subject admits and not to seek exactness where only an approximation is possible. —Aristotle
In regression, the predictors are (typically) assumed to have known values that are measured without error.
Practically, however, predictors are often measured with error. This has a profound (but predictable) effect on the estimates of relationships among variables – the so-called “error in variables”
Error in measuring the predictors is often ignored. In this column, we discuss when ignoring this error is harmless and when it can lead to large bias that can leads us to miss important effects.
Altman, N. & Krzywinski, M. (2024) Points of significance: Error in predictor variables. Nat. Methods 21:4–6.
Background reading
Altman, N. & Krzywinski, M. (2015) Points of significance: Simple linear regression. Nat. Methods 12:999–1000.
Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Logistic regression. Nat. Methods 13:541–542 (2016).
Das, K., Krzywinski, M. & Altman, N. (2019) Points of significance: Quantile regression. Nat. Methods 16:451–452.
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Circle - Alchetron, The Free Social Encyclopedia
The circle official trailer in theaters april 28 2017
A circle is a simple closed shape in Euclidean geometry. It is the set of all points in a plane that are at a given distance from a given point, the centre; equivalently it is the curve traced out by
a point that moves so that its distance from a given point is constant. The distance between any of the points and the centre is called the radius.
A circle is a simple closed curve which divides the plane into two regions: an interior and an exterior. In everyday use, the term "circle" may be used interchangeably to refer to either the boundary
of the figure, or to the whole figure including its interior; in strict technical usage, the circle is only the boundary and the whole figure is called a disc.
A circle may also be defined as a special kind of ellipse in which the two foci are coincident and the eccentricity is 0, or the two-dimensional shape enclosing the most area per unit perimeter
squared, using calculus of variations.
A circle is a plane figure bounded by one line, and such that all right lines drawn from a certain point within it to the bounding line, are equal. The bounding line is called its circumference and
the point, its centre.
The circle official trailer 1 2017 emma watson movie
Annulus: the ring-shaped object, the region bounded by two concentric circles.
Arc: any connected part of the circle.
Centre: the point equidistant from the points on the circle.
Chord: a line segment whose endpoints lie on the circle.
Circumference: the length of one circuit along the circle, or the distance around the circle.
Diameter: a line segment whose endpoints lie on the circle and which passes through the centre; or the length of such a line segment, which is the largest distance between any two points on the
circle. It is a special case of a chord, namely the longest chord, and it is twice the radius.
Disc: the region of the plane bounded by a circle.
Lens: the intersection of two discs.
Passant: a coplanar straight line that does not touch the circle.
Radius: a line segment joining the centre of the circle to any point on the circle itself; or the length of such a segment, which is half a diameter.
Sector: a region bounded by two radii and an arc lying between the radii.
Segment: a region, not containing the centre, bounded by a chord and an arc lying between the chord's endpoints.
Secant: an extended chord, a coplanar straight line cutting the circle at two points.
Semicircle: an arc that extends from one of a diameter's endpoints to the other. In non-technical common usage it may mean the diameter, arc, and its interior, a two dimensional region, that is
technically called a half-disc. A half-disc is a special case of a segment, namely the largest one.
Tangent: a coplanar straight line that touches the circle at a single point.
The word circle derives from the Greek κίρκος/κύκλος (kirkos/kuklos), itself a metathesis of the Homeric Greek κρίκος (krikos), meaning "hoop" or "ring". The origins of the words circus and circuit
are closely related.
The circle has been known since before the beginning of recorded history. Natural circles would have been observed, such as the Moon, Sun, and a short plant stalk blowing in the wind on sand, which
forms a circle shape in the sand. The circle is the basis for the wheel, which, with related inventions such as gears, makes much of modern machinery possible. In mathematics, the study of the circle
has helped inspire the development of geometry, astronomy and calculus.
Early science, particularly geometry and astrology and astronomy, was connected to the divine for most medieval scholars, and many believed that there was something intrinsically "divine" or
"perfect" that could be found in circles.
Some highlights in the history of the circle are:
1700 BCE – The Rhind papyrus gives a method to find the area of a circular field. The result corresponds to 256/81 (3.16049...) as an approximate value of π.
300 BCE – Book 3 of Euclid's Elements deals with the properties of circles.
In Plato's Seventh Letter there is a detailed definition and explanation of the circle. Plato explains the perfect circle, and how it is different from any drawing, words, definition or explanation.
1880 CE – Lindemann proves that π is transcendental, effectively settling the millennia-old problem of squaring the circle.
Length of circumference
The ratio of a circle's circumference to its diameter is π (pi), an irrational constant approximately equal to 3.141592654. Thus the length of the circumference C is related to the radius r and
diameter d by:
C = 2 π r = π d .
Area enclosed
As proved by Archimedes, in his Measurement of a Circle, the area enclosed by a circle is equal to that of a triangle whose base has the length of the circle's circumference and whose height equals
the circle's radius, which comes to π multiplied by the radius squared:
A r e a = π r 2 .
Equivalently, denoting diameter by d,
A r e a = π d 2 4 ≈ 0 . 7854 d 2 ,
that is, approximately 79% of the circumscribing square (whose side is of length d).
The circle is the plane curve enclosing the maximum area for a given arc length. This relates the circle to a problem in the calculus of variations, namely the isoperimetric inequality.
Cartesian coordinates
In an x–y Cartesian coordinate system, the circle with centre coordinates (a, b) and radius r is the set of all points (x, y) such that
( x − a ) 2 + ( y − b ) 2 = r 2 .
This equation, known as the Equation of the Circle, follows from the Pythagorean theorem applied to any point on the circle: as shown in the adjacent diagram, the radius is the hypotenuse of a
right-angled triangle whose other sides are of length |x − a| and |y − b|. If the circle is centred at the origin (0, 0), then the equation simplifies to
x 2 + y 2 = r 2 .
The equation can be written in parametric form using the trigonometric functions sine and cosine as
x = a + r cos t , y = b + r sin t
where t is a parametric variable in the range 0 to 2π, interpreted geometrically as the angle that the ray from (a, b) to (x, y) makes with the positive x-axis.
An alternative parametrisation of the circle is:
x = a + r 1 − t 2 1 + t 2 . y = b + r 2 t 1 + t 2
In this parametrisation, the ratio of t to r can be interpreted geometrically as the stereographic projection of the line passing through the centre parallel to the x-axis (see Tangent half-angle
substitution). However, this parametrisation works only if t is made to range not only through all reals but also to a point at infinity; otherwise, the bottom-most point of the circle would be
In homogeneous coordinates each conic section with the equation of a circle has the form
x 2 + y 2 − 2 a x z − 2 b y z + c z 2 = 0.
It can be proven that a conic section is a circle exactly when it contains (when extended to the complex projective plane) the points I(1: i: 0) and J(1: −i: 0). These points are called the circular
points at infinity.
Polar coordinates
In polar coordinates the equation of a circle is:
r 2 − 2 r r 0 cos ( θ − ϕ ) + r 0 2 = a 2
where a is the radius of the circle, ( r , θ ) is the polar coordinate of a generic point on the circle, and ( r 0 , ϕ ) is the polar coordinate of the centre of the circle (i.e., r[0] is the
distance from the origin to the centre of the circle, and φ is the anticlockwise angle from the positive x-axis to the line connecting the origin to the centre of the circle). For a circle centred at
the origin, i.e. r[0] = 0, this reduces to simply r = a. When r[0] = a, or when the origin lies on the circle, the equation becomes
r = 2 a cos ( θ − ϕ ) .
In the general case, the equation can be solved for r, giving
r = r 0 cos ( θ − ϕ ) ± a 2 − r 0 2 sin 2 ( θ − ϕ ) ,
Note that without the ± sign, the equation would in some cases describe only half a circle.
Complex plane
In the complex plane, a circle with a centre at c and radius (r) has the equation | z − c | = r . In parametric form this can be written z = r e i t + c .
The slightly generalised equation p z z ¯ + g z + g z ¯ = q for real p, q and complex g is sometimes called a generalised circle. This becomes the above equation for a circle with p = 1 , g = − c ¯ ,
q = r 2 − | c | 2 , since | z − c | 2 = z z ¯ − c ¯ z − c z ¯ + c c ¯ . Not all generalised circles are actually circles: a generalised circle is either a (true) circle or a line.
Tangent lines
The tangent line through a point P on the circle is perpendicular to the diameter passing through P. If P = (x[1], y[1]) and the circle has centre (a, b) and radius r, then the tangent line is
perpendicular to the line from (a, b) to (x[1], y[1]), so it has the form (x[1] − a)x + (y[1] – b)y = c. Evaluating at (x[1], y[1]) determines the value of c and the result is that the equation of
the tangent is
( x 1 − a ) x + ( y 1 − b ) y = ( x 1 − a ) x 1 + ( y 1 − b ) y 1
( x 1 − a ) ( x − a ) + ( y 1 − b ) ( y − b ) = r 2 .
If y[1] ≠ b then the slope of this line is
d y d x = − x 1 − a y 1 − b .
This can also be found using implicit differentiation.
When the centre of the circle is at the origin then the equation of the tangent line becomes
x 1 x + y 1 y = r 2 ,
and its slope is
d y d x = − x 1 y 1 .
The circle is the shape with the largest area for a given length of perimeter. (See Isoperimetric inequality.)
The circle is a highly symmetric shape: every line through the centre forms a line of reflection symmetry and it has rotational symmetry around the centre for every angle. Its symmetry group is the
orthogonal group O(2,R). The group of rotations alone is the circle group T.
All circles are similar.
A circle's circumference and radius are proportional.
The area enclosed and the square of its radius are proportional.
The constants of proportionality are 2π and π, respectively.
The circle which is centred at the origin with radius 1 is called the unit circle.
Thought of as a great circle of the unit sphere, it becomes the Riemannian circle.
Through any three points, not all on the same line, there lies a unique circle. In Cartesian coordinates, it is possible to give explicit formulae for the coordinates of the centre of the circle and
the radius in terms of the coordinates of the three given points. See circumcircle.
Chords are equidistant from the centre of a circle if and only if they are equal in length.
The perpendicular bisector of a chord passes through the centre of a circle; equivalent statements stemming from the uniqueness of the perpendicular bisector are:
A perpendicular line from the centre of a circle bisects the chord.
The line segment through the centre bisecting a chord is perpendicular to the chord.
If a central angle and an inscribed angle of a circle are subtended by the same chord and on the same side of the chord, then the central angle is twice the inscribed angle.
If two angles are inscribed on the same chord and on the same side of the chord, then they are equal.
If two angles are inscribed on the same chord and on opposite sides of the chord, then they are supplementary.
For a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
An inscribed angle subtended by a diameter is a right angle (see Thales' theorem).
The diameter is the longest chord of the circle.
If the intersection of any two chords divides one chord into lengths a and b and divides the other chord into lengths c and d, then ab = cd.
If the intersection of any two perpendicular chords divides one chord into lengths a and b and divides the other chord into lengths c and d, then a^2 + b^2 + c^2 + d^2 equals the square of the
The sum of the squared lengths of any two chords intersecting at right angles at a given point is the same as that of any other two perpendicular chords intersecting at the same point, and is given
by 8r ^2 – 4p ^2 (where r is the circle's radius and p is the distance from the centre point to the point of intersection).
The distance from a point on the circle to a given chord times the diameter of the circle equals the product of the distances from the point to the ends of the chord.
A line drawn perpendicular to a radius through the end point of the radius lying on the circle is a tangent to the circle.
A line drawn perpendicular to a tangent through the point of contact with a circle passes through the centre of the circle.
Two tangents can always be drawn to a circle from any point outside the circle, and these tangents are equal in length.
If a tangent at A and a tangent at B intersect at the exterior point P, then denoting the centre as O, the angles ∠BOA and ∠BPA are supplementary.
If AD is tangent to the circle at A and if AQ is a chord of the circle, then ∠DAQ = 1/2arc(AQ).
The chord theorem states that if two chords, CD and EB, intersect at A, then AC × AD = AB × AE.
If two secants, AE and AD, also cut the circle at B and C respectively, then AC × AD = AB × AE. (Corollary of the chord theorem.)
A tangent can be considered a limiting case of a secant whose ends are coincident. If a tangent from an external point A meets the circle at F and a secant from the external point A meets the circle
at C and D respectively, then AF^2 = AC × AD. (Tangent-secant theorem.)
The angle between a chord and the tangent at one of its endpoints is equal to one half the angle subtended at the centre of the circle, on the opposite side of the chord (Tangent Chord Angle).
If the angle subtended by the chord at the centre is 90 degrees then ℓ = r √2, where ℓ is the length of the chord and r is the radius of the circle.
If two secants are inscribed in the circle as shown at right, then the measurement of angle A is equal to one half the difference of the measurements of the enclosed arcs (DE and BC). I.e. 2∠CAB = ∠
DOE − ∠BOC, where O is the centre of the circle. This is the secant-secant theorem.
Inscribed angles
An inscribed angle (examples are the blue and green angles in the figure) is exactly half the corresponding central angle (red). Hence, all inscribed angles that subtend the same arc (pink) are
equal. Angles inscribed on the arc (brown) are supplementary. In particular, every inscribed angle that subtends a diameter is a right angle (since the central angle is 180 degrees).
The sagitta (also known as the versine) is a line segment drawn perpendicular to a chord, between the midpoint of that chord and the arc of the circle.
Given the length y of a chord, and the length x of the sagitta, the Pythagorean theorem can be used to calculate the radius of the unique circle which will fit around the two lines:
Another proof of this result which relies only on two chord properties given above is as follows. Given a chord of length y and with sagitta of length x, since the sagitta intersects the midpoint of
the chord, we know it is part of a diameter of the circle. Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord
times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2)^2. Solving for r, we find the required result.
Compass and straightedge constructions
There are many compass-and-straightedge constructions resulting in circles.
The simplest and most basic is the construction given the centre of the circle and a point on the circle. Place the fixed leg of the compass on the centre point, the movable leg on the point on the
circle and rotate the compass.
Construct a circle with a given diameter
Construct the midpoint M of the diameter.
Construct the circle with centre M passing through one of the endpoints of the diameter (it will also pass through the other endpoint).
Construct a circle through 3 noncollinear points
Name the points P, Q and R,
Construct the perpendicular bisector of the segment PQ.
Construct the perpendicular bisector of the segment PR.
Label the point of intersection of these two perpendicular bisectors M. (They meet because the points are not collinear).
Construct the circle with centre M passing through one of the points P, Q or R (it will also pass through the other two points).
Circle of Apollonius
Apollonius of Perga showed that a circle may also be defined as the set of points in a plane having a constant ratio (other than 1) of distances to two fixed foci, A and B. (The set of points where
the distances are equal is the perpendicular bisector of A and B, a line.) That circle is sometimes said to be drawn about two points.
The proof is in two parts. First, one must prove that, given two foci A and B and a ratio of distances, any point P satisfying the ratio of distances must fall on a particular circle. Let C be
another point, also satisfying the ratio and lying on segment AB. By the angle bisector theorem the line segment PC will bisect the interior angle APB, since the segments are similar:
A P B P = A C B C .
Analogously, a line segment PD through some point D on AB extended bisects the corresponding exterior angle BPQ where Q is on AP extended. Since the interior and exterior angles sum to 180 degrees,
the angle CPD is exactly 90 degrees, i.e., a right angle. The set of points P such that angle CPD is a right angle forms a circle, of which CD is a diameter.
Second, see for a proof that every point on the indicated circle satisfies the given ratio.
A closely related property of circles involves the geometry of the cross-ratio of points in the complex plane. If A, B, and C are as above, then the circle of Apollonius for these three points is the
collection of points P for which the absolute value of the cross-ratio is equal to one:
| [ A , B ; C , P ] | = 1.
Stated another way, P is a point on the circle of Apollonius if and only if the cross-ratio [A,B;C,P] is on the unit circle in the complex plane.
Generalised circles
If C is the midpoint of the segment AB, then the collection of points P satisfying the Apollonius condition
| A P | | B P | = | A C | | B C |
is not a circle, but rather a line.
Thus, if A, B, and C are given distinct points in the plane, then the locus of points P satisfying the above equation is called a "generalised circle." It may either be a true circle or a line. In
this sense a line is a generalised circle of infinite radius.
Circles inscribed in or circumscribed about other figures
In every triangle a unique circle, called the incircle, can be inscribed such that it is tangent to each of the three sides of the triangle.
About every triangle a unique circle, called the circumcircle, can be circumscribed such that it goes through each of the triangle's three vertices.
A tangential polygon, such as a tangential quadrilateral, is any convex polygon within which a circle can be inscribed that is tangent to each side of the polygon.
A cyclic polygon is any convex polygon about which a circle can be circumscribed, passing through each vertex. A well-studied example is the cyclic quadrilateral.
A hypocycloid is a curve that is inscribed in a given circle by tracing a fixed point on a smaller circle that rolls within and tangent to the given circle.
Circle as limiting case of other figures
The circle can be viewed as a limiting case of each of various other figures:
A Cartesian oval is a set of points such that a weighted sum of the distances from any of its points to two fixed points (foci) is a constant. An ellipse is the case in which the weights are equal. A
circle is an ellipse with an eccentricity of zero, meaning that the two foci coincide with each other as the centre of the circle. A circle is also a different special case of a Cartesian oval in
which one of the weights is zero.
A superellipse has an equation of the form | x a | n + | y b | n = 1 for positive a, b, and n. A supercircle has b = a. A circle is the special case of a supercircle in which n = 2.
A Cassini oval is a set of points such that the product of the distances from any of its points to two fixed points is a constant. When the two fixed points coincide, a circle results.
A curve of constant width is a figure whose width, defined as the perpendicular distance between two distinct parallel lines each intersecting its boundary in a single point, is the same regardless
of the direction of those two parallel lines. The circle is the simplest example of this type of figure.
Squaring the circle
Squaring the circle is the problem, proposed by ancient geometers, of constructing a square with the same area as a given circle by using only a finite number of steps with compass and straightedge.
In 1882, the task was proven to be impossible, as a consequence of the Lindemann–Weierstrass theorem which proves that pi (π) is a transcendental number, rather than an algebraic irrational number;
that is, it is not the root of any polynomial with rational coefficients.
Circle Wikipedia
(Text) CC BY-SA
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5,757 research outputs found
Numerical analysis is conducted for a generalized particle method for a Poisson equation. Unique solvability is derived for the discretized Poisson equation by introducing a connectivity condition
for particle distributions. Moreover, by introducing discrete Sobolev norms and a semi-regularity of a family of discrete parameters, stability is obtained for the discretized Poisson equation based
on the norms.Comment: 7 pages, 1 figur
The amount of entanglement necessary to teleport quantum states drawn from general ensemble $\{p_i,\rho_i\}$ is derived. The case of perfect transmission of individual states and that of
asymptotically faithful transmission are discussed. Using the latter result, we also derive the optimum compression rate when the ensemble is compressed into qubits and bits.Comment: 9 pages, 1 figur
We propose a usage of a weak value for a quantum processing between preselection and postselection. While the weak value of a projector of 1 provides a process with certainty like the probability of
1, the weak value of -1 negates the process completely. Their mutually opposite effect is approved without a conventional `weak' condition. In addition the quantum process is not limited to be
unitary; in particular we consider a loss of photons and experimentally demonstrate the negation of the photon loss by using the negative weak value of -1 against the positive weak value of
1.Comment: 12 pages, 6 figures, close to published versio
We consider a case where a weak value is introduced as a physical quantity rather than an average of weak measurements. The case we treat is a time evolution of a particle by 1+1 dimensional Dirac
equation. Particularly in a spontaneous pair production via a supercritical step potential, a quantitative explanation can be given by a weak value for the group velocity of the particle. We also
show the condition for the pair production (supercriticality) corresponds to the condition when the weak value takes a strange value (superluminal velocity).Comment: 12 pages, 3 figures, close to
published versio
From the Hamiltonian connecting the inside and outside of an Fabry-Perot cavity, which is derived from the Maxwell boundary conditions at a mirror of the cavity, a master equation of a non-Lindblad
form is derived when the cavity embeds matters, although we can transform it to the Lindblad form by performing the rotating-wave approximation to the connecting Hamiltonian. We calculate absorption
spectra by these Lindblad and non-Lindblad master equations and also by the Maxwell boundary conditions in the framework of the classical electrodynamics, which we consider the most reliable
approach. We found that, compared to the Lindblad master equation, the absorption spectra by the non-Lindblad one agree better with those by the Maxwell boundary conditions. Although the discrepancy
is highlighted only in the ultra-strong light-matter interaction regime with a relatively large broadening, the master equation of the non-Lindblad form is preferable rather than of the Lindblad one
for pursuing the consistency with the classical electrodynamics.Comment: 22 pages, 9 figure
Consider a situation in which a quantum system is secretly prepared in a state chosen from the known set of states. We present a principle that gives a definite distinction between the operations
that preserve the states of the system and those that disturb the states. The principle is derived by alternately applying a fundamental property of classical signals and a fundamental property of
quantum ones. The principle can be cast into a simple form by using a decomposition of the relevant Hilbert space, which is uniquely determined by the set of possible states. The decomposition
implies the classification of the degrees of freedom of the system into three parts depending on how they store the information on the initially chosen state: one storing it classically, one storing
it nonclassically, and the other one storing no information. Then the principle states that the nonclassical part is inaccessible and the classical part is read-only if we are to preserve the state
of the system. From this principle, many types of no-cloning, no-broadcasting, and no-imprinting conditions can easily be derived in general forms including mixed states. It also gives a unified view
on how various schemes of quantum cryptography work. The principle helps to derive optimum amount of resources (bits, qubits, and ebits) required in data compression or in quantum teleportation of
mixed-state ensembles.Comment: 24 pages, no fogur
We propose a model for the electric current in graphene in which electric carriers are supplied by virtual particles allowed by the uncertainty relations. The process to make a virtual particle real
is described by a weak value of a group velocity: the velocity is requisite for the electric field to give the virtual particle the appropriate changes of both energy and momentum. With the weak
value, we approximately estimate the electric current, considering the ballistic transport of the electric carriers. The current shows the quasi-Ohimic with the minimal conductivity of the order of e
^2/h per channel. Crossing a certain ballistic time scale, it is brought to obey the Schwinger mechanism.Comment: 15 pages, 3 figures, close to published versio
When the weak value of a projector is 1, a quantum system behaves as in that eigenstate with probability 1. By definition, however, the weak value may take an anomalous value lying outside the range
of probability like -1. From the viewpoint of a physical effect, we show that such a negative weak value of -1 can be regarded as the counterpart of the ordinary value of 1. Using photons, we
experimentally verify it as the symmetrical shift in polarization depending on the weak value given by pre-postselection of the path state. Unlike observation of a weak value as an ensemble average
via weak measurements, the effect of a weak value is definitely confirmed in two photon interference: the symmetrical shift corresponding to the weak value can be directly observed as the rotation
angle of a half wave plate.Comment: 10 pages, 5 figures, close to published versio
According to the Schwinger mechanism, a uniform electric field brings about pair productions in vacuum; the relationship between the production rate and the electric field is different, depending on
the dimension of the system. In this paper, we make an offer of another model for the pair productions, in which weak values are incorporated: energy fluctuations trigger the pair production, and a
weak value appears as the velocity of a particle there. Although our model is only available for the approximation of the pair production rates, the weak value reveals a new aspect of the pair
production. Especially, within the first order, our estimation approximately agrees with the exponential decreasing rate of the Landau-Zener tunneling through the mass energy gap. In other words,
such tunneling can be associated with energy fluctuations via the weak value, when the tunneling gap can be regarded as so small due to the high electric field.Comment: 15 pages, 2 figure
Several superconducting circuit configurations are examined on the existence of super-radiant phase transitions (SRPTs) in thermal equilibrium. For some configurations consisting of artificial atoms,
whose circuit diagrams are however not specified, and an LC resonator or a transmission line, we confirm the absence of SRPTs in the thermal equilibrium following the similar analysis as the no-go
theorem for atomic systems. We also show some other configurations where the absence of SRPTs cannot be confirmed.Comment: 12 pages, 6 figure
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Towing Force Calculator - Savvy Calculator
Towing Force Calculator
About Towing Force Calculator (Formula)
A Towing Force Calculator is an invaluable tool used to determine the amount of force required to tow an object, such as a vehicle or trailer. This calculation is crucial for various applications,
including automotive, transportation, and logistics industries, where safety and efficiency are paramount. By understanding the towing force, users can ensure they select the appropriate equipment
and techniques for towing, thus minimizing risks and optimizing performance.
The formula to calculate the towing force (TF) is:
TF = BF * EF
• TF is the towing force.
• BF is the base force required to initiate towing.
• EF is the efficiency factor that accounts for various conditions affecting towing.
How to Use
Using the Towing Force Calculator involves the following steps:
1. Determine the Base Force (BF): Assess the base force required to tow the object, which depends on its weight, type, and terrain conditions.
2. Identify the Efficiency Factor (EF): Consider factors such as friction, incline, and the condition of the towing vehicle and towed object. This will give you the efficiency factor.
3. Input Values: Enter the values for the base force (BF) and efficiency factor (EF) into the calculator.
4. Calculate Towing Force: Click the calculate button to find the towing force required. This result will help you understand the necessary force for safe towing.
Let’s look at a practical example:
• Base Force (BF): 3000 pounds
• Efficiency Factor (EF): 0.85
Using the formula:
TF = BF * EF
TF = 3000 * 0.85
TF = 2550 pounds
In this example, the towing force required is 2550 pounds, indicating the force needed to tow the object effectively.
1. What is towing force?
Towing force is the amount of force required to pull an object, such as a trailer or vehicle, using a towing vehicle.
2. Why is calculating towing force important?
Calculating towing force is crucial for ensuring safety and efficiency in towing operations and preventing equipment damage or accidents.
3. What factors influence towing force?
Factors such as weight, terrain, incline, and vehicle condition significantly affect the required towing force.
4. How do I determine the base force (BF)?
The base force can be determined by assessing the weight of the object being towed and considering additional factors like friction and drag.
5. What is the efficiency factor (EF)?
The efficiency factor accounts for various conditions affecting towing, such as friction, incline, and mechanical inefficiencies.
6. Can I use the calculator for different types of vehicles?
Yes, the Towing Force Calculator can be used for various vehicles, including cars, trucks, and trailers.
7. What happens if the towing force is too low?
If the towing force is insufficient, it may result in difficulty towing the object, potential equipment failure, and safety hazards.
8. Is there an acceptable range for towing force?
Acceptable towing force varies based on the weight of the object and the capabilities of the towing vehicle; always consult manufacturer specifications.
9. How can I improve towing efficiency?
To improve towing efficiency, ensure proper vehicle maintenance, reduce excess weight, and choose optimal towing techniques.
10. What is the best way to secure a load for towing?
Use appropriate towing equipment, such as straps, chains, or hitches, and ensure they are rated for the load’s weight.
11. Can I tow on inclines?
Yes, but towing on inclines requires additional force. Consider this when calculating your towing force.
12. How does tire pressure affect towing force?
Proper tire pressure ensures optimal traction and reduces rolling resistance, thereby influencing the overall towing force.
13. What types of vehicles are best for towing?
Trucks, SUVs, and certain vans are typically designed for towing due to their power and structural integrity.
14. Do environmental conditions affect towing?
Yes, factors like wind resistance, weather, and road conditions can impact the required towing force.
15. How can I estimate the weight of my load?
You can weigh your load using a scale or estimate it based on manufacturer specifications and known weights of similar items.
16. What equipment is necessary for safe towing?
Essential equipment includes a towing hitch, safety chains, brake lights, and potentially a weight distribution system.
17. Can towing force be calculated for off-road conditions?
Yes, but additional factors such as terrain type and traction must be considered in the calculations.
18. Is it safe to exceed the calculated towing force?
Exceeding the calculated towing force can be dangerous and should be avoided to prevent accidents and equipment damage.
19. How often should I check my towing setup?
Regular checks before towing—especially if changing loads or conditions—are essential for safety.
20. What resources can help me learn more about towing?
Consider industry guides, manufacturer manuals, and online tutorials that provide valuable information on safe towing practices.
The Towing Force Calculator is an essential tool for anyone involved in towing operations, providing critical insights into the forces at play. By understanding and calculating the towing force
accurately, users can ensure safe and effective towing, ultimately preventing accidents and equipment damage. Regularly using this calculator and adhering to best practices can lead to enhanced
safety and operational efficiency in various towing scenarios.
Leave a Comment
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Calculus at 34
I started this journey with a the following concern: would I be able to learn Calculus, or is it something beyond my reach?
I think I have answered that question this year. I am quite able to learn Calculus and I have actually learned some. My statement of intent was "to learn and master Calculus", which now sounds like a
very bold statement. I believe I would have needed many more years to master a discipline as vast as Calculus. However, my new understanding of it fills me with satisfaction. I have indeed attained
closure on a topic that was as personal as it was academic.
What do I take with me? Quite a lot actually, from the importance of starting with a good base, to the realization that I learn best with the structure of a course than on my own. I also take limits
both as a mathematical concept as well as life concept. One helps you explore infinity, the other, I realized, is mainly in your mind. I take differentiation and integration, since at times is
important to know the guiding essence of a thing and at others the power of togetherness.
I also learned a lot about change.
Change of rates as well as change of heart. Change as the definition of what we are and were we are going.
I am happy I did this project. It has been hard, it took a lot of time and effort. I made some hard sacrifices as time. But I made it.
I want to thank my wife María del Carmen for her love and support, and also my daughter Amanda for putting up with me calculating away some Saturday mornings. I could not have done it without you.
You are, ahem, integral to me. I love you.
So here it is, the end of this blog, I guess. The last entry. Thank you Calculus, it has been a blast.
Here I am. At the end of my project. My next to last post. And my topic is the fundamental theorem of Calculus. Like the name says, this theorem is a pretty big deal. Let's start by what the Theorem
says according to Wolfram Alpha:
I can proudly say I actually understood some of that, but at first could not yet grasp the implications that statement had on all I had studied so far.
Here is Dr. Fowler from Mooculus.com explaining this theorem and it's implications brilliantly:
Now here is what (I think) the fundamental theorem of calculus means in my own words. If you want to integrate acontinuous function on a closed interval, instead of doing the limits of the Riemman
Sum applicable, just find the anti-derivative of the function you need to integrate and substract the result of the evaluation of that anti-derivative at the befining and end points of your closed
Or in better words, forget about integrating, just anti-diferentiate!
In my last two posts I had been searching for the area under the curve of x^2 from the interval 0 to 2. And I had to do a bunch of sigma calculations and set up Riemman Sums and then even take a
limit in order to get to 2.667 square units which is 8/3.
The fundamental theorem of calculus is, as I see it, a reward for all my efforts. It is a way of saying:"Fernando, you have toiled and fret, and sweated over these sums and spend countless pages
calculating and recalculating all these stuff. You have earned a shortcut." Why thank you very much calculus!
Do you want to check out my new super power? Ok.
The FTOC is telling me that to integrate from the interval a=0 to b=2 of the function x^2 all I need is an anti-derivative of that formula which I will proceed to evaluate at both points of my
interval and then subtract. What is an antiderivative of x^2?
Technically there is a + C after that formula but I am assuming it is 0, check Dr. Fowler's video again for that to make sense.
Ok, so now I have my antiderivative x^3/[3] and I am ready to evaluate it at 2 (b=2) and at 0 (a=0) and then take the difference .
Now let's plug it back into the formula for the fundamental theorem of calculus.
And finally we get:
There you have it, the elusive 2.667, the area under the curve of x^2 From the interval starting at 0 ending at 2.
Wow, that took way less effort than before. However, If I had not passed through all those previous steps, all that trouble and effort, I would not have appreciated the beauty, simplicity and deeper
meaning of what I have accomplished.
This is my next to last post in this blog and it is a fitting one. I cannot begin to express the emotions I feel at the moment. I have attained great insights in this journey. I now have a deeper
understanding of math and the world around me.
Let me know your thoughts in the comments section.
I wish I could explain how satisfying it is to finally learn the concepts of integration in calculus. For years I have seen how people would represent calculus with a picture of a curved function
with the area under it shaded, and for the life of me, I could not figure out how they would calculate that. Now I know. The relative simplicity of the process has a ring of truth, and symmetry, and
beauty that threw me back to discovering geometrical theorems in high school.
Well, here it goes: Integrals as I understood them.
Imagine you have a formula that is continuous at least for a given interval [a,b]. I chose again x^2 in the closed interval form 1 to 2. I already tried to find the area under that curve in my post
on Sigma.
By using 10 rectangles of width 1/5 and height x^2, I was able to approximate the area under the curve to be 2.28 square units an underestimation of the true area. However, I posted that I could get
better approximations if my rectangles had been infinitely small.
To get those better approximations I need to improve the sum I used before, pictured to the right. And in order to do that, I would need to convert it into a Riemman Sum.
The toughest part of doing integration is to set up the correct Riemman Sum for the purposes intended. I struggled so hard with this part that I want to give you this video to follow just in case I
mess up. Here is the general formula for a Riemman Sum:
Since I am doing a right Riemman Sum, I will use this version of the formula. Where a and b are the interval of my function, and n represents the times I will be cutting that interview. A right
Riemman sum will give me an overestimate of the area under the curve, which will complement the underestimate of 2.28 I got
The first thing you need to do to set up a Riemman sum is deciding what your interval is (in this case is from 0 to 2) and then decide how wide you want your divisions within that interval to be. I
want my intervals to be 1/5 units wide, but that is not important right now, just remember there are ten 1/5 divisions from the interval 0 to 2. Then for the height of my rectangles I chose to
evaluate the function x^2 on the right hand side of those intervals. With those steps selected then I follow the x^2 rules for Riemman Sums.
A Riemman Sum is the addition of the formula I want to evaluate (x^2) at the specific cuts I made. The first point on my interval is a=0 and the last point is b=2. I want to divide that interval in n
cuts of a certain size. The formula for the cut size is:
Now here is the formula for the height of my rectangles.
So putting all the steps together here is the formula for my Riemman Sum
And that is actually the hard part, for me at least. The rest is arithmetic.
So the answer I got was: 8/3 + 4/n + 4/3n^2
But what does that mean? Remember when I said I wanted the width of my rectangles to be 1/5 units and that it meant I would get 10 segments from interval 0 to 2? Well, if you substitute n=10. The
area under the curve it gives me is 3.08 square units.
Now since this is a right Riemman sum I know it is an overestimate. My last attempt in the Sigma post was equivalent to a left Riemman sum which gave me an underestimate of 2.28.
If I take the average of these two numbers I should be able to get a better estimate of the area under the curve: (3.08 + 2.28)/2 = 2.68.
And 2.68 is very close to the true area under the curve which is approximately 2.667. Now, how can we get there?
Well supposed that instead of splitting my interval of this Riemman sum into 10 pieces I split it into 100, n=100, what happens then? We get 2.708 instead of 3.08. And what if n=10,000, that would
make our rectangles very, very small, we then get 2.670! That is very, very close to 2.667
And what if n=infinity?
Then the n in this formula 8/3 + 4/n + 4/3n^2 would be so small (and therefore the width of the rectangles would be also so small) that the only effect relevant in 8/3, and guess what 8/3 comes down
to: 2.667 approximately.
And what did we just do? We just took a limit.
Whoah! What?!!
Yes, we took the limit of our formula to get the true area under the curve. And that my friends is called integration.
To integrate is to do the following:
To take the limit of the Riemman Sum you are working on as n approaches infinity.
In fact the definite integral is a normally written as a variation of the formula above.
The elongated S just means, take the limit of the sum of f(x) times the change of x from the interval from a to be, as that change of x gets infinetly small.
And there you have it. Integration via the sum of infinite rectangles.
This one was a tough one and there are some considerations to this integrations stuff, but you can review them here.
Let me know what you think in the comments.
At last I reached the integral part of Calculus. I was eager to get there since it dealt with a topic which I find fascinating, how can I find the area of a curved object.
You see, area for me,and for the rest of the world, is width x height. Which is pretty simple if you are dealing with rectangles. However, what happens when you are not dealing with straight lines
and need the area of a curved region. Well, apparently you just part from what you know and build a bridge!
What I mean by that is that if I wanted to find the area under a curve like the one in the picture opposite. I can draw rectangles under the curved region up to the function and approximate it's area
by summing the areas of all the rectangles I drew. That is a neat trick that has served humanity for centuries. And the cool part is that if I can make the rectangles thin enough, I can get a better
approximation of the area I am looking for.
If the concept of getting better approximations by making an interval smaller and smaller a sounds familiar it's because we saw it back in limits and derivatives.
So how do I go about summing all these rectangles? I mean of they are 5 or 10 it's simple enough to do it by hand, but what is they are 100 or a 1000...or n rectangles!
That is were sigma comes in.
Have you met sigma? That weird looking capital E that might hunt your math nightmares. It turns out it is tame enough once you get to know it.
Sigma is just notation for adding something over and over again. I loved it when Professor Fowler at Mooculus compared it to using a "loop" in programming. It basically encodes and bounds a sum for a
specified interval. One of the things I remembered having problems with was the notation used in sigma. For some reason back in college I found it difficult and unapproachable. Now it looks quite
Let's imagine I want to add 1+2+3+4+5+6+7+8+9+10. If I have to write that sum over and over again it gets kinda tiresome. So I will use sigma as shorthand for that sum. But first, let's think about
what I am summing: the number (whole numbers; integers; in order) form 1 to 10. So if I had a machine that would spit out a one then a plus sign then the next number after one and so on, I would
replicate that string of numbers above. Well that is what sigma does, let's see:
Whenever you see that weird E, just imagine it is saying "please add the result of whatever formula is after me, starting with the number that is under me in the variable and repeating the process as
many times as the number above me while making n to be a whole number greater after every repetition." Well, at least it said please.
Okay, first we have that n is the formula we are summing. Then under Sigma is an n=1, meaning the first result is 1. The it is asking us to repeat the process 10 times, that is the number over sigma.
However, after each repetition we must make n a number greater than before.
So we start with n=1, then add n=2, then add n=3 and so on until we add n=10, 1+2+3+4+5+6+7+8+9+10! For a better (much better) explanation you can go here.
So if I take the example above, and lets say I divide the area under the curve into 10 sections of 1/5 square units, whose height is the formula f(x)=x^2 evaluated at those cut points.
Therefore, the first rectangle would have area 0^2 times1/5=0, the next would have area (1/5)^2 times 1/5=.008, the one after would be (2/5)^2 times 1/5=.032 and so on. We could represent that sum
this way, at least I think we could:
I Sigmalize we are saying square the 5th of whatever n is from 0 to 9 (the height of our rectangles), then multiply it by 1/5 ( their width) and Sum all 10 results.
This will give you an answer that the area under the curve is approximately 2.28 square units.
That is not bad as approximations go. If you see the two pictures bellow, I shaded the 2.28 square units and pasted them to cover my rectangles.
But as you can see, I could not cover all the area under the curve. There are some triangles that go from the base of the rectangles up to the curve that I cold not cover. To cover all the area under
the curve I would need to make smaller and smaller rectangles. In fact if I could make those rectangles infinetly small, I could approximate almos exactly the area under the curve.
I'll give you a preview. The area under the curve is actually closer to 2.666, but we will need to use something I have been looking forward to learning for a long time: Integrals.
Let me know if I got this right in the comments.
As I near the end marker of this journey of learning calculus I meet anti-derivatives. And I am glad I did. Somehow this concept helped me put into perspective all the concepts that have come before
it and makes me feel I am back on track again. Just in case you were wondering where I lost my way, it was somewhere into L'hopital's Rule.
First things first: What is an Anti-derivative?
Well an anti-derivative answers the question of: what is this formula a derivative of? Or from what original formula could we have gotten this derivative. For example, if we have x^2 (the squaring
function), it's derivative is 2x. Therefore, x^2 is an anti-derivative of 2x. (For formal definition go here)
Notice that I wrote that x^2 is an anti-derivative of 2x. This is important because there can be many anti-derivatives for a given function. If we consider this formula for an anti-derivative: ^x^n+1
/[n+1]+C, (which looks a lot prettier in pictures, see bellow) there is a constant C that is introduced. The way I understand it is that there is only so much information a anti-derivative can give
you. In order to recover a specific formula from it's derivative you need to know where that functions "started".
In the anti derivative formula, if x=0 the all we get is C. For example, imagine that x is time, then x=0 is time 0 or your starting point. At that starting point then f(0)=C.
To complete the argument above, let now imagine this scenario. I know that -2x+5 is a derivative of a function I am interested in knowing. Using the ati-derivative formula I get that the function I
am interested in is:
But what is C? I have no clue with the information I was given. Let's say I know C is a whole number between 1 and 4. A graph can show me what to expect the graph to be.
But until I know what that constant actually is, I will not know the original formula.
In the picture opposite I have 4 possible graphs, each passing through the y (vertical axis) at 1,2,3 and 4. All 4 graphs are exact copies of the formula I am looking for, but they "start" at
different points C.
For a great explanation about how C is relates to anti-differntiation in terms of position and velocity. Check out this video from my Mooculus course. I liked this video not only because Dr. Fowler
seemed to have had had too much coffee, but because his explanation incorporates the steps to solve an anti-differentiation equation that has "physical" applications.
Anti-differentiation is an important bridge in my road to understand calculus. At least that is the promise that was made by my professor when he introduced the topic. Whether that is the case or
not, I find it fascinating that having information about a function, I can derive other functions that are related, and give me additional information about the original one.
Let me know what you think in the comments.
A few weeks ago I was introduced to the chain rule in my Coursera/Mooculus course. And I found it to be one of those concepts that is straight forward to understand but hard to our into practice.
The chain rule deals with composition of functions. In other words, it deals with the derivative of functions within functions.
The best example I can think of to explain this is,the following: Imagine you are a salesperson whose job is to call customers, set up an appointment for a consult, give them a sales presentation and
close a sale.
That sale depends on how many presentations are given, which are dependent of how many appointments were made, which are in turn dependent on how many calls were made. If you had a formula that
described this process and you wanted to know how do a change in calls affect sales you might need to use the chain rule.
Let's imagine that such a formula exists. We will use the following variables for it: Sale (S), Presentation (P), Appointments (A), and Calls will be our (x). The formula is S(x)= P(A(x)), in order
to know how changes in x affect S(x) we will need to differentiate the function with using the chain rule:
S'(x) = P'(A(x))(A'(x)) which is the same as saying we are taking the derivative of the outside function evaluated at the inside option and we multiply that with the derivative if the inside option.
Let's imagine than in the example above these formulas can be substituted for S(x)= √(3/4x). How will a change in x affect S(x)?
S'(x)= [1/2(3/4x)^-1/2](3/4) and simplifying we get:
S'(x)= 3
Let's imagine the sales person makes 100 calls a day according to the original formula he will get approximately 8.66 sales. What if the sales person makes 50 extra calls by what amount would sales
S'(150)= 3
According to the derivative sales would change .03535 times multiplied by the 50 extra calls which would yield approximately 1.76 extra sales. Now that's the way I understood it. But, since I might
be wrong, here are five videos explaining the chain rule.
Professor Jim Fowler produced this amazing video explaining the concept. It is 10 minutes long but for someone strugling to understand what the chain rule is and how it works, its worth the time.
Here is the chain rule introduction by Khan Academy.
I liked this example from That Tutor Guy
Another straightforward example from justmathtutoring.com
This one is from the IntegralCalc channel in YouTube
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All About Quantity of cement sand and aggregate for1500 sq ft slab
When it comes to constructing a solid and sturdy slab, the quantity of cement, sand, and aggregate used plays a crucial role. These materials are the building blocks of any concrete structure, and
achieving the right proportion is essential for a successful project. In this article, we will delve into the specifics of determining the quantity of cement, sand, and aggregate required for a 1500
sq ft slab, taking into account various factors such as type of construction, strength of the concrete, and more. So whether you’re a DIY enthusiast or a professional contractor, read on to learn all
about calculating the optimal amount of cement, sand, and aggregate for your next project.
Quantity of cement sand and aggregate for1500 sq ft slab
In order to calculate the quantity of cement, sand, and aggregate required for a 1500 sq ft slab, we first need to determine the thickness of the slab.
Assuming a slab thickness of 4 inches (0.33 feet), the total volume of concrete required will be:
Volume = Area x Thickness
= 1500 sq ft x 0.33 ft
= 500 cubic feet
Now, we need to take into account the ratio of cement, sand, and aggregate used in concrete. The most commonly used ratio is 1:2:4 (1 part cement : 2 parts sand : 4 parts aggregate).
Therefore, the quantity of cement required will be:
Cement = (1/7) x Volume
= (1/7) x 500
= 71.43 cubic feet
Next, we calculate the quantity of sand required:
Sand = (2/7) x Volume
= (2/7) x 500
= 142.86 cubic feet
Lastly, we calculate the quantity of aggregate required:
Aggregate = (4/7) x Volume
= (4/7) x 500
= 285.72 cubic feet
These quantities can also be converted into cubic meters by dividing them by 35.315 (1 cubic feet = 0.0283 cubic meters).
Therefore, the quantity of cement, sand, and aggregate required for a 1500 sq ft slab with a thickness of 4 inches will be approximately:
– Cement – 2.026 cubic meters (71.43/35.315)
– Sand – 4.043 cubic meters (142.86/35.315)
– Aggregate – 8.105 cubic meters (285.72/35.315)
It is important to note that these quantities are approximate and may vary depending on the quality and density of the materials used. It is always recommended to add a small percentage of extra
material to account for any wastage during the construction process.
In conclusion, for a 1500 sq ft slab with a thickness of 4 inches, approximately 71.43 cubic feet of cement, 142.86 cubic feet of sand, and 285.72 cubic feet of aggregate will be required. Proper
calculation and estimation of these quantities are crucial for a successful and cost-effective construction project.
What is M20 mix grade of Concrete
M20 mix grade of Concrete is a type of concrete grade that falls under the category of medium-strength concrete. It is commonly used in construction projects where a moderate amount of strength is
The term M20 is derived from the mix ratio of cement, sand, and aggregate used in the concrete mix. It means that M20 grade concrete has a mix ratio of 1 part cement, 1.5 parts sand, and 3 parts
The strength of concrete is measured in megapascals (MPa) after 28 days of curing. The compressive strength of M20 concrete is usually around 20 MPa, which makes it suitable for a wide range of
applications such as foundations, footing, slabs, and beams.
The components used in M20 grade concrete are Portland cement, fine and coarse aggregates, and water. Portland cement is the main binder used in concrete and is responsible for its strength and
durability. The fine and coarse aggregates provide volume and stability to the concrete mix.
The water-cement ratio (w/c ratio) is an essential factor in the strength and durability of concrete. In M20 grade concrete, the ideal w/c ratio should be 0.5. This means that for every 1 kg of
cement, 0.5 kg of water is added.
The use of admixtures in M20 grade concrete is common to enhance certain properties such as workability, durability, and setting time. Common admixtures used in M20 concrete include plasticizers,
superplasticizers, and air-entraining agents.
M20 grade concrete can be produced using different methods such as hand mixing, machine mixing, and ready-mix concrete. The most commonly used method is machine mixing as it ensures a uniform and
consistent mix.
It is essential to follow proper curing techniques for M20 grade concrete to achieve its full strength potential. Curing involves keeping the concrete moist and at a controlled temperature for at
least 28 days. It helps prevent shrinkage, cracking, and ensures the concrete reaches its desired strength.
In conclusion, M20 mix grade of Concrete is a commonly used grade for medium-strength concrete in construction. Its mix ratio, components, and curing techniques play a crucial role in its strength
and durability. As a civil engineer, understanding the properties and applications of M20 grade concrete is essential for ensuring the safety and stability of any construction project.
Construction cost of RCC slab in India
The cost of constructing a reinforced cement concrete (RCC) slab in India can vary depending on a number of factors such as the size of the slab, type of reinforcement, local market rates, and labor
costs. However, on average, it can range from Rs. 2000 to Rs. 2500 per square meter for a basic RCC slab.
Here is a breakdown of the cost of constructing an RCC slab in India:
1. Material Cost: The main materials required for constructing an RCC slab are cement, sand, aggregates, and steel reinforcement. The cost of these materials can vary depending on the quality and
location. On average, the material cost can range from Rs. 800 to Rs. 1000 per square meter.
2. Labor Cost: The cost of labor can vary depending on the location and the skill level of the workers. In India, the labor cost for constructing an RCC slab can range from Rs. 300 to Rs. 500 per
square meter.
3. Shuttering and Formwork: Shuttering and formwork are essential for providing support and molding the concrete slab. The cost of shuttering and formwork can range from Rs. 100 to Rs. 200 per square
4. Reinforcement Cost: The cost of steel reinforcement used in the RCC slab can vary depending on the type of reinforcement and the quantity required. On average, it can range from Rs. 400 to Rs. 500
per square meter.
5. Excavation and Foundation Cost: The excavation and foundation costs are additional expenses that are required for constructing an RCC slab. The cost of excavation can range from Rs. 200 to Rs. 300
per square meter, while the foundation cost can range from Rs. 500 to Rs. 700 per square meter.
6. Additional Costs: Other factors such as the cost of transportation, taxes, and overheads may also add to the overall cost of constructing an RCC slab. These costs can vary and are usually
calculated as a percentage of the total material and labor costs.
In addition to these costs, the type of reinforcement used in the RCC slab can also have an impact on the overall cost. For example, using a lower-grade steel reinforcement can reduce the cost, but
it may compromise the strength and durability of the slab. Similarly, using precast concrete slabs instead of casting on site can also affect the cost.
Overall, the cost of constructing an RCC slab in India can vary depending on various factors. It is essential to consider quality and safety while also keeping an eye on the budget. Hiring an
experienced and reputable contractor can help ensure that the construction cost is reasonable and the RCC slab is built to meet all safety and quality standards.
How to calculate Steel for RCC slab
Steel is an essential component in the construction of reinforced concrete (RCC) structures, including slabs. In RCC slabs, steel reinforcement is used to resist tension and distribute load evenly,
making the structure more durable and stable. In this article, we will discuss how to calculate the amount of steel required for an RCC slab.
1. Understand the design requirements:
Before starting the calculation, it is crucial to thoroughly understand the design requirements of the slab. This includes the type of slab (one-way or two-way), the dimensions, the loading
condition, and the design code to be followed.
2. Determine the grade of steel:
The first step is to determine the grade of steel to be used. The standard grades for steel reinforcement are Fe415, Fe500, and Fe550, where the number represents the minimum yield strength of the
steel in MPa.
3. Calculate the design bending moment:
The design bending moment is a crucial parameter in the calculation of steel for an RCC slab. It is the maximum bending moment that the slab can resist without cracking. The design bending moment can
be calculated using the formula:
M = (q x l^2) / 8
Where M is the design bending moment, q is the uniformly distributed load, and l is the span of the slab.
4. Determine the modular ratio:
The modular ratio is the ratio of the elastic modulus of concrete to the elastic modulus of steel. It is used to convert the design bending moment into the required reinforcement area. The value of
the modular ratio depends on the grade of concrete and steel and can be obtained from design codes.
5. Find the main reinforcement:
The main reinforcement is the steel bars placed along the longer direction of the slab, also known as the longer span. The spacing of these bars depends on several factors such as slab thickness,
grade of steel, and design code. The required area of reinforcement for main bars can be calculated using the formula:
Ast = (M x 10^6) / (0.87 x fy x d)
Where Ast is the required area of steel, M is the design bending moment, fy is the yield strength of steel, and d is the effective depth of the slab.
6. Calculate the distribution reinforcement:
Distribution reinforcement is used to resist the shear forces in the slab, which are higher along the shorter direction. Its spacing is usually less than that of the main reinforcement and is placed
perpendicular to the main bars. The required area of steel for distribution bars can be calculated using the formula:
Asd = (V x 10^3) / (0.87 x fy x d)
Where Asd is the required area of distribution steel, V is the design shear force, fy is the yield strength of steel, and d is the effective depth of the slab.
7. Add the main reinforcement and distribution reinforcement:
The total reinforcement area is the sum of the required steel areas for main and distribution bars, which can be calculated as Ast + Asd.
8. Calculate the number of steel bars:
To determine the number of bars, divide the total reinforcement area by the area of a single bar. Depending on the design requirements, the diameter of the steel bars can also be varied.
9. Check for minimum reinforcement:
Design codes specify the minimum reinforcement required in the slab. This value is usually a percentage of the total cross-sectional area of the slab. Check whether the calculated reinforcement is
greater than the minimum reinforcement required. If not,
How much quantity of sand required for Roof cast of rcc slab
The quantity of sand required for casting a RCC (Reinforced Cement Concrete) roof slab can vary depending on the design and thickness of the slab. Generally, the minimum amount of sand required for a
roof cast of RCC slab is about 0.5 cubic meters for every 100 square meters of slab area.
To calculate the amount of sand required for a specific roof slab, we need to consider the following factors:
1. Thickness of the slab: The thickness of the slab is an important factor in determining the quantity of sand required. A thicker slab will require more sand to achieve the desired strength and
2. Design of the slab: The design of the slab also plays a crucial role in determining the quantity of sand required. The design includes the type of reinforcement, spacing and arrangement of bars,
and load-bearing capacity.
3. Type of sand: The type of sand used for the roof cast also affects the quantity required. Generally, river sand is preferred as it has a better bonding strength and is free from impurities.
4. Wastage factor: A wastage factor of 5-7% should be considered for the quantity of sand required. This accounts for spillage, uneven distribution, and any additional requirements during the casting
Based on these factors, a standard calculation can be used to determine the approximate amount of sand required for a specific roof cast.
For example, if the thickness of the slab is 0.15 meters and the area of the slab is 100 square meters, the total volume of the slab will be:
100 x 0.15 = 15 cubic meters.
Assuming a wastage factor of 5%, the total volume of sand required for a 100-square-meter roof cast will be:
15 x 0.05= 0.75 cubic meters.
Therefore, at least 0.75 cubic meters of sand is required for a 100-square-meter roof cast of RCC slab. However, this quantity may vary depending on other factors as mentioned above.
It is crucial to ensure the proper quality and quantity of sand is used for the construction of a roof cast of RCC slab. Using the wrong type or inadequate amount of sand can compromise the
structural integrity and stability of the slab. It is advisable to consult a structural engineer or a construction professional for accurate calculations and recommendations for the sand quantity
required for a specific roof cast.
How much cement is required for roof cast of 1500 square feet RCC slab
To calculate the amount of cement required for a 1500 square feet RCC (Reinforced Cement Concrete) slab, we first need to determine the thickness of the slab. The thickness of a slab can vary
depending on the design and load requirements, but typically, a residential building may have a slab thickness of 4-6 inches.
Assuming a thickness of 6 inches, the volume of concrete required for the roof cast can be calculated as follows:
Volume of concrete = Area of slab x Thickness
= 1500 sq. ft. x (6 inches/12) ft.
= 750 cubic feet
Next, we need to determine the proportion of cement in the concrete mix, which is generally expressed as a ratio of cement to other materials such as sand and aggregate. The most common mix ratio
used for residential buildings is 1:2:4, which means 1 part cement, 2 parts sand, and 4 parts aggregate.
Therefore, the amount of cement required for the roof cast can be calculated as:
Mass of cement = (Proportion of cement / Total proportion) x Volume of concrete x Density of cement
= (1/7) x 750 cubic feet x 1440 kg/cubic meter
= 154285.71 kg
To convert the mass of cement into bags, we need to divide it by the weight of one bag of cement. As 1 bag of cement weighs 50 kg, the number of bags required for the roof cast of 1500 square feet
would be:
Number of cement bags = Mass of cement / Weight of one bag of cement
= 154285.71 kg / 50 kg
= 3085.71 bags
Hence, approximately 3086 bags of cement would be required for the roof cast of 1500 square feet RCC slab.
It is important to note that this is an approximate calculation and may vary depending on the quality of materials and the concrete mix design used. It is always recommended to consult a structural
engineer for a more accurate estimation. Additionally, it is essential to follow proper construction guidelines and techniques to ensure a strong and durable roof cast.
How much course aggregate is required for 1500 RCC roof slab
A reinforced concrete (RCC) roof slab is a structural element used in buildings to provide a stable and durable cover over the top of a space. It is commonly used in residential, commercial, and
industrial buildings. The course aggregate, also known as coarse aggregate, is an essential component in the construction of an RCC roof slab as it provides strength and stability to the structure.
The amount of course aggregate required for a 1500 RCC roof slab depends on various factors such as the size of the slab, the concrete mix design, and the type of aggregate used. However, the basic
principles and guidelines of the American Concrete Institute (ACI) can be followed to estimate the required quantity of course aggregate for the slab.
The first step is to determine the volume of the slab. For a 1500 square feet RCC roof slab, the volume can be calculated by multiplying the length by the width by the thickness. Assuming a standard
thickness of 6 inches, the volume of the slab would be 1500 x 0.5 = 750 cubic feet.
Next, the amount of concrete needed is calculated by multiplying the volume of the slab by the concrete density. The density of concrete varies depending on the composition, but an average value of
150 pounds per cubic foot can be used. Therefore, the amount of concrete needed for the 1500 RCC roof slab would be 750 x 150 = 112,500 pounds.
Now, to calculate the amount of course aggregate required for the slab, the total volume of concrete needs to be multiplied by the percentage of aggregate in the mix. The ACI recommends a minimum of
60% to 75% of course aggregate to be used in a concrete mix. Let’s consider a mix with 60% course aggregate. The amount of course aggregate required would be 112,500 x 0.6 = 67,500 pounds.
In addition to the guidelines from the ACI, it is important to consider the grading or particle size distribution of the course aggregate. The grading of the aggregate affects the workability,
strength, and durability of the concrete. Too much or too little of certain particle sizes can lead to a weaker concrete mix. Therefore, it is essential to ensure that the course aggregate used for
the RCC roof slab is well-graded and within the recommended limits.
In conclusion, for a 1500 square feet RCC roof slab, approximately 67,500 pounds of the course aggregate would be required. However, it is important to note that this is an estimate, and the actual
amount required may vary depending on the specific project requirements and design. It is always recommended to consult with a qualified structural engineer to determine the exact amount of course
aggregate required for a specific RCC roof slab.
Total cost of material required for 1500 square feet RCC slab
To determine the total cost of materials required for a 1500 square feet Reinforced Concrete (RCC) slab, we need to consider various factors such as the type and quantity of materials, quality of
materials, location, and market prices.
The most common materials used in constructing an RCC slab include cement, coarse aggregates (such as gravel or crushed stones), fine aggregates (like sand), and steel reinforcement bars. Other
materials may include formwork, curing compounds, waterproofing materials, and other construction supplies.
The total cost of materials for an RCC slab can be calculated by estimating the cost of each material and adding them together. Let us consider the average current market prices for materials and
assume a basic quality of materials for this calculation.
1. Cement: The amount of cement needed for a 1500 square feet RCC slab will depend on the thickness of the slab. For a 4-inch thick slab, approximately 7.5 bags (94lbs each) of cement will be
required. The current average market price for cement is around $10 per bag. So, the total cost of cement will be around $75.
2. Coarse Aggregates: The coarse aggregates, which make up the bulk of the concrete, are usually sold by the cubic yard. For a 4-inch thick slab, approximately 1.5 cubic yards of aggregates will be
required. At an average market price of $40 per cubic yard, the total cost of coarse aggregates will be around $60.
3. Fine Aggregates: The amount of fine aggregates (sand) needed for a 1500 square feet RCC slab will depend on the ratio of cement to sand used in the concrete mix. Assuming a 1:2:4 mix ratio (1 part
cement, 2 parts sand, 4 parts coarse aggregates), around 1.5 cubic yards of sand will be required. At an average market price of $30 per cubic yard, the total cost of fine aggregates will be around
4. Steel Reinforcement: The steel reinforcement bars are measured in linear feet (lf). The amount of steel reinforcement required for an RCC slab will depend on the size, spacing, and design of the
bars. For a basic 6 inches grid spacing, around 4000 lf of steel bars (assuming 4 bars of #4 steel bars per foot) will be needed. The current average market price for #4 steel bars is around $0.50
per linear foot. So, the total cost of steel reinforcement will be approximately $2,000.
5. Formwork: The cost of formwork will depend on the complexity of the slab design. For a simple rectangular 1500 square feet slab, it can vary from $1,000 to $2,000.
6. Curing Compounds and Waterproofing Materials: These materials are not usually included in the concrete mix but are needed for proper curing and waterproofing of the slab. The total cost of these
materials can vary from $500 to $1000.
Considering all the above factors, the total cost of materials required for a 1500 square feet RCC slab can range from $3,680 to $4,180. However, it is important to note that the cost may vary
depending on the location, quality of materials, and other market factors.
In conclusion, the total cost of materials required for a 1500 square feet RCC slab can be estimated by considering the type and quantity of materials
How much does a 1500 square foot concrete slab cost?
Building a concrete slab is a common method used in construction for various projects such as garages, patios, and foundations. If you are planning to construct a 1500 square foot concrete slab, one
of the essential things to consider is the cost. The total cost of a concrete slab will depend on various factors, including the materials used, labor, and location.
The materials used to build a concrete slab include concrete, reinforcement materials, and formwork. The cost of concrete varies depending on the type and quality. On average, the cost of concrete
ranges from $3 to $10 per square foot. For a 1500 square foot slab, you will need approximately 20 cubic yards of concrete, which can cost between $1200 to $4000.
Reinforcement materials, such as rebar and wire mesh, are used to strengthen the concrete and prevent cracking. The cost of reinforcement materials can be around $0.15 to $0.75 per square foot. For a
1500 square foot slab, you may need around 200 to 300 pounds of reinforcement materials, which can cost approximately $100 to $225.
Formwork is necessary to hold the concrete in place while it sets and hardens. The cost of formwork will depend on the type and size of formwork used. On average, the cost of formwork can range from
$0.50 to $2 per square foot. For a 1500 square foot slab, you may need around $750 to $3000 for formwork materials.
The cost of labor will also affect the total cost of a 1500 square foot concrete slab. The labor cost can vary depending on the complexity of the project, the location, and the experience of the
workers. On average, the labor cost for pouring and finishing a concrete slab can range from $6 to $10 per square foot. For a 1500 square foot slab, the labor cost can be between $9,000 to $15,000.
The location of the project can also affect the cost of a concrete slab. Factors such as accessibility, soil condition, and local building codes can impact the overall cost. In some areas, there may
be additional fees or permits required, which can add to the total cost.
Other Considerations
In addition to the materials, labor, and location, there may be other expenses to consider when calculating the cost of a 1500 square foot concrete slab. These may include excavation and site
preparation, grading and leveling, and any additional finishing touches such as stamping or coloring the concrete.
On average, the total cost of a 1500 square foot concrete slab can range from $12,000 to $25,000. It is essential to get quotes from different contractors and carefully consider the materials and
labor costs to ensure you get the best value for your money.
In conclusion, the cost of a 1500 square foot concrete slab can vary depending on several factors, including materials, labor, location, and additional expenses. It is crucial to do your research and
carefully consider all these factors to get an accurate estimate of the cost before starting the project.
In conclusion, determining the quantity of cement, sand, and aggregate for a 1500 sq ft slab is a crucial step in the construction process. It is essential to accurately calculate the proportions to
ensure a strong and durable slab. By following the steps mentioned in this article, one can easily calculate the required quantity of materials for their project. It is important to note that factors
such as the type of concrete mix, the thickness of the slab, and the available construction tools can affect the calculations. Therefore, it is advisable to consult a professional engineer or
contractor for a more accurate estimation. Ultimately, understanding the quantity of cement, sand, and aggregate for a 1500 sq ft slab will lead to a successful and cost-effective construction
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How to use Average Function and Different Ways to Use It | Easy Excel Tips | Excel Tutorial | Free Excel Help | Excel IF | Easy Excel No 1 Excel tutorial on the internet
Home Excel Functions Average Function and Different Ways to Use It
Average Function and Different Ways to Use It
Average Function and Different Ways to Use It
In this lesson you can learn how to use Average function. The average function in Excel is a feature that would use the arguments to return with average of those arguments. However, it would only use
the cells that are containing numbers to calculate the average. The advantage of this function is that it can enter addresses of individual cells and also the whole range. Average is also available
in the status bar. Parameters (Syntax) The syntax of average has one obligatory argument, while others are optional.The syntax looks like this: =AVERAGE(number1; [number2];…).Number1: This is a
required argument, and could be cell reference, a range that you would like to have the average of.
Number2: This is optional. It could also contain reference, range, and these ones reach maximum of 255.
Take a look at the picture above.
Function AVERAGE doesn’t care about empty cells and text in the cell. In the first and third table function AVERAGE there is a different average than in the second table.
It is very easy to make a mistake. Remember about zeros in cells.
Example 1 Average Between Two Values
B1 and B2 are the values between which you want to calculate average.
This is an array formula so accept it by CTRL + SHIFT + ENTER.
Example 2 To avoid errors
Use this formula when you can haven’t got any data in your worksheet.
=IF(ISERROR(AVERAGE(A1:A5)),”No Data”,AVERAGE(A1:A5))
If your data in A1:A5 are missing, Excel will show “No data”. If it is ok with your data there will be average calculated.
Example 3 Average of 10 largest values
This formula gives you the average of 10 largest values in A1:A50 range as the result.
This is an array formula so accept it by CTRL + SHIFT + ENTER.
Example 4 Average with some data missing
How to calculate values average in case of zeros or missing data. Use that formula:
This is an array formula so accept it by CTRL + SHIFT + ENTER.
This formula doesn’t count blanks and zeros. In this case average is (2+5+6+3):4. Formula doesn’t calculate value in A2 cell. There will be the same when A2 will be blank or text.
Example 5: Simple Average Function Usage
For the past six months, we have been making sales, and we have decided to find out how much we have averagely made under those six months, so we could find a perfectly suitable ways of closing. This
is because we want to reach a goal.
Example 6: Customized Average Usage
The same data as the previous one, but this time I am looking at a specific period. It becomes necessary to know when the average was highest, which helps the comprehension of company’s performance.
Example 7: Average Functions Multiplied
This is the example that allows us to multiply the average of both sales and products. However, this only works under a whole new scenario. It addresses the issues that the first column is the price,
while the other column is the quantity of products we sold.
Example 8: AVERAGE Evaluation
We have our business going, but we would like to make sure that we are having the appropriate value, because it has been determined that at any given time, the average should be above the expenses.
We are now going to use the average function, in consideration to evaluating the financial performance of the company on average level.
Example 9: Double Average Evaluation
This is the perfect example where as we are solely focusing on our average performance when making the evaluation. The business has the very same data as previous example. But, this time the
situation is very different, because we are solely evaluating average on both columns.
Example 10: Average with Text
We have made a mistake of labelling all our data with text, and we really need to do the average of our sales.
Example 11: Evaluating with Text
With the same situation as the previous one, it is necessary to use the text to find out the average financial performance.
Example 12: Double Evaluation with Text
This is using the text to do another double evaluation. We are going to evaluate this in form of average.
Example 13: Another Spreadsheet
Just as the previous example, but this time the information is in another spreadsheet.
Example 14: Average from Multiple Files
The business is three years old, we have our data spread in three different documents, including our current file. Our business is now in need of knowledge, and we would like to know how the business
is doing on an average level. We are therefore taking data from those other two files, in other to get the answer that we are looking for.
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Stellar: Is it worth investing in a clone of the popular Ripple now?
The crypto project Stellar is an open-source payment network. Now, 7 years after the start of this blockchain project, its currency – Stellar Lumens (XLM) – steadily occupies a place in the top of
the most promising and popular altcoins. Is it worth investing in?
The history of Stellar began in 2014 when its founder – Jed McCaleb – left the Ripple platform, on which he had been working since 2011, for a new project. More precisely, he decided to do the same
thing, but with a different vector.
The thing is that in its concept Stellar is almost identical to Ripple – both systems:
• Do not support mining: unlike bitcoin, their tokens cannot be mined, all coins are immediately put into circulation by the system itself.
• Provide virtually free instant transactions.
• Designed to allow users to buy and sell different currencies at minimal cost.
The key difference is the audience: while Ripple is aimed at large banks and consortiums, Stellar is designed for individuals and businesses – including those from developing countries. In essence,
McCaleb took the off-the-shelf Ripple model and then created a more affordable clone of it.
This decision looks ambiguous: on the one hand, Ripple is incomparably superior to Stellar in scale, but on the other hand, Stellar does not attract as much attention from regulators. While Ripple is
suing the SEC, Stellar is developing, and that is a big plus.
In addition, Ripple’s successes play into the hands of its junior counterpart, confirming the system’s reliability and potential. Practice shows that market players care about this:
• In the first year of its existence, Stellar received 3 million users, and the capitalization of the project came close to $15 million.
• In 2015, the company released integration with Vumi, the popular messenger of the Praekelt Foundation (South Africa).
• In 2016, integration with Deloitte took place, and then Coins.ph, ICICI Indian Bank and several other large companies joined the Stellar payment network.
• In 2017, KickEx and IBM, interested in facilitating cross-border transactions in the South Pacific region, became partners of the project. During this period, Stellar’s token, Lumens, ranked 13th
by market capitalization – 30 banks confirmed their cooperation with the platform.
And so on. By now, Stellar Lumens is among the top 30 altcoins, and does so quite deservedly. At the moment, the market capitalization of XLM is $8.3 billion, and the daily trading volume is around
$700 million – more than impressive figures. But where is the demand coming from?
If Lumens can’t be mined, why are they needed?
The demand for XLM token is caused by the peculiarities of the marketplace itself. In fact, it is Lumens that allows Stellar users to trade currencies with minimal fees, tokens are indispensable for
cross-border transactions.
For example, you need to transfer money from China (yuan) to the United States (U.S. dollars). If you use a regular bank transfer with currency conversion at the exchange rate, the commission can be
up to 7%. In the case of transfer via Stellar you technically perform two operations:
1. First, you buy Lumens for yuan;
2. Then you buy dollars for those Lumens, which go to the recipient (you or another person).
The final commission will be less than a percentage, and the transaction itself will take seconds. Direct savings.
With its own tokens, the platform does not allocate any currency as the main one. Stellar financial system includes money from most countries of the world, their availability to users – in the form
of digital tokens, the rate of which is pegged to the real currency – is exactly the same. Transparency, decentralization, independence, reliability and universality: Stellar’s success is based on
the fact that this financial system is extremely convenient for all its participants. And this is a very powerful foundation.
Is this a good time to invest in Stellar?
The position of the coin is very strong at the moment – since the beginning of the year Lumens grew almost three times: if on January 1 the coin was trading at $0.132, at the moment of writing this
article the value of altcoin is $0.34. In May, at its peak it was offered $0.73 for one XLM.
However, such a sharp upsurge is unlikely to happen again in the near future. Moreover, in the summer of 2021 the rate of XLM has fallen almost to $0.2 per coin. And though since that time Stellar
Lumen strengthened almost 1.5 times, the predictions about the rate of $0.6-0.75 by the end of 2021 are obviously a pipe dream. As boring as it sounds, a rate of $0.33-0.39 per coin is most likely in
2022. On the other hand, it is hardly a good idea to invest in cryptocurrency at the peak of its price. Perhaps, given the relatively low rate of XLM, there is still a lucrative opportunity to buy it
The next question is whether there is currently a possibility of a significant loss of funds when investing in XLM? From our perspective, the likelihood of that happening is minimal. If we study the
history of the coin’s prices, it becomes obvious: despite the ups and downs, there was not a single moment in the history of XLM when the token became unclaimed. This means that against the
background of the general strengthening of cryptocurrency prices, which will resume sooner or later, XLM has a chance to grow.
Another important factor affecting the price of XLM is the conclusion of cooperation agreements with major players by Stellar. If there are media reports about the likelihood of a deal like the one
with IBM, a sharp rise in Lumen is very likely. Therefore, when investing in Stellar, it is important to “keep your hand on the pulse”.
The same goes for investing in XRP, though. Right now, the fate of this asset largely depends on the outcome of the case with the SEC. If it ends up in Ripple’s favor, its coin has a good chance of
increasing in value.
That is the nature of all crypto assets – they rise sharply on a wave of hype and positive news, and collapse just as sharply when the wind blows in the other direction.
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EE 396: Lecture 4 - P.PDFKUL.COM
EE 396: Lecture 4 Ganesh Sundaramoorthi Feb. 19, 2011 In the last lectures, we constructed by using the additive noise model and MAP estimation, the following energy to be minimized : Z Z 2 (1) E(u)
= − log p(u|I) = (I(x) − u(x)) dx + α |∇u(x)|2 dx, Ω
where E : U → R, I : Ω → R is the observed or measured image, and α > 0. The u ∈ U that minimizes E is our estimate of the denoised image. We also derived the Euler-Lagrange equations for the energy
above, and they are ( u(x) − α∆u(x) = I(x) x ∈ int(Ω) . (2) ∂u x ∈ ∂Ω ∂n (x) = 0 Euler-Lagrange equations are in general necessary conditions for a local minimum/maximum. However, last time we have
shown that E is convex and thus, we know that if there exists a solution to the EulerLagrange equation then it must be a global minimizer. Today, we are going to understand if a solution to the
Euler-Lagrange equations exists, and if so, we are going to develop algorithms to numerically solve (2).
Analytic Solution to the Euler-Lagrange Equation
The left hand side of (2) is a linear operator acting on u: indeed, we have that u − α∆u = (Id − α∆)u
where Id : U → U is the identity map, and so if u1 , u2 ∈ U, then (Id − α∆)(u1 + u2 ) = Id(u1 + u2 ) − α∆(u1 + u2 ) = u1 + u2 − α u1 − α
n X ∂ 2 u1 i=1
(x) + u2 − α
n X ∂ 2 u2 i=1
n X ∂ 2 (u1 + u2 ) i=1
(x) =
(x) = (Id − α∆)u1 + (Id − α∆)u2 ,
which implies linearity of the operator. Let us define for convenience the (linear) operator L : U → U:
Then we have the PDE:
L = Id − α∆.
( (Lu)(x) = I(x) x ∈ int(Ω) ∂u x ∈ ∂Ω ∂n = 0
Let us now try to solve the above PDE. We define a kernel on the Ω; the kernel is a map K : Ω × Ω → R. The arguments of K are going to be x and y, respectively. Let us now apply the Divergence
Theorem that we have seen in the last lecture to the vector field U (y) = K(x, y)∇y u(y) − ∇y K(x, y)u(y) defined for each x ∈ Ω : Z Z div [K(x, y)∇y u(y) − ∇y K(x, y)u(y)] dy; (K(x, y)∇y u(y) − ∇y K
(x, y)u(y)) · N (y) dS(y) = Ω
(6) note that K and u must be smooth, i.e., have continuous partials, for the above Divergence Theorem to hold. Also note that the above divergence is taken with respect to y. We see that div [K(x,
y)∇y u(y)] = ∇y K(x, y) · ∇y u(y) + K(x, y)∆y u(y)
div [∇y K(x, y)u(y)] = ∆y K(x, y)u(y) + ∇y K(x, y) · ∇y u(y).
Thus (6) becomes Z Z ∂u ∂K K(x, y) (y) − (x, y)u(y) dS(y) = (K(x, y)∆y u(y) − ∆y K(x, y)u(y)) dy. (9) ∂n ∂n ∂Ω Ω R R Now multiplying the above equation by α and adding 0 = Ω K(x, y)u(y) dy − Ω K(x,
y)u(y) dy to the right hand side, we have: ∂K ∂u (x, y)u(y) dS(y) = α K(x, y) (y) − ∂n ∂n ∂Ω Z [K(x, y)(u(y) − α∆u(y)) − u(y)(K(x, y) − α∆y K(x, y))] dy = Ω Z [K(x, y)Lu(y) − u(y)Ly K(x, y)] dy (10)
Let us be a little informal (but later we will see that this informal argument can be justified) : suppose that we can find a kernel K such that the following holds : ( Ly K(x, y) = δ(x − y) for all
x, y ∈ int(Ω) (11) ∂K for x ∈ int(Ω), y ∈ ∂Ω ∂n (x, y) = 0 where δ indicates the (two-dimensional) Dirac delta function. Also, suppose u solves (5). Then (10) becomes Z 0= K(x, y)I(y) dy − u(x); (12)
that is
Z u(x) =
K(x, y)I(y) dy;
K is known as the Green’s function for L, and we sometimes write (with abuse of notation) K = L−1 . Remark 1. Recall from signal processing, that if we have a linear system that is also
shift-invariant (sometimes known as time-invariant), then you may recall that the output of the system is the input signal convolved
with the impulse response. In other words, if x : R → R is the input signal, and if h : R → R is the impulse response of a linear shift invariant system, S, i.e., S(h)(t) = δ(t), then Z h(t − τ )x(τ
) dτ = (h ∗ x)(t). (14) S(x)(t) = R
We can verify this by writing
Z x(t) =
x(τ )δ(t − τ ) dτ ;
now applying S to both sides of the above equation and using linearity and shift-invariance of S, we have Z Z x(τ )h(t − τ ) dτ. (16) x(τ )Sδ(t − τ ) dτ = Sx(t) = R
Note the resemblance of the formula to (13). Because in general L is not shift-invariant (since Ω is in general not shift-invariant), the kernel will not depend on the difference x − y and so it
cannot be in general written as a convolution. In the case Ω = R2 , though, L is shift-invariant and thus, K depends on x − y, and the result is that u is the convolution of K with the image! Remark
2. You may be wondering why we studied all this mathematics of the Bayesian approach, MAP estimation, calculus of variations, and PDE, and then in the end we arrive at result that is well known in
signal processing and to all of you : to reduce the noise, convolve the image with a low-pass filter! Why the need for talking about models, noise and prior statistics, etc... ? The answer is that
the whole framework we have developed now allows us to make better assumptions about the image formation process, the noise statistics, and the prior class of denoised images, and then we have the
techniques already developed will allow us to develop algorithms in the case of different assumptions. As you can imagine, the assumptions we have already made are not very good! As we shall soon see
in the subsequent lectures, when we choose different priors on the class of denoised images or the different noise statistics, etc.. we will see that we do not get linear systems, and in particular,
the methods of filtering developed in signal processing are simply inadequate! We now need to calculate the Green’s function for the operator L defined above. Unfortunately, there is no closed form
solution for K (at least that I know of) for general Ω or even when Ω = R2 . To get some idea of how these kernels look like, we derive the Green’s function for the case when α is large and Ω = Rn .
Note that when α is large then Lu(x) ≈ −α∆u(x). Because, Ω = Rn is shift invariant, we know that the kernel will depend only on the difference x − y. Therefore, we solve −α∆K(x) = δ(x).
Note that ∆ is rotationally invariant, i.e., if K solves the above then K ◦R(x) = K(Rx) (where RT R = Idn is the n × n identitiy matrix) is also a solution to the above. Therefore, K should be
radially symmetric, i.e., K(x) = ν(|x|) for some function ν : R → R. Computing the gradient of K, we have that ∇K(x) = ∇ν(|x|) = ν(|x|) and
x |x|
X n n X ∂ xi x2 |x| − x2i /|x| n−1 0 ∆K(x) = ν(|x|) = ν 00 (|x|) i2 +ν 0 (|x|) = ν 00 (|x|)+ ν (|x|). (19) 2 ∂xi |x| |x| |x| |x| i=1
For x 6= 0, we solve ∆K(x) = 0, that is ν 00 (|x|) +
n−1 0 ν (|x|) = 0 |x|
which is equivalent to n−1 C d log ν 0 (r) = − =⇒ log ν 0 (r) = −(n − 1) log r + log C =⇒ ν 0 (r) = n−1 dr r r where r = |x| and C > 0 is some constant. Integrating again, we find that ( C log r + B
n = 2 ν(r) = C +B n ≥ 2. rn−2 Using the above reasoning, we choose K to be ( 1 log |x| − 2π K(x) = 1
n(n−2)a(n) |x|n−2
n=2 n≥3
where a(n) is the volume of the unit ball B1 (0) = {x ∈ Rn : |x| ≤ 1} in Rn . The above is know as the fundamental solution of the Laplace equation. We now prove the following theorem that is taken
from [1] : Theorem 1. Let I : R2 → R be an image and further assume that I ∈ C 2 (R2 , R) and I and its derivatives are nonzero only on Ω, which bounded and a closed set (we often write I ∈ Cc2 (R2 ,
R)). Then Z 1 u(x) = − K(x − y)I(y) dy (24) α R2 solves −α∆u = I. Proof. In class; see handout! Note that showing the existence of the equations (5) for second order operators requires the uses of
the theory of Sobolev spaces, and that is out of the scope of this course. But, indeed the solution of (2) exists. The kernel formulation of the solution of the PDE is not particularly useful for
numerical implementation since the kernel is typically singular (i.e., it blows up at the origin), and thus we develop numeric solutions directly from the PDE. However, the kernel formulation aids in
our understanding of the solution of the PDE.
References [1] Lawrence C. Evans. Partial Differential Equations, volume 19 of Graduate Studies in Mathematics. American Mathematical Society, 1997.
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A detailed, end-to-end assessment of a quantum algorithm for portfolio optimization, released by Goldman Sachs and AWS | David A. Bader
A detailed, end-to-end assessment of a quantum algorithm for portfolio optimization, released by Goldman Sachs and AWS
AWS Quantum Technologies Blog
To date, the financial services industry has been a pioneer in quantum technology, owing to its plethora of computationally hard use cases and the potential for significant impact from quantum
solutions. Daily operations at financial firms such as banks, insurance companies, and hedge funds require solving well-defined computational problems on underlying data that is constantly changing.
If quantum computers can perform the same calculations faster or with better accuracy than existing solutions, these customers stand to reap significant benefit.
In this post we’ll walk you through some key takeaways from a body of work by scientists from Goldman Sachs and AWS that was published today in the journal PRX Quantum ^1.
A commonly studied use case in quantitative finance is the task of portfolio optimization — consequently, customers are eager to learn what impacts quantum computing can have on this task, if any.
Several distinct quantum algorithms have been proposed for portfolio optimization, each of which leverage quantum effects to provide a theoretical speedup over comparable classical algorithms.
However, it has been difficult to assess whether these theoretical speedups can be realized in practice. Such an assessment requires an end-to-end resource estimation of the quantum algorithm — that
is, a calculation of the total number of qubits and quantum gates (or other relevant metrics) required to solve the actual problem of interest, without sweeping any costs or caveats under the rug.
End-to-end assessments are challenging because certain elements of the quantum algorithm are heuristic, and there are caveats related to the way the quantum computer accesses the data. Thus, it can
be difficult to make apples-to-apples comparisons with existing methods
Despite the challenges, scientists from Goldman Sachs and AWS took on the task of performing an end-to-end resource estimate for a leading quantum approach to portfolio optimization — the quantum
interior point method (QIPM), which we will discuss in this post. You can read the full paper in the journal PRX Quantum, but let’s walk through some of the main takeaways of that work ^1.
How do we determine if a quantum algorithm is practical?
For quantum computers to provide value over classical algorithms, we need to develop practical, end-to-end quantum algorithms, for which we identify the following criteria:
1. The quantum algorithm produces a classical output that allows for benchmarking against classical methods. This requirement rules out some quantum “algorithms” that efficiently produce quantum
states as output, but are much less efficient when forced to turn that quantum state into an answer to the original (classical) computational problem.
2. The quantum algorithm relies on a reasonable input model. In particular, it does not utilize any unspecified black-box subroutines (often called “oracles”) for which the costs are not counted,
nor does it assume fast data access without accounting for the true cost of quantum access to the classical data. For example, several quantum algorithms for Machine Learning were thought to
offer exponential advantages over classical methods until it was pointed out that this was primarily an artifact of unreasonable assumptions about the input model ^2.
3. The quantum algorithm has a plausible case for asymptotic quantum speedup — that is, for sufficiently large instance sizes, the quantum algorithm requires substantially fewer elementary
operations than the classical algorithm. Since classical computers can perform elementary operations at a much faster rate than quantum computers, without an asymptotic speedup, the quantum
computer will always lose.
4. For classically challenging problem sizes, the quantum algorithm can solve the same end-to-end problem in a reasonable run-time. Asymptotic speedups are of no value if the crossover point where
quantum computers outperform classical computers occurs at instance sizes that are too large to be useful to customers.
Portfolio optimization
Portfolio optimization is a well-defined computational task that appears in many fields, but its utility is particularly clear for financial operations. Suppose you have a fixed budget to invest in
the stock market. You’d like your investments to provide predictable growth; in particular, you’d like to construct a portfolio that maximizes the expected return of the portfolio, while minimizing
the risk, that is, the amount by which the return might deviate from its expectation.
The model
Part of the problem is thus determining which stocks are expected to perform well in a given timeframe, based on historical data — we assume this has already been done. If there are n stocks to
invest in, we are provided as input with a length-n vector containing the expected return for each of the stocks, as well as an n x n matrix 𝛴 containing the historical volatilities of each stock and
the covariance of each pair of stocks — the performance of different stocks is correlated, and the optimal portfolio will exploit these correlations to minimize portfolio risk.
Let w be the length-n vector representing the fraction of our budget we allocate to each stock. In general, entries of w can be negative, which represents a short sale. Here, though, we will require
that the portfolio contain only “long” assets — i.e., no short sales: w[i] ≥ 0. The portfolio optimization problem is then formulated as a convex optimization problem:
The parameter q is called the “risk-aversion parameter”, and it is set by the user according to how much risk they are willing to take on to gain additional return.
The first constraint is a total budget constraint. We’ve normalized w so that the total budget available is just 1, but without normalizing this sum would equal the total amount of money available to
The second constraint enforces the “long-only” condition above; if we allow short selling, this can be relaxed. Additional constraints, such as minimum or maximum investments in certain stock
sectors, cardinality constraints (i.e., the maximum allowed number of assets), or transaction costs, can also be incorporated into this framework (see ^3).
The set of portfolios that obey all the constraints is known as the feasible region. The feasible region can be divided into the boundary, where at least one of the inequality constraints holds as an
equality (e.g. w[i] = 0), and the interior, where all inequality constraints are strict (w[i] > 0 for all i).
One widely used classical approach for solving portfolio optimization is to map the convex optimization problem above to a type of optimization problem known as a second-order cone program (SOCP),
and then solve it with an interior point method (IPM). The basic idea of an IPM is to start at some portfolio in the interior of the feasible region and iteratively update it in a way that is
guaranteed to eventually approach the optimal portfolio, which lies on the boundary of the feasible region, as depicted in Figure 1.
Figure 1: A toy example depicting how an Interior Point Method (IPM) solves portfolio optimization. The hexagonal region represents the feasible set of portfolios. The IPM generates a sequence of
points (known as the central path) that follows a path from the interior of the region to the optimal point, which lies on the boundary. The color represents the value of the objective function,
including barrier functions that implement constraints (darker is better). The black dots represent the central path, which is defined to be the optimal point of the objective function, including the
barrier. As the barrier functions are relaxed, the objective function approaches the true objective function, and the central path approaches the optimal solution to the problem on the boundary of
the feasible space.
Performing a single iteration of the IPM requires solving a linear system of equations of size L x L, where (in our formulation), L ≈ 14n and n is the number of stocks in which we could invest. The
output of the IPM is a portfolio w that obeys all the constraints and for which the objective function is within 𝜖 of the optimal value. Many flavors of IPM exist, and developing better classical
IPMs is an active area of research.
Here we focus on the IPMs that are most comparable with quantum algorithms. These classical IPMs solve the PO problem in time scaling as n^3.5 log(1/𝜖). This runtime is efficient in theory (i.e., it
scales polynomially with n) and also in practice (IPMs are a leading method underlying many open-source and commercial solvers). Using IPM-based, off-the-shelf solvers, one can solve the PO problem
on portfolios with thousands of stocks reasonably quickly on standard compute resources. However, the n^3.5 scaling is steep, and pushing the number of assets into the tens of thousands presents an
opportunity for alternative solutions (such as quantum computing) to provide value. In the next section, we explore a proposal to do exactly that.
Quantum interior point methods
Quantum interior point methods (QIPMs) aim to accelerate classical IPMs by replacing certain algorithmic steps with quantum subroutines. The fact that IPMs are efficient both in theory and in
practice gives hope that, if indeed these subroutines can be completed more quickly, QIPMs could satisfy the four criteria of a practical quantum algorithm and deliver practical utility in customer
use cases.
QIPMs preserve the main loop of the IPM (which iteratively generates a sequence of better and better portfolios), and replaces a subroutine in the IPM that solves linear systems of equations with a
quantum algorithm for completing this task. To do this, QIPMs leverage three quantum ingredients — Quantum Random Access Memory, quantum linear system solvers, and quantum state tomography — in each
iteration of the main loop. These ingredients perform the following roles:
• Quantum Random Access Memory (QRAM) allows the quantum algorithm to access the data that defines the instance, namely u and 𝛴, in an efficient and quantum mechanical way. Theoretical analyses of
QRAM-based algorithms often merely report the number of queries to the QRAM data structure and assume the query time is very fast. To account for criterion 2 (reasonable input model), a full
end-to-end analysis must go further and compute the exact resources required to perform the QRAM queries. Specifically, our resource analysis for the QIPM utilizes the resource estimates for
implementing a block-encoding of an arbitrary matrix — a primitive that is in some sense equivalent to QRAM — that we developed in separate, parallel work ^4.
• Quantum linear system solvers (QLSSs), as their name suggests, are quantum subroutines for solving linear systems of the form Ax = b, where A is an invertible L x L matrix, and x and b are
length-L vectors. In the PO problem, L ≈ 14n. The catch is that, unlike classical linear system solvers, QLSSs do not output the vector x; rather, they output a quantum state |x⟩ whose amplitudes
are proportional to the entries of x.
• Quantum state tomography is precisely the subroutine that turns the quantum state |x⟩ into a classical description of the vector x, which is then used to iterate the IPM. In essence, tomography
is performed just by preparing many copies of |x⟩, measuring them, and gathering statistics on the frequency of each measurement outcome to estimate the entries of x. The need for tomography is
related to criterion 1, above; the QLSS is efficient, but it outputs a quantum state rather than classical data, and this quantum state is not the object that is immediately useful to the rest of
the algorithm.
The full end-to-end analysis takes care to account for errors incurred by each of these primitive ingredients. The largest of these errors is incurred by tomography, due to the statistical noise of
estimating an entry of x from measurement outcomes of |x⟩. An additional caveat is that these small errors prevent the QIPM from exactly following the same path as the classical IPM, and they also
cause the QIPM to leave the feasible region. Luckily, the IPM is designed to be self-correcting, in the sense that it partially fixes any errors it makes in future iterations.
Ultimately, through this effect or other workarounds we explore in the full paper, the QIPM will solve the same problem as the classical IPM: it returns an ϵ-optimal solution in an amount of time
that essentially scales as:
The appearance of new parameters ⲕF and ξ represent additional technical complications, and need to be unpacked. The parameter ⲕF is the “Frobenius condition number” — it represents the difficulty of
inverting the relevant matrices with the QLSS. The parameter ξ is the maximum error on quantum state tomography that does not break the algorithm (i.e., how poorly we can estimate the state |x⟩) —
small errors are okay as long as subsequent iterations of the IPM find portfolios that are close enough to those of an ideal IPM.
These are instance-specific parameters, meaning that two different instances of the PO problem with stocks will have different values for ⲕF and ξ, which makes the magnitude and scaling of these
parameters difficult to study. This is an issue, because it is clear that the effectiveness of the algorithm critically hinges on how large they are! If ⲕF and 1/ξ are small and constant, one is
tempted to declare that the quantum algorithm has a large n^3.5 → n^1.5 speedup over the classical IPM. However, it is expected that these parameters will also have an n dependence, cutting into this
speedup. Work by Kerenidis, Prakash, and Szilágyi ^5 studied one version of the QIPM for portfolio optimization and gave some preliminary numerical evidence that the QIPM still offers a speedup after
accounting for the dependence on ⲕF and ξ.
The expression above omits some additional logarithmic factors: we computed the full expression of the runtime to be proportional to:
These log factors are typically ignored in asymptotic analyses (such as that of Kerenidis, Prakash, and Szilágyi ^5), where the goal is to judge criterion 3 of a practical end-to-end algorithm, as
these factors have a relatively small contribution in the limit where n and other parameters are large. However, to judge criterion 4, it is essential to keep track of all contributions, as they can
make a big difference on the runtime estimate for specific choices of n.
Resource estimate for the QIPM
To judge criterion 4, we need to know how many quantum resources the QIPM consumes for values of n relevant to customers. We compute the number of logical qubits, the number of computationally
expensive T gates (i.e., the “T-count”), and the number of layers of T gates (i.e., “T-depth”) required by the algorithm. A T gate is a commonly studied quantum gate that, when coupled with other
easy-to-implement quantum gates, equips quantum computers with their expressive power. We only count the T gates because they are substantially more expensive than the other gates in popular
approaches to fault-tolerant quantum computation, and they dominate the runtime of the algorithm. The number of logical qubits and the total number of T gates are then used to determine the overall
physical footprint of the quantum algorithm — this will also depend on details of the physical hardware, such as the physical error rate.
Meanwhile, the T-depth determines the runtime in an architecture where T-gates on disjoint qubits can be implemented in parallel. These metrics have been computed for other quantum algorithms,
allowing for reasonable comparisons.
For a version of the PO problem like the one above, with a couple of additional, common constraints, we find that optimizing an n-qubit portfolio to tolerance ϵ requires
where we omit subleading terms. Notably, the order-n^2 overhead for the T-count vs. the T-depth, and the order-n^2 logical qubit dependence reflects the fact that QRAM can be implemented at shallow
depth albeit, with large qubit and gate cost.
Figure 2: Median value of the Frobenius condition number as a function of the number of stocks in the portfolio optimization instance, from numerical simulations. Error bars represent the 16th to
84th percentile of observed instances, and the dashed line is a power-law fit showing that the growth of the Frobenius condition number is nearly linear in n.
The above expressions are not particularly illuminating, as they still depend on instance-specific parameters ⲕF and ξ. To probe these parameters, we simulated random instances of portfolio
optimization with n varying from 10 to 120. For each instance, we chose a random set of n stocks from the Dow Jones U.S. Total Stock Market Index (DWCF) and constructed the expected daily return
vector u and covariance matrix 𝛴 using historical stock data collected over 2n time epochs. We found that both ⲕF and ξ^-2 appear to grow with n, on average, and were each typically on the order of
104 or 105 for n = 100. Overall, we evaluated the average cost of the algorithm at n=100 to be:
• Number of logical qubits: 8 million
• T-depth: 2 x 1024
• T-count: 7 x 1029
Even if layers of T-gates are optimistically implemented at the GHz speeds of classical processors (in reality they are likely to be 2–4 orders of magnitude slower than that) these estimates suggest
that the runtime of the quantum algorithm would still be millions of years, even for an instance size that is already classically tractable on a laptop.
This outcome allows us to confidently say that the QIPM, in its current form, does not satisfy criterion 4 of the end-to-end practical algorithm. The reason this occurred was due to a confluence of
several independent factors: a large constant pre-factor coming mainly from the QLSS, a large condition number ⲕF, a significant number of samples needed for tomography, and even a non-negligible
contribution from logarithmic factors (about a factor of 100,000 for the T-depth) that are traditionally ignored in asymptotic analyses.
These estimates also incorporate a few improvements we made to the algorithm, including an adaptive approach to tomography, and preconditioning the linear systems. However, there are several ways one
could try to improve the algorithm further, including using newer, more advanced versions of quantum state tomography ^6 with ξ^-1 rather than ξ^-2 dependence, additional preconditioning, or
improvements to the QLSS. Nonetheless, the resource estimate we arrive at is so far from practicality that multiple improvements to various parts of the algorithm would be needed to make the
algorithm practical.
In retrospect, we can also consider whether the algorithm meets criterion 3, i.e., if it has an asymptotic speedup. The numerical simulations of Kerenidis, Prakash, and Szilágyi ^5 suggested it does,
but our numerical simulations fail to corroborate this, since we find that the scaling of the algorithmic runtime in the regime 10 ≤ n ≤ 120 is roughly the same as the classical algorithm. However,
this scaling does not have a robust trend, and cannot confidently extrapolate to larger industrially relevant choices of n.
Conclusion and takeaways
Our end-to-end resource analysis of the QIPM for portfolio optimization is a revealing case study on the utility of quantum algorithms. Even when an algorithm presents signals of utility, closer
inspection can reveal a drastically different picture when all factors are considered more fairly. This was the case for the QIPM—our analysis strongly suggests it will not be useful to customers,
barring significant improvements to the underlying algorithm. However, the insights gleaned go beyond the QIPM; the core subroutines of the QIPM (QRAM, QLSS, tomography) are common to many other
quantum algorithms as well. Parts of our detailed cost analysis could thus be re-utilized in those algorithms. Indeed, one specific lesson learned is the great cost of implementing quantum circuits
for the data-access component of the quantum algorithm (QRAM).
The QRAM accounts for most of the logical qubits in our accounting, and contributes a considerable factor to the gate depth and gate count. To improve the QIPM to the point of practicality, one
ingredient which would likely be necessary is a dedicated QRAM hardware element that can leverage the specialized, non-universal nature of the QRAM operation to perform the operation more efficiently
than our general analysis accounts for. This hardware element would also improve the practicality of many other quantum algorithms in many domains such as quantum machine learning, quantum chemistry,
and quantum optimization.
With our analysis revealing that QIPMs face significant challenges toward practicality, where does this leave portfolio optimization in the landscape of quantum computing applications? Fortunately,
there are other proposed methods for solving portfolio optimization on quantum computers (e.g., ^7), and some of these are even amenable to near-term experimentation.
These approaches operate by reformulating the convex problem above as a binary optimization problem and attacking them with variational quantum algorithms, such as the quantum approximate
optimization algorithm (QAOA), or using specialized hardware such as quantum annealers.
The upshot is that some of these variational approaches can already be run on near-term quantum devices through Amazon Braket, but one loses the theoretical asymptotic success guarantees afforded by
the QIPM. As such, it is unclear whether these near-term solutions satisfy the third criterion above (asymptotic speedup), and determining whether this is the case will require further empirical
investigation using current hardware and future generations of devices.
Finance remains a rich space for potential quantum algorithms, and perhaps some of the ideas in the QIPM will prove fruitful toward developing practical algorithms in the future. We hope that this
example inspires researchers to search for the next generation of innovative, practical quantum algorithms in a systematic, end-to-end way.
Alexander Dalzell
Alex Dalzell is a Research Scientist at the AWS Center for Quantum Computing. Alex joined the team in 2021 after completing his PhD in Physics at Caltech, where he studied the complexity theory of
quantum advantage experiments using noisy devices. His current research interests lie primarily in quantum algorithms and applications, but also quantum computation and quantum information more
Mario Berta
Mario Berta is a Professor of Physics at the Institute for Quantum Information RWTH Aachen University and holds a position as a Visiting Reader at the Department of Computing Imperial College London.
He is working on the theory of quantum information science.
Dave Clader
Dave Clader is the founder of BQP Advisors, LLC providing quantum computing technical consulting services. His expertise includes quantum algorithms, error correction, and noise mitigation. Prior to
founding BQP Advisors, he was a Vice President in the Quantum Research team at Goldman Sachs where he focused on financial applications of quantum computing. Prior to this, he was as a Principal
Research Scientist at the Johns Hopkins University Applied Physics Lab. He completed his PhD in Physics at the University of Rochester.
David Bader
David A. Bader is a research scientist in Research & Development at Goldman Sachs; and a Distinguished Professor and founder of the Department of Data Science in the Ying Wu College of Computing and
Director of the Institute for Data Science at New Jersey Institute of Technology. Prior to this, he served as founding Professor and Chair of the School of Computational Science and Engineering,
College of Computing, at Georgia Institute of Technology. He is a Fellow of the IEEE, ACM, AAAS, and SIAM; a recipient of the IEEE Sidney Fernbach Award; and the 2022 Innovation Hall of Fame inductee
of the University of Maryland’s A. James School of Engineering. In 1998, Bader built the first Linux supercomputer that led to a high-performance computing (HPC) revolution.
Helmut Katzgraber
Dr. Helmut Katzgraber earned a diploma in physics from ETH Zurich, and his master’s degree and PhD in physics at the University of California Santa Cruz. After postdoc positions at the University of
California Davis and ETH Zurich, he was awarded a Swiss National Science Foundation professorship. In 2009, he joined Texas A&M University as an assistant professor, and became a full professor in
2015. Katzgraber joined Microsoft as a principal research manager in 2018 before joining Amazon in 2020. He was elected Fellow of the American Physical Society in 2021 and leads the Quantum Solutions
Lab at AWS.
Cedric Lin
Cedric Lin is a Sr. Applied Scientist at Amazon Braket. He previously worked at Google as a software engineer, where he primarily designed and built data pipelines and algorithms for optimization.
Cedric has a PhD in Physics from the Massachusetts Institute of Technology; he spent several years developing quantum algorithms, and understanding their limits through the tools of quantum
Martin Schuetz
Martin Schuetz is a Principal Research Scientist at the Amazon Quantum Solutions Lab. Martin has worked several years as an academic researcher with a focus on quantum simulation and computing, at
ETH Zurich, the Max-Planck-Institute for Quantum Optics and Harvard University. Today Martin is working with customers to help solve some of their hardest problems, designing and building quantum
computing, machine learning and optimization solutions on AWS.
Nikitas Stamatopoulos
Nikitas Stamatopoulos is the Head of Quantum Computing Research at Goldman Sachs, with the goal of identifying and developing applications of quantum computing in finance. After working for the
better part of a decade as a quantitative researcher with focus on classical HPC solutions for quantitative problems in derivative pricing/hedging and portfolio optimization, he began exploring how
quantum computers could be harnessed to enhance computationally intensive problems typically encountered in finance. He completed his PhD in Physics (high energy - theory) at Dartmouth College.
Grant Salton
Grant Salton is a senior research scientist in the Amazon Quantum Solutions Lab. Grant obtained his PhD from Stanford University, and his research interests include quantum information theory,
quantum algorithms, quantum error correction, applications of near-term quantum devices, and the role of quantum information in other areas of fundamental physics (e.g., gravity). Prior to joining
Amazon, Grant was a postdoctoral fellow at the IQIM, Caltech.
William Zeng
Dr. William Zeng is a partner at Quantonation, investing in early stage quantum and deep physics technology companies. He is also founder and President of the Unitary Fund, a non-profit dedicated to
developing the quantum ecosystem to benefit the most people. His research focuses on quantum computer architecture, algorithms and software. He previously led a quantum computing research group at
Goldman Sachs and initial development of Rigetti Computing’s quantum cloud platform. He received his PhD in quantum algorithms from Oxford University and his BSc. in Physics from Yale University.
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Passage I
The three equations of rotational motion are ω=ω0+αt... | Filo
Passage I The three equations of rotational motion are and , where the symbols have their usual meanings. Also, are the known standard relations. Use them to answer the following questions Torque of
equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to
rotate about an axis passing through its centre. Which of the two will aquire a greater angular speed after a given time?
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Let and be the mass of radius of the solid sphere and hollow cylinder. Moment of inertid of the hollow cylinder about its axis of symmetry, Moment of linear of the solid sphere about its diameter Let
torque of triangle magnitude be applied on hollow cylinder and solid sphere. the angular acceleration produced in it are and respectively. or Let after time and be the angular speeds of the hollow
cylinder and solid sphere respectively. and From Eqs. (ii) and (iii), we get Therefore, solid sphere will acquire a greater angular speed after a given time.
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Passage I The three equations of rotational motion are and , where the symbols have their usual meanings. Also, are the known standard relations. Use them to answer the following questions
Question Torque of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and
Text the sphere is free to rotate about an axis passing through its centre. Which of the two will aquire a greater angular speed after a given time?
Topic System of Particles and Rotational Motion
Subject Physics
Class Class 11
Answer Text solution:1
Upvotes 79
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Rectangular Load Soil Stress Tool
Geometry of Uniform Load (plan view)
• This tool is for calculating vertical stress increment in soil induced by the applied uniform load, as shown in the figure above.
• The soil layer is assumed as homogeneous, elastic and isotropic.
• For the Boussinesq method, the vertical stress increment calculated is for the position at the depth z below the point P (x[p],y[p]).
• For the approximate 2:1 method, the vertical stress increment calculated is only for the position at the depth z below the center point (0,0).
• All the input parameters except the coordinates of point P should be positive.
• For "nan", "0" or "inf" displayed in Results, please check your input parameters.
Input Parameters
Output Results
Welcome! Please input the parameters!
• Vertical soil stress incerment
Boussinesq Method (any point)
Δσ[z] (kN/m^2) =
Approximate 2:1 Method (center point)
Δσ[z] (kN/m^2) =
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tonum( value, [format_flags] )
Attempts to return the given value as a number. Result is empty on failure.
The value to be converted: strings, booleans, and numbers are accepted.
A bitfield which allows different operations to occur in the conversion.
The provided value converted to a native number format, if possible; otherwise an empty result.
Returns a number if value can be converted to a number, or value already is a number. If not, the result will be empty, which for most purposes will work as if nil had been returned.
If value is a string, tonum() will try to parse it as one of these supported number formats:
• Decimal, e.g.: "255", "3.14159"
• Hexadecimal, e.g.: "0xff", "0x3.243f"
• Binary, e.g.: "0b11111111", "0b11.0010010000111111"
• Scientific/exponential, e.g.: "2.55e2", "3.14159e0"
Note that leading or trailing 0s are ignored, as one would expect, which may be useful when extracting fixed-width fields from a string.
If value is a boolean, it will return 0 for false and 1 for true, thus providing a simple way to convert booleans to numbers.
If value is already a number, it will be returned as-is.
The format_flags parameter is composed of individual bits that control the conversion process:
0x1: Read using hexadecimal notation, without requiring the "0x" prefix.
Note: Non-hexadecimal characters, including '.' and '-', are taken to be '0'.
0x2: Shift the value right 16 bits to create a 16.16 fixed-point number.
This works with all formats, even booleans: true becomes 0x.0001.
0x4: When value cannot be converted to a number, return 0 instead of nothing.
Technical Notes[ ]
• If the string represents a number outside of the range of PICO-8 numbers, it wraps the number to that range before returning. Effectively, the result is bitwise-anded with 0xffff.ffff to fit in
the PICO-8 16.16 fixed-point number format.
• When presented with a non-number, e.g. tonum("xyz"), the result will technically be empty to indicate that the string can't be parsed as a number. In Lua, the concept of empty is technically
distinct from nil, but when empty is assigned to a variable, the variable becomes nil, and if empty is directly tested (e.g. if tonum("xyz") then ... end) it will be considered false.
Examples[ ]
x = tonum('12345') -- 12345
x = tonum('-12345.67') -- -12345.67
x = tonum('-1.23456789e4') -- -12345.6789
x = tonum('0x0f') -- 15
x = tonum('0x0f.abc') -- 15.6709
x = tonum('0b1001') -- 9
x = tonum('32767') -- 32767
x = tonum('99999') -- -31073 (wrapped)
x = tonum('xyz') -- nil
-- Examples with format_flags bitfield
x = tonum("ff", 0x1) -- 255
x = tonum("1146880", 0x2) -- 17.5
x = tonum("1234abcd", 0x3) -- 0x1234.abcd
x = tonum("xyz", 0x4) -- 0 (instead of nil)
See also[ ]
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알리페이, 알리바바 한국 공식 파트너 (주)에스큐이모션
A projectile is fired with speed v0 at an angle theta from the horizontal Find the highest point in the trajec Express the highest point in terms of the magnitude of the acceleration due to gravity
g, the initial velocity v0, and the angle theta.
the horizontal belocity of the plane equals the vertical velocity of the bomb when it hits the target a stone with a mass M is dropped from an air plane that has horizontal velocity V at a height H
above a lake. if air resistance is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point …
It hits ground with 45° angle when the vertical velocity, vy, is the same as the horizontal velocity, vx, which equals u (initial velocity) if the projectile is fired horizontally and ignoring
air-resistance. vx=u Let H=given height of cliff vy²-0²=2gH vy²=2gH since vy=u, u²=2gH u=√(2gH)
The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a
projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.
m per second at an angle of 30 above the horizontal. A projectile is fired with an initial velocity of 120. meters per second at an angle (theta) above the horizontal. If the projectile's initial
horizontal speed is 55 meters per second, then angle (theta) measures approx.
A cannonball explodes into two pieces at a height of h = 100 m when it has a horizontal velocity Vx=24 m/s. The masses of the pieces are 3 kg and 2 kg. The 3-kg piece falls vertically to the ground 4
Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal. Such a projectile begins its motion with a horizontal velocity of 25 m/s and a
vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity.
Example John kicks the ball and ball does projectile motion with an angle of 53º to horizontal. Its initial velocity is 10 m/s, find the maximum height it can reach, horizontal displacement and total
time required for this motion. Example In the given picture you see the motion path of cannonball.
Projectile Motion "The path that a … larger initial horizontal velocity. … If the rifle is fired directly at the target in a horizontal direction, will the bullet …
Kinematics in 2-D (and 3-D) … the initial velocity components are then Vx = v0 cosθ and Vy = v0 sinθ, … if the projectile returns to the height from which it …
H velocity is constant vx = v i V velocity is changing vy = – gt H range or displacement: dx = vi t 2 V distance: h = gt Calculation of Projectile Motion A projectile was fired with initial velocity
vi horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile. vi
Projectile Motion. Projectile motion occurs when objects are fired at some initial velocity or dropped and move under the influence of gravity. One of the most important things to remember about
projectile motion is that the effect of gravity is independent on the horizontal motion of the object.
Kinetic projectiles. A kinetic projectile can also be dropped from aircraft. This is applied by replacing the explosives of a regular bomb with a non-explosive material (e.g. concrete), for a
precision hit with less collateral damage. A typical bomb has a mass of 900 kg and a speed of impact of 800 km/h (220 m/s).
Projectile motion has both vertical and horizontal motion. However, we will neglect air resistance in this case. As for what this means, the horizontal component of the initial velocity remains
constant, and the vertical component follows the same rules as a rising/falling body, i.e., has an acceleration of 9.81 m/s^2
9 m/s, Vx is constant Ex) A wild bowler releases the his bowling ball at a speed of 15 m/s and an angle of 34.0 degrees above the horizontal. Calculate the horizontal distance that the bowling ball
travels if it leaves the bowlers hand at a height of 2.20 m above the ground .
A projectile is fired with initial speed V(subscript 0) m/s from a height of h meters at an angle of θ above the horizontal. Assuming that the only force acting on the object is gravity, find the
maximum altitude, horizontal range and speed at impact. V(subscript 0) = 98, h=0, θ= π/6
The initial velocity components are V0x = 100 m/s and V0y = 65 m/s. The … A projectile is fired from the origin (at y = 0 m) as shown in the figure. The initial velocity components are V0x = 100 m/s
and V0y = 65 m/s. The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q.
A projectile fired from a height of 19.6m with an initial velocity of 10m/sec parallel to the ground will? A projectile is fired with an initial velocity of 75.2m/s at an angle of 34.5 above the
A projectile is fired with an initial speed of 37.7 m/s at an angle of 44.2 ∘ above the horizontal on a long flat firing range. A) Determine the maximum height reached by the projectile. B) Determine
the total time in the air.
A projectile is fired at such an angle from the horizontal that the vertical component of its velocity is 49 m/s. The horizontal component of its velocity is 61 m/s.
A projectile fired from a height of 19.6m with an initial velocity of 10m/sec parallel to the ground will? A projectile is fired with an initial velocity of 75.2m/s at an angle of 34.5 above the
Exactly 2.7s after a projectile is fired into the air from the ground, it is observed to have a velocity = (8.1 i+4.6 j)m/s, where the axis is horizontal and the y axis is positive upward.
The initial velocity components are V0x = 100 m/s and V0y = 65 m/s. The … A projectile is fired from the origin (at y = 0 m) as shown in the figure. The initial velocity components are V0x = 100 m/s
and V0y = 65 m/s. The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q.
Question: A projectile is fired from the origin (at y = 0 m) as shown inthe figure below. The initial veloc… A projectile is fired from the origin (at y = 0 m) as shown inthe figure below. The
initial velocity components areV 0x = 940 m/s and V 0y = 96 m/s. Theprojectile reaches maximum height at point P, then it falls andstrikes the ground at point Q.
ground level with an initial velocity of 28 m s–1 at an angle of 30° to the horizontal. Q3. Calculate the horizontal component of the velocity of the ball: a initially b after 1.0 s c after 2.0 s.
A3. a vx = (28 m s–1) cos 30° = 24.2 m s–1 north and remains constant throughout the flight. b 24.2 m s–1 north c 24.2 m s–1 north Q4.
Determine: a. the maximum height it can reach b. the time it takes to reach this height c. the instantaneous velocities at the end of 40 and 60 seconds 2. A football is kicked at a certain angle
above the horizontal. The vertical component of its initial velocity is 40m/s and the the horizontal component id 50m/s.
The height H depends only on the y variables; the same height would have been reached had the ball been thrown straight up with an initial velocity of v 0 y = +14 m/s. It is also possible to find the
total time or “hang time” during which the football in Figure 3.12 is in the air.
Four cannonballss, each with a mass M and initial velocity v, are fired from a cannnon at different angles relative to the Earth. If air friction is upward, which angular direction of the cannon
produces the greatest projectile range? 35.) A ball projected horizontally with an initial velocity of 20 meters per second east, off a cliff 100 meters high.
A projectile is fired with an initial velocity of 120 meters per second at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then θ measures
approximately (a) 13° (b) 63° (c) 27° (d) 75° 5. A golf ball is hit at an angle of 45° above the horizontal.
The ball has an initial vertical velocity of and accelerates uniformly over to reach a final vertical speed of . The area under the graph is the vertical distance travelled: So the height of the …
Im attempting to find the angle to fire a projectile at such that it lands at a specific point. I have only done (in fact, I havnt even completed) high school level physics and calculus, so please
bear with me.
I'm not asking for a complete answer, any help is appreciated – really stuck on where to begin. We will look at a projectile (an object that is given an initial velocity at an angle with the
horizontal). Our assumptions include that it moves only under the influence of gravity (after the initial velocity) and wind resistance which we will neglect.
Projectile Motion: Motion through the air without a propulsion Examples:. … PROJECTILE MOTION Senior High School Physics … h – initial height, v0 – initial …
Here we’re solving a problem where a ball is projected horizontally from a height of h=1.8 m with a horizontal velocity of Vx. At the impact with the ground, the ball has travelled 0.5 m
If the projectile is aimed at the target and fired at t = 0, then motion with constant velocity v 0 will bring the projectile to the initial position of the target at some later time t. In the time
interval between 0 and t the downward motion with constant acceleration carries the projectile downward by an amount ½gt 2 .
Basically, a projectile is an object that has an initial velocity and follows a specific path, factoring in gravitational acceleration and air resistance. Projectile motion has both vertical and
horizontal motion.
A projectile is launched with 200 kg m s of momentum with solution. A projectile is fired from the origin (at y = 0 m) as shown in the figure. the initial velocity components are = 840 and = 47 . the
projectile reac
at 25o with the horizontal. The mass of the sled is 80.0 kg and there is negligible friction between the … A projectile is fired with an initial speed of 40.2 …
A bullet of mass m = 50g is fired at a block of … plus bullet system to swing up a height h = 0.45 m. What is … with mass m, traveling with velocity -1 m/s. Find the
Since the angle of firing of the projectile is 45 degrees, therefore, at the time of firing the projectile, V0x = V0y, where V0x = initial horizontal velocity of projectile and V0y = initial vertical
velocity of projectile. Now V0y = gt (g = gravitational acceleration = 9.81 m/s2 and t = time from start to highest point).
What equation/equations would be used to solve this problem? A projectile, fired with unknown initial velocity, lands 24.0 s later on the side of a hill, 3350 m away horizontally and 454 m vertically
above its starting point.
Vx be the horizontal velocity. Vx 2t = 600 m t is the time to reach the topmost point along y or g direction. We know Vy^2 = gh Vy = root(gh) = v sin theta Vy = g t Because the final velocity =
initial velocity – g t Final velocity in the y direction is 0. This gives t= Vy/g Horizontal range will be Vx 2 t = V cos theta * 2* Vy/g = V^2 cos …
LMGHS. Name:_____ Band:_____ H. W # 13. Projectile Motion additional Problems with answers. 1. A golfer practicing on a range with an elevated tee 4.9 m above the fairway is able to strike a ball so
that it leaves the club with a horizontal velocity of 20 m s–1.
The T = 2t = 2(51.02 s) = 102 s projectile strikes the target after 4.0 s. b At maximum height the projectile only has the a Determine the horizontal velocity of the horizontal component of velocity.
projectile. Then speed = the initial horizontal component b Calculate the value of θ.
If you know the velocity, vx, of a ball launched horizontally from a table and the ball’s initial height, y, above the floor, the equation of projectile motion can be used to predict where the ball
will land. Recall that the horizontal displacement or range, x of an object with horizontal velocity, vx, at time, t is
Topic: Linear Motion with a Constant Velocity or a Constant Acceleration; Topic: Linear Motion under Gravity … Topic: Two dimensioal motion of a projectile; Topic …
A projectile is fired with an initial speed of 65.8 at an angle of 39.6 above the horizontal on a long flat firing range. Determine the speed and direction (angle) of the projectile 1.41 seconds
after firing.
ground. The horizontal component of the projectile's velocity. vx, is initially 40. meters per second. The vertical component of the projectile's velocity, vy, is initially 30. meters per second.
What are the components of the projectile's velocity after 2.0 seconds of flight? [Neglect friction.] A) 0.654 s B) 1.53 s C) 3.06 s D) 6.12
(A) 6,000 m (B) 7,000 m (C) 9,000 m (D) 10,000 m. 11. A projectile is fired with an initial velocity of 100 m/s at an angle above the horizontal. If the projectile's initial horizontal speed is 60 m/
s, then angle measures approximately (A) 30o (B) 37o (C) 40o (D) 53o. 12.
A projectile moves at a constant speed in the horizontal direction while experiencing a constant acceleration of 9.8 m/s2 downwards in the vertical direction. To be consistent, we define the up or
upwards direction to be the positive direction. Therefore the acceleration of gravity is, -9.8 m/s2.
A projectile is fired at an angle of 45 degree with horizontal . eleavation angle of d projectile at its high? I fire a cannon with muzzle velocity 120m/s at an angle of 45 degrees. (Projectile
Using the average horizontal distance and the time from 2) above, determine the initial (muzzle) velocity of the bearing. Record below. Note that there is no acceleration in the horizontal direction,
ax = 0, so the equation for horizontal motion is simply
PROJECTILE MOTION Honors Physics … initial height, v0 – initial horizontal velocity, g = -9.81m/s2 … constant forward velocity. cp projectile test.
Q4.20 A projectile is fired at an angle of 30 o from the horizontal with some initial speed. Firing at what other projectile angle results in the same range if the initial speed is the same in both
cases? Neglect air resistance. Any angle and 90 o minus that angle have the same range.
Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum
height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays,…
For a general velocity problem you can simply write an equation using "V" for velocity, such as V = a × t. However, to write a motion equation that treats horizontal and vertical velocity separately,
you must distinguish the two by using Vx and Vy, for horizontal and vertical velocity, respectively.
Examples of conversion factors are: 1min 60s 100cm 1m 1yr 365.25day! 1m 3.28ft 1.1.3 Density A quantity which will be encountered in your study of liquids and solids is the density of a sample. It is
usually denoted by ρ and is defined as the ratio of mass to volume: ρ = m V (1.1) The SI units of density are kg m3 but you often see it expressed in g cm3.
Learn how to simplify vectors by breaking them into parts.
A projectile is fired with an initial speed of 13.0m/s at an angle of 35.0? above the horizontal. … Its initial velocity is vx = 5 m/s. … Janet jumps off a high …
If we call the horizontal displacement dx and the initial horizontal velocity vx then, at time t, (Note: vxf = vxi) dx = vxt. The equations for an object falling with constant acceleration, g,
describe the vertical motion. If dy is the vertical displacement, the initial vertical velocity of the object is vy. At time t, the vertical displacement is
A projectile is launched from ground level at 36.3 m/s at an angle of 26.1 ° above horizontal. Use the launch? A projectile is fired from ground level at an angle of 40.0° above horizontal at a speed
of 30.0 m/s.
A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5 degree above the horizontal on a long flat firing range. Determine: a) the max. height reached by the projectile b) the
total time in the air, c) the total horizontal distance covered (that is the firing range), d) the velocity of the projectile 1.50s after firing.
I need to write the equations of a ball in projectile motion (ignoring air friction) with an initial velocity of 40 m/s at an angle of 40° with respect to the horizontal. Specifically, I need to
write: x (n) and y (n) for the projectile (where n represents the nth evaluation point) The x and y components of velocity (Vx(n) and Vy(n))
Since the angle of firing of the projectile is 45 degrees, therefore, at the time of firing the projectile, V0x = V0y, where V0x = initial horizontal velocity of projectile and V0y = initial vertical
velocity of projectile. Now V0y = gt (g = gravitational acceleration = 9.81 m/s2 and t = time from start to highest point).
(a) we solve for y = h: which yields h = 51.8 m for y 0 = 0, v 0 = 42.0 m/s, q 0 = 60.0° and t = 5.50 s. − 0 0y − 2 (b) The horizontal motion is steady, so v x = v 0x = v 0 cos θ 0, but the vertical
component of velocity varies according the equations before. Thus, the speedi id at impact is v = ()v 0 cosθ 0 2 + v 0 sinθ 0 −gt 2 =27 …
So this 30 is my initial velocity I can think of this as my initial velocity in the x axis because it is entirely on the x axis or I can think of this as my Vx which never changes. These are all the
OBJECTIVES : 1. To determine the range (horizontal displacement) as a function of the projectile angle. 2. At maximum height At the top of its path, the projectile no longer is going up and hasn't
started down, yet. Its vertical velocity is zero ( vy = 0 ). The only velocity it has is just its horizontal velocity, vx.
We solved the wind-influenced projectile motion problem with the same initial and final heights and obtained exact analytical expressions for the shape of the trajectory, range, maximum height …
This graph shows the parabolic relationship between the change in the initial height and how it affects the horizontal range. … Momentum = mass x velocity p = m v …
The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. If the force acting on a particle is oblique with initial velocity then the motion of
particle is called projectile motion. 3.2 Projectile.
12.0 times the mass of the neutron.) • b) The initial kinetic energy of the neutron is 1:1010 13 J. Find its nal kinetic energy and the kinetic energy of the carbon nucleus after the collision. a
Let’s adopt the following notations : • for the neutron, mass m, v i and v f are the initial and nal velocity respectively. • for the atom …
Let the initial velocity of the ball be v, The initial kinetic energy, E=(1/2 )mv^2 the horizontal velocity: v×cos45°=v/√2 When the ball reaches the highest point, it’s vertical velocity becomes zero
and it’s horizontal velocity remains the same. Thus, it’s velocity at the highest point is equal to its horizontal velocity.
Q)A projectile is launched with an initial velocity of 75.2m/s at an angle of 34.5 above the horizontal on a long flat firing range. Determine a) the max height reach by the projectile b) the total
time in the air c) the total horizontal distance covered (range) d) the velocity of the projectile 1.5s after firin k so far — for a) Viy = Vi(sin )
Click Velocity to obtain data for row 9-11 – note that magnitude of velocity is what we call _____ EXPERIMENT 1. Control 1A 1B 1C 1D Initial Height 0 0 0 0 0 Initial . Speed 10.00 m/s 10.00 m/s 10.00
m/s 10.00 m/s 10.00 m/s Angle of inclination 0 30 45 60 90 Mass
With this projectile range calculator, you'll quickly find out how far the object can be thrown. All you need to do is enter the three parameters of projectile motion – velocity, angle, and height
from which the projectile is launched. In no time you'll find the horizontal displacement of your …
diver runs horizontally with a speed of 1.2 m/s off a platform that is 10.0 m above the Practice Problems – Projectile Motion 2 Answers mr. talboo – physics projectile motion practice problems 2 1. a
ball is thrown in such a way that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s, respectively.
16. A projectile is fired from a gun and has initial horizontal and vertical components of velocity equal to 30.0 m/s and 40.0 m/s respectively. Assuming air resistance is negligible, approximately
how long does it take the projectile to reach the highest point in its trajectory? a. 1.0 s b. 2.0 s c. 4.0 s d. 8.0 s e. 16.0 s 17.
Time to reach maximum height. It is symbolized as (t), which is the time taken for the projectile to reach the maximum height from the plane of projection. Mathematically, it is give as t=USin(teta)/
g Where g=acceleration due to gravity(app 10m/s²) U= initial velocity (m/s) teta= angle made by the projectile with the horizontal axis.
Determine the maximum height and range of a projectile fired at height of 3 m above the ground with initial? Plus de questions Determine the maximum height and range of a projectile fired at a height
of 3 feet above the ground with an in?
• At maximum height, the velocity equals to 0 ms … a 25 m building and is thrown with initial horizontal velocity of 8.25 ms … the product of mass and velocity …
v0= velocity initial vf=velocity final A. Find the horizontal velocity first. You have break the 20 m/s into two components the x and the y by making a right triangle with the 20 m/s being the
A projectile is fired with a velocity of 45 m/s at an angle of 32. What is the horizontal component of the velocity? … Vx = 38.16 m/s.
28 An object is launched with an initial velocity of 10 meters per second from the ground level at an angle of 53° above the horizontal. What are the horizontal and vertical components of the ball’s
velocity at the apex of its flight? A vx = 6 m/s, vy = 8 m/s B vx = 6 m/s , vy = 0 m/s C vx = 0 m/s , vy = 0 m/s
An object of mass m is dropped from the roof of a building of height h. While the object is falling, a wind bl? An object of mass m is dropped from the roof of a building of height h. While the
object is falling, a wind blowing parallel to the face of the building exerts a constant force F on the object.
Decomposition of velocity into initial horizontal velocity (Vx) and initial vertical velocity (Vy). Horizontal velocity remains constant during the projectile motion. Vertical velocity can be
calculated using the suvat equations, where the acceleration is acceleration of free-fall ( g ) and the displacement is height ( h ).
e. What will be the direction and magnitude of the velocity of the marble as it reaches the floor?6. A projectile is fired from the ground with a velocity of 96.0 m/s at an angle of 35.0 O above the
horizontal; a. What will be the vertical and horizontal components of the initial velocity of this projectile? b.
Vi Vyinitial t ½ay t2 * Dr. Sasho MacKenzie – HK 376 * Tips and Equation Rearrangements If the initial vertical velocity is zero, then If the object’s initial vertical position = the final vertical
position, then * Dr. Sasho MacKenzie – HK 376 * Shot Put Example A shot put is released from a height of 2 m with a velocity of 15 m/s at an …
A bullet is fired horizontally at 575 m/s from a height of 1.75 m. How far from the gun ( horizontally ) will the bullet hit the ground? y = 1.75m X y = vo t + ½ a t2 t = x = vx t = vx x = 575 m/s 2y
/g 2y / g 2(-1.75 m) / ( -9.81 ms-2 ) = 343 m A bullet is fired horizontally at 575 m/s from a height of 1.75 m.
Physics question on 2D projectile motion? The distance between the striker and midfielder is 20.0m. The midfielder passes the ball towards the striker with an inital speed of 22.1m/s, 25.0° above the
AP Physics 1 Investigation 1: … horizontal track, and finally as a projectile off the end of the ramp onto the floor. … how the initial velocity of the ball in …
A test Rocket is fired vertically upward from a well. A catapult gives it an initial velocity of 80 m/s at ground level. Subsequently, its engines fire and it accelerates upward at 4 m/s 2 until it
reaches an altitude of 1000 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.8 m/s 2.
1. Time to reach maximum height. It is symbolized as (t), which is the time taken for the projectile to reach the maximum height from the plane of projection. Mathematically, it is give as t=USin
(teta)/g. Where g=acceleration due to gravity(app 10m/s²) U= initial velocity (m/s) teta= angle made by the projectile with the horizontal axis. 2.
A ball of mass m falls down without initial velocity from a height h over the Earth's surface. Find the increment of the ball's angular momentum vector picked up during the time of falling (relative
to the point O of the reference frame moving translationally in a horizontal direction with a velocity V). The ball starts falling from the point O.
The range of projectile is given by the formula – R = u²Sin2q /g where u is the initial velocity of the projectile, q is the angle with which the body is projected and g is the acceleration due to
gravity and R is the range or the horizontal distance that the projectile covers.
A projectile is fired with initial velocity vo = 20 m/s at an angle o = 30o with the horizontal and follows the trajectory shown above. 1. The speed of the ball when it reaches its highest point is
closest to (A) 20 m/s (B) 17 m/s (C) 10 m/s (D) 0 m/s . 2.
The first method will apply the principles of uniformly accelerated motion to treat the ball as a projectile. Measuring d and h 2 as illustrated in Figure 2 will be enough to calculate the velocity
of the ball at the time it leaves the ramp. As a reminder, the uniformly accelerated motion equations are reproduced below.
Exam 1 Review Questions PHY 2425 – Exam 1 . … A velocity vector has an x component of +5.5 m/s and a y component of –3.5 m/s. The … on the vertical axis and …
The horizontal component of the velocity, Vx is constant. The horizontal component of acceleration in a projectile motion is equal to zero. Hence, motion along the horizontal is uniform. The vertical
component of velocity, Vy, is increasing. It has a constant acceleration along the vertical axis equal to g, the acceleration due to gravity.
Mass (m) is a measure of a body's? … Acceleration in the opposite direction of the initial velocity may be called? Deceleration . … but Vx will remain constant …
A) 64.3 m B) 100. m C) 40.0m D) 76.6m approximately A) 130 B) 750 C) 270 26) D) 630 A projectile is fired from a gun near the surface Of Earth. The initial velocity of the projectile has a vertical
component of 98 meters per second and a horizontal component Of 49 meters per second. How long will it take the projectile to reach the highest
Ignore the horizontal velocity . The horizontal velocity is irrelevant, so ignore it. It doesn't matter whether it's going 4 m/s, 40 m/s or 4000 m/s.
(a) Calculate: (i) the initial vertical component of the projectile’s velocity; (ii) the initial horizontal component of the projectile’s… show more the question is. A projectile is fired with a
velocity of 20 m s–1 at an angle of 25 ° above the horizontal. Any effect due to air resistance can be ignored.
Find the y component of the motion of the projectile, which is Total velocity times the sine of the angle it makes with the x axis = 25sin(65) = 22.66 m/s. From there, use kinematic equations for
motion in one direction (the y direction).
Projectile Motion Calculations You must be able to calculate the following quantities: horizontal and vertical components of velocity. time of flight. range. maximum height. velocity at any point
These can all be found using the equations of motion. When using these equations, the substituted quantities . must either be all horizontal or all vertical values.
Projectile Motion – Practice Questions – Download as PDF File (.pdf), Text File (.txt) or read online.
The distance traveled (d) of a projectile over a given period of time (dt) is dependent on its: launch angle (A) initial velocity (V) Initial velocity is split into its vertical and horizontal
components (Vx,Vz) based on the launch angle.
(b)Calculate the velocity of the dart as it leaves the gun (give answer in m/s). 9.A projectile is shot from the ground with an initial velocity of 100 m/s at an angle of 40 above the horizontal. It
follows a parabolic path and hits the ground.
Consider a body of mass ‘m’ moving with a velocity Vl. A net force F acts on it for a time ‘t’. … Initial Horizontal Velocity … At a height ‘h’ above …
projectile. The user is prompted to enter values for mass, energy, angle, and the initial height of the projectile, and the program will calculate and plot the projectile's trajectory. This
simulation will ignore complicating factors such as friction, air resistance, spin, and rebound. A trajectory is the path followed by a projectile.
Horizontal Velocity Vx Vertical Velocity Vy Solution Because the plane is flying horizontally, the intitial velocity vectors of the bomb are: Horizontal, Ux= 50.0ms-1, Vertical, Uy= zero a) Time to
hit the ground We know the vertical distance to fall (-700m (down)), the acceleration rate (g= -9.81ms-2) and that Uy=0.
Once the height (1.027m) and the time (.457s) were found, it was then required to find the value of Vx or horizontal velocity (Vx=.0376m/.457s= .082m/s). Vy or the velocity in the vertical direction
was found (9.8= (Vy/.457)= 4.47m/s).
A cannon works very similar to how a gun works. A charge is loaded into the cannon (such as gunpowder) and then the cannonball is loaded in on top of the charge.
Joe now throws the javelin into the air at an angle of 40o above the horizontal at an initial velocity of 30 ms-1. Joe now throws the javelin into the air at an angle of 40° above the horizontal at
an initial velocity of 30 m s–1 37. (c) Show that the horizontal component of the initial velocity of the javelin is 23 ms-1.
Edexcel AS Physics in 100 Pages. V. kg ms 2 m A s. kgm 2 s 3 A 1 . Also, remember the following scale. It will make your life easier while doing unit conversions. 7 Edexcel AS Physics in 100 Pages.
Chapter 1 Mechanics. 8. Edexcel AS Physics in 100 Pages. 1.1 Motion in one dimension Speed, velocity, distance and displacement
A catapult gives it an initial velocity of 80 m/s at ground level. Subsequently, its engines fire and it accelerates upward at 4 m/s 2 until it reaches an altitude of 1000 m. At that point its
engines fail, and the rocket goes into free fall, with an acceleration of -9.8 m/s 2 .
A bullet is fired with a horizontal velocity of 330 m/s from a height of 1.6 m above the ground. … The arrow was fired with an initial vertical velocity of 49 m/s relative to the truck …
Displacement of projectile in launcher, s = 0.284 m Angle of launch, = 31 Mass of projectile, m = 0.015 kg. Discussion: Initial horizontal velocity, ux = 5.6 ms-1 Initial vertical velocity, uy = 3.4
ms-1 Initial resultant velocity, v = 6.6 ms-1, 031 above horizontal Average acceleration of projectile while in launcher, aav = 76.7 ms-2 Net Force …
independent of the path taken between the initial and final point. An example of this is the force of gravity (weight): work by gravity is always equal to -mg times the change in height, ∆h. The
amount of work done does not depend on how the object in question gets from one height to another, only on the final change in height.
The initial position for the projectile is taken to be the origin (x=0, y=0), and the magnitude and direction for the initial velocity is supplied by the user with the direction being the angle above
the horizontal measured in degrees.
The initial horizontal and vertical speed of the decoy can be had by resolving the plane's speed into components. Let these be Vv and Vh. Let the time of fall be T, and the height at release be H.
diver runs horizontally with a speed of 1.2 m/s off a platform that is 10.0 m above the Practice Problems – Projectile Motion 2 Answers mr. talboo – physics projectile motion practice problems 2 1. a
ball is thrown in such a way that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s, respectively.
Solution Below: An arrow is shot at an angle of θ = 45° above the horizontal. The arrow hits a tree a horizontal distance D = 220 m away, at the same height above the ground as it was shot. Use g =
9.8 m/s 2 for the magnitude of the acceleration due to gravity. Find the time that the arrow spends in the air.
Best answer: initial vertical velocity=15 sin 40 initial horizontal velocity=15 cos 40 total flight time=2*initial vertical velocity/g=2*15 sin 40/g=30 sin 40/g maximum height reached by the ball=(
initial vertical velocity)^2/(2*g) =225*(sin 40)^2/(2*g) horizontal distance traveled by the…
the mass of water (m), the force of gravity (g) and the height above the tap (h 1). 2. Water leaving the tank through the tap has Kinetic Energy. Give an expression for this in terms of the mass of
the water (m) and the velocity (v) at which it 3. Equate these two expressions (why?) and solve to find the exit velocity of the water as it leaves …
g = 9.8 m/s2 = (3.0)(9.8)(42.0) h = 42.0 m KE = 1235 J. KE = ? 75. What is momentum? What is impulse? Momentum is mass times velocity or an object’s inertia in motion. Impulse is the change in
momentum caused by a force being applied to the object for. so period of time. 76.
By measuring the vertical height climbed and knowing your mass, the change in your gravitational potential energy can be found with the formula: ∆ PE = mgh (where m is the mass, g the acceleration of
gravity, and h is the vertical height gained) Your power output can be determined by Power = ∆ PE/∆t (where ∆t is the time to climb the …
The horizontal component of the motion is irrelevant. The ball starts with a vertical velocity of 0 metres per second and has an acceleration of 9.8 metres per second squared … It travels 50 metres
so s = 1/2*g*t^2 that is, 50 = 1/2*g*t^2 therefore t^2 = 10.2 and so t = 3.19 seconds.
5 i- z 2 -_ Y Forces developed during the ricochet of projectiles of spherical and other shapes 5~– 2' 5 Fx =.035 Vx 1V l .2 L1 50 100 200 VX m sec-1 la) J 2I ~ ~n i 500 50 100 200 500 Vx m.sec 1 lb)
FIG. 7.
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A bullet is fired with a horizontal velocity of 330 m/s from a height of 1.6 m above the ground. … The arrow was fired with an initial vertical velocity of 49 m/s relative to the truck …
The other input parameters are the same for these two cases and the values are as follows: (1) Mass of the projectile 33.4 kg (2) Dia of projectile 0.13 m (3) Angle of elevation 7.5° (4) Wind
velocity 0 The Y-distance (height), Y-velocity, X-distance (range), X-velocity values for initial (muzzle) velocities 996.1 m/s and 1003.1 m/s were found …
A glider with a mass of 0.355 kg moves along a friction-free air track with a velocity of 0.095 m/s. It collides with a second glider with a mass of 0.710 kg moving in the same direction at a speed
of 0.045 m/s.
The range is the horizontal velocity times the time in the air: x = vxt = (275)(5) = 1390 m. 5. A rope is tied to the handle of a bucket, which is then whirled in a vertical circle of radius 60.0 cm.
(4) increase the launch angle and increase the ball’s initial speed 6 A ball attached to a string is moved at constant speed in a horizontal circular path. A target is located near the path of the
ball as shown in the diagram.
Consider a body of mass 'm' placed initially at a height h(i), from the surface of the earth. … Initial Horizontal Velocity … There is no net force acting on the …
Using the values given, we know the initial velocity is 40.0 m/s and the angle of the motion is 30.0 degrees. We then use cosθ to get horizontal velocity (vx), which is 36.64 m/s, and the vertical
velocity (vy1), which is 20 m/s. Since we all know distance equals time multiplying speed, and horizontal velocity is a constant value.
mass M, as shown in the figure. Initially, the unwrapped portion of the rope is vertical and the cylinder is horizontal. The linear acceleration of the cylinder is a. (2/3)g b. (1/2)g c. (1/3)g d. (1
/6)g e. (5/6)g a 14. A pendulum bob of mass m is set into motion in a circular path in a horizontal plane as shown in the figure.
We then used the equation Vx= ∆x/t to solve for the horizontal velocity, Vx. Vx = 13.99 m/s. … velocity in the horizontal direction since there are not any forces …
A 15-kg projectile is fired with an initial speed of 75 m/ s at an angle of 350 above the horizontal on a long, flat firing range. A 25-m-high wall is located 590 m downrange.
Time in air calculated Initial velocity calculated 1 2 3 Average initial Velocity Show all calculations: Using your initial velocity average from above as the initial velocity for the cannon below.
Now shoot your cannon at 15 , 40 , 60 and record the d x (range).
Author: Topic: Projectile Orbits and Satellite orbits (Read 226931 times) 0 Members and 2 Guests are viewing this topic. Click to toggle author information(expand …
U θ angle of launch An example which is NOT a Projectile: Maximum Height Uy Photo: Keith Syvinski Horizontal Velocity Vx Vertical Velocity Vy Ux “Range” = Total Horizontal Displacement A rocket or
guided missile, while still under power, is NOT a projectile. Equations for Projectile Motion 1.
A 2.30 kg mass is suspended from the ceiling and a 1.70 kg mass is suspended from the 2.30 kg mass, as shown. The tensions in the strings … A 10-kg block on a rough horizontal surface is attached to
a light spring (force constant = 1.4 kN/m).
The velocity vector of a projectile with a vertical velocity of 25.0 m/s and a horizontal velocity of 18.0 m/s is ___ m/s. a. 7.00 / b. 21.5 / c. 30.8 / d. 35.8 16.
1.2 m/s. Cart 2 has a mass of 0.61 kg. … vertically upward with a velocity (18 m/s ) … the ball to rest horizontally but gives it an initial horizontal speed …
Introduction: In this experiment a steel ball will be shot into the bob of a pendulum and the height, h, to which the pendulum bob moves, as shown in Figure 1, will determine the initial velocity, V,
of the bob after it receives the moving ball.
The initial velocity has magnitude v0 and because it is horizontal, it is equal to vx the horizontal component of velocity at impact. Thus, the speed at impact is where and we have used Eq. 2-16 with
x replaced with h = 20 m.
Velocity of a Ball When it Hits the Ground. … because the ball falls starting from its maximum height, the initial velocity is $0$. Therefore, the equation becomes …
initial population of N 0 muons where N= N 0e t=˝ = N 0e 10:53=7:046 = 0:225N 0 (12) Length Contraction and Rotation Problem 1.15, page 46 A rod of length L 0 moves with speed valong the horizontal
direction. The rod makes an angle 0 with respect to the x0axis. Determine the length of the rod as measured by a stationary observer.
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The dog runs horizontally off the end of the dock, so the initial components of the velocity are (vx)i = 8.5 m/s and (vy)i = 0 m/s. We can treat this as a projectile motion problem, so we can use the
details and equations presented in Synthesis 3.1 above.
Would it possible to show the initial position of the projectile, its velocity and the angle the velocity makes with the line joining it and the center of the earth and then give the option to
Chapter 6 Work and Kinetic Energy … horizontal table. The box starts at rest and ends at rest. … 12 •• A hockey puck has an initial velocity in the +x …
Calculate the horizontal & vertical components of the . initial. velocity. Calculate the stone’s maximum height . above the top of the building. Calculate the time the stone takes to reach this
height. Calculate the time it takes to go up, come down & again reach the height it. started from (45 m . above the ground; where the dashed curve …
Learn about them by typing C-h m when the cursor is in the Bufer List window. Recover data from an edited buffer: If Emacs crashed, do not despair. Start a new Emacs and type M-x recover-file and
follow the instructions. The command M-x recover-session recovers all unsaved buffers.
Secret is isolating the object and the forces on it. Consider a block of mass m on a frictionless horizontal surface being pulled with a string by a force F. Forces on block: 1. weight of the block,
w. 2. Contact force Fn exerted by table onto the block. (friction = 0, therefore Fn is perpendicular to table. 3.
A ball is thrown horizontally from the top of a tower with a velocity 10 m/s. the height of the tower is 45 m . Calculate (i) time to reach ground, (ii) horizontal distance covered by the body (iii)
the direction of the ball when it just hits the ground. Hints: (i) Consider vertical downward motion and determine time.
Velocity and direction of motion at a given height: At a height h, Vx = ucos And Vy = 2 Resultant velocity v = + ; v = 2 Note that this is the velocity that a particle would have at height h if it is
projected vertically from ground with u.
As the angle increases, the range (horizontal distance that the shuttlecock travels) increases and vice versa. However, the speed of release has to stays the same. For example, if the angle of
release is 30 degrees, the initial velocity is 15 m/s and the time is 2 seconds
Problem 2: A projectile is launched from point O at an angle of 22 with an initial velocity of 15 m/s up an incline plane that makes an angle of 10 with the horizontal. The projectile hits the
incline plane at point M. a) Find the time it takes for the projectile to hit the incline plane. b)Find the distance OM. Solution to Problem 2.
Velocity of a Ball When it Hits the Ground. … because the ball falls starting from its maximum height, the initial velocity is $0$. Therefore, the equation becomes …
(b)Calculate the velocity of the dart as it leaves the gun (give answer in m/s). 7.3 m/s 12.A projectile is shot from the ground with an initial velocity of 100 m/s at an angle of 40 above the
Therefore, its velocity just before landing is ˆ + ( – 4 m/s ) y ˆ. v = ( 2 m/s ) x Maximum height depends on the initial speed squared. Therefore, to reach twice the height, projectile 1 must have
an initial speed that is the square root of 2 times greater than the initial speed of projectile 2.
C 47. Instantaneous speed is the slope of the line at that point. B 48. A non-zero accleeration is inidcated by a curve in the line D 2 2 49. Maximum height of a projectile is found from vy = 0 m/s
at max height and (0 m/s) = v + 2gh and gives h = v2/2g.
physics-linear kinematics. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads.
Linear motion topics 44 A cannonball of mass 5 kg is fired into the air at an angle of 60o with an initial velocity of 500.0 m/s. Exactly ten seconds later, it reaches its maximum height of 400 m at
point A with a velocity of 100.0 m/s to the right.
Determine the range of a projectile that has an initial speed of vo = 100m/s and is fired at angles of a. 10° b. 20° c. 30° d. 40° e. 50° f. 60° g. 70° h. 80° 20.
velocity of the projectile. The projectile is propelled by a pressurized gas chamber. If the gas expands isothermally the pressure p in the tank (together with the portion of the barrel behind the
projectile) is related to its volume V by Boyle’s law . pV =constant . The external air has pressure . p. a. Friction can be neglected. 1.1.
Velocity, V(t) is the derivative of position (height, in this problem), and acceleration, A(t), is the derivative of velocity. Thus Thus The graphs of the yo-yo’s height, velocity, and acceleration
functions from 0 to 4 seconds.
Analysis of Projectile Motion Horizontal motion No horizontal acceleration Horizontal velocity (vx) is constant. How would the horizontal distance traveled change during successive time intervals of
0.1 s each? Horizontal motion of a projectile launched at an angle: Analysis of Projectile Motion Vertical motion is simple free fall. Acceleration (ag) is a constant -9.81 m/s2 . Vertical velocity
A cannon ball is shot from the ground with an initial velocity . v0 = 42 m/s. at an angle . θ0 = 55° with the horizontal. It lands on top of a nearby building of height . h = 52 m . above the ground.
Neglect air resistance. To answer this, take . x = y = 0. where the ball is shot. It is probably best to take the upward direction as positive …
What is the horizontal distance which the projectile travels before it hits the ground What is the direction of the takeoff velocity vector Suppose a 64.0 kg boy and a 48 kg girl use a massless rope
in a tug-of-war on an icy, resistance-free surface.
Incidently, on the second impact I will add on a horizontal velocity of Cr.Vx2 where Vx2 is the velocity when the pendulum returns to the new part of the wedge. Cr is the coefficient of restitution.
I expect the new Vx = VyTan(Q) + Cr.Vx2
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) Its initial velocity and (b)
The height it reaches.
(4 ed) 8.2 A 3.0-kg block starts at a height h = 60 cm = 0.60 m on a plane that has an inclination angle of 30 o as in Figure P8.20. Upon reaching the bottom, the block slides along a horizontal
Vertical launch Horizontal launch Non-horizontal but only if we are given horiz and vertical velocities Our equations are strictly x direction, strictly y direction How can we figure out the x and y
initial velocities of a potato launched with some initial velocity at an angle? Velocity in x dir vx = ? Initial vel in y dir vy i = ?
Physics > General Physics: $9.00: Past due foundations of physics An object starts from rest at x=0 m at time t=0 s. Five seconds later the object is observed to be at x=40.0 m and to have velocity
vx = m/s a) Was the object's acceleration uniform or nonuniform? explain your reasoning. b) Sketch the velocity vs time graph implied by these data.
at which it was hit. It lands with a velocity of 36 m/s at an angle of 28° below the horizontal. Ignoring air resistance, find the initial velocity with which the ball leaves the bat. This has two
dimensions – the ball changes height during the movement as well as “covers ground”
Fig 2.1 shows a ball kicked from the top of a cliff with a horizontal velocity of 5.6ms-1. Air resistance can be neglected. i) Show that after 0.90s the vertical component of the velocity is 8.8ms-1.
VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST These equations assume that air resistance is negligible, and apply only when the initial vertical velocity is zero. On Earth's surface, ay=-g=
-9.81 mil. HORIZONTAL MOTION OF A PROJECTILE These equations assume that air resistance is negligible. Vy,f = ay~t 2 Vy,f = 2ay~Y ~y = ~ay(~t)2
A projectile with an initial velocity can be written as The horizontal motion has zero acceleration, and the vertical motion has a constant downward acceleration of – g. jv iv v y 0 x 0 0 + = 0v0 0 0
0 0x0 sin v v and cos v v u u = = y27 The range R is the horizontal distance the projectile has traveled when it returns to its launch height. 28 …
Now we can readily write down their positions at any time, given the starting height h of the first ball and the initial velocity v0 of the second: 1 x1 (t) = h − gt2 2 1 x2 (t) = v0 t − gt2 2 First
we can find the time when the second ball is at rest (23) (24) v2 (t) = =⇒ t= v0 g dx2 = v0 − gt = 0 dt (25) (26) At this time, the …
Initial Horizontal Velocity. … The total distance covered by the projectile in horizontal direction (X-axis) is called is range … Consider a body of mass 'm …
A projectile was launched at 10 m/s at an angle of 55 deg from ground level (h = 0). A second projectile was fired with the same speed from the same location at a different angle and landed at the
same position.
height. Find (a) the ball’s initial velocity and (b) the height it reaches. g. Why is the following situation impossible? A freight train is lumbering along at a constant speed of 16.0 m/s. Behind
the freight train on the same track is a passenger train traveling in the same direction at 40.0 m/s. When the front
Physics > General Physics: $9.00: Past due foundations of physics An object starts from rest at x=0 m at time t=0 s. Five seconds later the object is observed to be at x=40.0 m and to have velocity
vx = m/s a) Was the object's acceleration uniform or nonuniform? explain your reasoning. b) Sketch the velocity vs time graph implied by these data.
S In the horizontal direction, vx 0 1.8 m sand ax 0 In the vertical direction t 3.0 s vy 0 0 ay 9.80 m s2 y0 0 y y0 v y 0 t 12 a y t 2 y 0 0 1 2 9.80 m s 3.0 s 2 2 44 m The distance from the base of
the cliff to where the diver hits the water is found from the horizontal motion at constant velocity: x vxt 1.8m s 3 s 5.4 m S A ball thrown horizontally at 22.2 m/s from the roof of a building lands
36.0 m from the base of the building.
Chapter 23 Solutions … directed to the right about 30.0° below the horizontal. O: … where m is the mass of the object with charge …
(ii) Consider horizontal motion and determine horizontal range.(iii) Determine vertical velocity Vy and horizontal velocity Vx separately and angle with the horizontal line is given by . 45. A
projectile is fired with a velocity 320 m / s at an angle 300 to the horizon.
So the initial velocity is given by 0.137 m u 0.023ms 1 0.32 s 5.67 s Comment: To solve projectile motions problems the first step is to resolve the motion into horizontal and vertical directions.
Then for the motion in each direction, just use the three equations to find the unkown.
Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move
the man for you.
Velocity and direction of motion at a given height: At a height ‘h’, Vx = ucos … mass 2 kg has an initial velocity of 3 m/s … A projectile is fired at 30o to …
3. Construct a table or spreadsheet for recording data from all the trial throws. Record all your calculations in the table. Data and Observations Range (R) Time (t) Horizontal Vertical Initial
(meters) (seconds) Velocity Velocity Velocity (vx ) (m/s) (vy ) (m/s) (v0) (m/s) Trial 1 Trial 2 162 Forces and Motions in Two Dimensions Apply 1.
Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 04 . 20 232 0. TÀI LIỆU 123 Gửi tin nhắn Báo tài liệu vi phạm.
0.85 m/s 0.89 m/s 0.77 m/s 0.64 m/s 0.52 m/s A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 1.2 kN/m). The block is initially at rest at its
equilibrium position when a force (magnitude P) acting parallel to the surface is applied to the block, as shown.
37 The Trajectories of Large Fire Fighting Jets A. P. Hattont and M. J. Osborne:]: This article describes a computer simulation of the trajectories of large water jets which allow the effects of
changes in initial velocity, elevation, nozzle diameter, and head and tail winds to be examined.
5 randall d knight physics for scientists and engineers a strategic approach with modern physics 05 … mass, and vx is in m/s That is, the square of the car’s …
(b) In part (a) of this problem, the initial horizontal velocity was determined to be 37.751 m/s. For projectiles, this horizontal velocity does not change during the flight of the projectile. Thus,
the projectile strikes the balcony moving with a final horizontal velocity (vfx) of 37.751 m/s.
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Model Math
Computers do differentiation much better than us, here’s why
The more the merrier (especially dimensions)
f(Linear Algebra) -> Neural Networks
Going deeper into neural networks
Convolutional Nerual Networks (math that gives vision)
It just got more convoluted
The problem of memory loss, and it’s cure
Pay attention, from the start
A match made in heaven, keys and values
From discrete -> continuous
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Center of Gravity
Placement of Center of Gravity
for Best Spin and Launch Angle
Dave Tutelman -- August 16, 2013
Here is the analysis behind the spreadsheet. Feel free to skip this if you're not interested in the physics and math.
Calculations for spin
x (horizontal along the target line) and z (vertical). We will set the origin at the point on the clubface where the nominal loft is measured. If we talk about a 10.5º clubhead, then [0,0] is the
point on the clubface where the loft is actually 10.5º. (Because of face roll, there is only one height on the clubface where the loft is exactly 10.5º.)
Within this coordinate system, we can begin to describe the parameters of our analysis.
• We designate nominal loft as L[o]. The [0,0] point is where the black line at an angle of 90º-L[o] is tangent to the blue clubface.
• The placement of the CG is at [X, Z]. X is a depth from the face, and Z is height above the nominal-loft height. A low CG (as in the diagram) corresponds to a negative Z.
• Impact occurs h above the nominal loft, measured along the clubface. If it occurs lower on the clubface than nominal loft, h is negative.
• The effective loft where impact occurs is L[i]. The diagram shows L[i] as incorporating the effect of face roll, but it will also include clubhead rotation due to shaft bend. The face roll is
specified by a radius of R. In most driver heads, R is between 10 and 14 inches.
C and y, the quantities that will determine the spin. Our strategy will be to find the [x,z] coordinates of two points in space, [i,j] and [s,t]. Then:
• C is the distance between [i,j] and [s,t].
• y is the distance between [s,t] and [X,Z].
Let's go through the steps.
First compute L[i].
L[i] = L[o] + arcsin (h/R) + 1.5 (X-F)
The arcsin(h/R) accounts for face roll for impact h inches above nominal. The "1.5" factor accounts for loft change due to shaft bend, explained later on this page.
Once we have the effective loft, we use the "standard" formula to get the launch angle.
A = L[i] (0.96 - .0071 L[i])
We can also get good values for [i,j].
i = h sin (L[o] + 1.5(X-F)) + h^2/2R
j = h cos (L[o] + 1.5(X-F))
In each case, the sine-cosine factor accounts for position along the black line. The h^2 factor for i accounts for the small space between the black line and the blue curve at impact. (We aren't
going to worry about the tiny difference in j due to the launch angle.)
Now let's get values for [s,t]. This is a little harder, but we can use the fact that C is measured along the launch line and y is perpendicular to the launch line. Let's start by defining the slope
of the launch angle as m, then use the point-slope method (remember, back in high school algebra?) to define the lines of C and y. This is why we want a coordinate system; we can treat the whole
diagram like a graph.
m = tan (A)
Now we can write the equations for the two green lines. The equations are easily found using the point-slope form. The line for C (along the launch line) goes through the point [i,j] at slope -m. The
line for y goes through the point [X,Z] at slope 1/m (the perpendicular to -m). So the lines in point-slope form are:
z - j = -m (x - i)
z - Z = 1/m (x - X)
Next we need to find the point [s,t]. We know that the two green lines intersect at [s,t]. So let's solve the two equations for x and z, and set those values to s and t. The solution is:
s = (m / (m^2+1)) ( j + mi - Z + X/m )
t = m (i-s) + j
So now we know the coordinates of the points [i,j], [s,t], and [X,Z]. What we really want is C and y, but those are just distances between the points. We find those distances with the Pythagorean
C^2 = (s-i)^2 + (t-j)^2
y^2 = (s-X)^2 + (t-Z)^2
However, y can be negative as well as positive. The Pythagorean Theorem gives the magnitude but not the sign. So let's use trigonometry instead.
y = (t-Z) / cos(A)
Now we have everything we need to find the spins, using formulas from basic impact and gear effect.
V[ball] = V[club] * SmashFactor * cos (L[i])
SPIN[GE] = 58830 * V[ball] * C * y / I[v]
SPIN[loft] = 160 * V[club] * sin (L[i])
SPIN = SPIN[loft] - SPIN[GE]
This is very close to the calculations in the spreadsheet. (We took a few shortcuts and consolidations in the spreadsheet.)
Calculations for launch angle
Head rotation from gear effect
(This is the weakest rationale for any of the calculations in this article. I assert here that the mechanism at work is the change in loft due to the rotation of the clubhead during impact. If there
is some other mechanism at work, the conclusions here may be wrong.)
big gear that drives the little gear -- the ball. So the head will spin at a rate proportional to the ball's spin rate (just the gear effect spin) and also proportional to the "gear ratio", the ratio
of diameters. Expressed as an equation:
BallSpin * BallRadius = HeadSpin * HeadRadius
What are these equal to?
• BallSpin: we computed that already as S[GE]. We'll use the number we computed.
• BallRadius: 0.84 inches, controlled by the Rules of Golf. (Actually, the Rules controls the diameter at 1.68 inches.)
• HeadSpin: what we want to find.
• HeadRadius: by definition, the distance from the center of rotation (the CG) to the face. We recognize it as the computed value of C.
So the equation becomes:
The unit is RPM. We're looking for loft change, and we measure loft in degrees. Actually, degrees per second would be a much more convenient unit for HeadSpin. So we want to multiply by 360 (to get
to degrees) and divide by 60 (to get to seconds). The equation becomes:
S[h] = 0.84 * 360 S[GE] = 5.04 S[GE]
60 C C
S[h] is a spin rate, an angular velocity, measured in degrees per second. But what we really want is a total rotation in degrees. That is the amount that the loft can be considered to change, thus
changing the launch angle. Just as distance is the accumulation of velocity ("integration" in calculus terms), the rotation is the accumulation of angular velocity. So we need to know the angular
velocity function during impact, and integrate it to get the rotation of the clubhead.
The angular velocity does not jump to S[h] from the beginning of impact, just as ball spin doesn't jump instantly to S[GE]. Let's assume it climbs on a straight line. This is a plausible
approximation, though it is more likely a slight S-curve; that is, the angular acceleration on the gear pair -- the clubhead and ball -- probably builds as the ball gradually compresses and falls off
as the ball releases. But let's keep things simple and assume a straight-line increase.
S[h] when the ball releases .0004 seconds later. The equation for the straight-line angular velocity function is:
In order to find the angle of rotation at any time T after the beginning of impact, we have to integrate the angular velocity.
θ(T) = ∫ S[h].0004 t dt
= S[h].0008 T^2
The function for angle of rotation θ(T) is the red curve. We can see from the equation and the graph that it is a square-law parabola. At the moment of separation (T=.0004 sec), the clubhead has
rotated .0002S[h] degrees.
But is that what we really want? The loft as the ball leaves is not likely to be the angle that most affected the launch angle; the effective loft is probably somewhere in the middle of impact, e.g.
the point of maximum compression of the ball on the clubface. So what is the clubhead rotation in the middle of impact? That is T=.0002 seconds. We evaluate it as:
θ(T=.0002) = S[h].0008 (.0002)^[2] = .00005 S[h]
This is only a quarter of the rotation we found for separation. Not a half, but a quarter. If you're familiar with quadratic functions -- or even take a good look at the red curve in the graph --
this is not surprising.
We will use this value as the loft change due to gear effect, as we estimate the change in launch angle. If you want to see what a higher fraction of the head rotation would do, the spreadsheet
allows the fraction as a parameter; the default is 0.25, but you can adjust it.
Head rotation from centrifugal pull
Jeff Summitt has given me a rule of thumb that he says works over a variety of clubheads and shafts. The clubhead rotates to increase the loft by 0.15º for each 0.1 inch increase in the moment arm.
That is a rate of 1.5º per inch, though Jeff is quick to remind me that moving the CG by an inch is huge. Yes, that rate varies a bit with the velocity, clubhead mass, and shaft stiffness, but the
difference is small enough that you can ignore the variation.
LoftChange = 1.5º * MomentArm
• LoftChange in degrees
• MomentArm in inches
• v in miles per hour
If we want more resolution than this, the formula has to account for:
• Velocity, which is a square-law factor in centrifugal force.
• Radius of the clubhead's path at the moment of impact. That is typically somewhat more than the club length -- maybe 25% more for a full swing with good release.
• Bending properties of the shaft. Not just the overall stiffness, but the flex profile over the length of the shaft. This can be characterized by an EI profile, a frequency profile, or a
deflection profile -- and each has its own computational difficulties. The EI profile is probably the most elementary flex profile for use in a computation, but is rather rare to have around for
a given shaft. (Russ Ryden has been characterizing shafts by EI profile.)
So there you have it, a set of calculations embodied in a spreadsheet, to compute the effect of CG placement on spin and launch angle.
Last updated - Mar 10, 2014
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Explain how filter and find functions work on lists in functional programming. | TutorChase
Explain how filter and find functions work on lists in functional programming.
In functional programming, filter and find functions are used to manipulate lists based on certain conditions.
The filter function is a higher-order function that processes a data list based on a given condition. It takes two arguments: a function and a list. The function is applied to each element of the
list. If the function returns true for an element, that element is included in the new list. If the function returns false, the element is excluded. The filter function returns a new list and does
not modify the original list. This is in line with the principles of functional programming, where data is immutable and functions have no side effects.
For example, consider a list of numbers and a function that checks if a number is even. If you apply the filter function with this condition to the list, you will get a new list that only contains
the even numbers from the original list.
The find function, on the other hand, is used to find the first element in a list that satisfies a certain condition. Like the filter function, it also takes a function and a list as arguments. The
function is applied to each element of the list in order, and as soon as an element is found for which the function returns true, that element is returned by the find function. If no such element is
found, the find function returns a special value, often null or undefined.
For instance, if you have a list of numbers and a function that checks if a number is greater than 10, the find function will return the first number in the list that is greater than 10. If there are
no numbers greater than 10 in the list, the function will return null or undefined.
In summary, filter and find are powerful functions in functional programming that allow you to manipulate lists based on certain conditions. They embody the principles of functional programming by
being stateless and producing no side effects.
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Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 2)
You're reading: Integer Sequence Review
Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 2)
Last week A002210, the decimal expansion of Khintchine’s constant, emerged victorious from Bracket 1. Now, get ready for round 2 of…
Here are the rules: we’re judging each sequence on four axes: Aesthetics, Completeness, Explicability, and Novelty. We’re reviewing six sequences each week for four weeks, picking a winner from each.
Then, we’ll pick one sequence from the ones we reviewed individually before this thing started, plus a wildcard. Finally, a single sequence will be crowned the Integest Sequence 2013!
Weird numbers: abundant (A005101) but not pseudoperfect (A005835).
70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, 13790, 13930, 14770, 15610, 15890, 16030, 16310, 16730, 16870, 17272, 17570, 17990, 18410, 18830, 18970, 19390, 19670, ...
Christian: Lots of mumbo-jumbo. What are abundant and pseudoperfect, apart from my virtues, and imitations of myself, respectively?
David: I… do you want me to take a guess, or do you want me to look it up?
Christian: Take a guess and we’ll see if you’re right.
David: I would say that they were numbers the sum of whose divisors is more than one greater than themselves. Because abundant for numbers the sum is just greater than the original, while for
pseudoperfect numbers it’s exactly one more.
Christian: Survey says…
David: Ooh! A number is pseudoperfect if the sum of some of its divisors is equal to the original number. I like pseudoperfect numbers. But unfortunately, this sequence doesn’t contain any. I’m not a
fan of this sequence. And abundant numbers are just greedy.
Christian: I’m fairly sure greedy numbers already exist. Anyway, we’ve spent too long on this sequence already. Explicability: 2?
David: It’s not too bad, actually. 3.
Christian: Aesthetics?
David: I know I say it a lot, but the last digit of each number is really annoying me. It’s pretty much a zero all the time.
Christian: I don’t like it much at all. I can go 2.
David: Yeah.
Christian: Completeness and Novelty both 3. Happy?
Aesthetics $\frac{2}{5}$
Completeness $\frac{3}{5}$
Explicability $\frac{3}{5}$
Novelty $\frac{3}{5}$
TOTAL $\frac{11}{20}$
Decimal expansion of number with continued fraction expansion 2, 3, 5, 7, 11, 13, 17, 19, … = 2.3130367364335829063839516 …
2, 3, 1, 3, 0, 3, 6, 7, 3, 6, 4, 3, 3, 5, 8, 2, 9, 0, 6, 3, 8, 3, 9, 5, 1, 6, 0, 2, 6, 4, 1, 7, 8, 2, 4, 7, 6, 3, 9, 6, 6, 8, 9, 7, 7, 1, 8, 0, 3, 2, 5, 6, 3, 4, 0, 2, 1, 0, 1, 2, 4, 4, 4, 2, 1, 4, 4, 5, 6, 4, 7, 3, 1, 7, 7, 6, 2, 7, 2, 2, 4, 3, 6, 9, 5, 3, 2, 2, 0, 1, 7, 2, 3, 8, 3, 2, 8, 1, 7, 5, ...
Christian: ANOTHER decimal expansion. Zero for everything.
David: It’s a pretty cool constant though, isn’t it?
Christian: Yes, it’s eerily close to Rényi’s parking constant.
David: What’s that and why aren’t we reviewing it?
Christian: It’s the proportion of space typically left empty on a street when it fill up with parked cars.
David: Does this sequence have any cool facts like that?
Christian: I actually have a paper about this saved somewhere… “Continued fractions constructed from prime numbers” by Marek Wolf.
David: Should we read it or give this a rubbish score?
Christian: I’ll just check… Oh! That paper gives a different number! In this one the two is an integer, while –
Aesthetics $\frac{3}{5}$
Completeness $\frac{4}{5}$
Explicability $\frac{4}{5}$
Novelty $\frac{1}{5}$
TOTAL $\frac{12}{20} = \frac{3}{5}$
Wieferich primes: primes p such that $p^2$ divides $2^{p-1} – 1$.
1093, 3511, ...?
Christian: WHAT. IS. THIS.
David: From Fermat’s little theorem, every prime $p$ divides $2^{p-1}-1$. So, it’s a natural question to ask when $p^2$ does. Turns out, not very often.
Christian: Do you know if this is finite?
David: No. In fact, we don’t even know if there’s a point after which every prime belongs to the sequence.
Christian: Does the OEIS say how far has been checked?
David: $6.7 \times 10^{15}$.
(ring ring)
Christian: Hold on, it’s Doron Zeilberger on the phone… he says that’s proof enough that the list is finite.
David: It’s conjectured that the number of Wieferich primes less than $x$ is approximately $\log \log x$. And $\log \log (10^{15})$ is approximately… 3.5. So we aren’t too far out. The point is they
don’t appear very often, but they do appear. Possibly.
Christian: SCORES DAVID.
David: LOTS AND LOTS OF SCORES CHRISTIAN.
Aesthetics $\frac{5}{5}$
Completeness $\frac{1}{5}$
Explicability $\frac{5}{5}$
Novelty $\frac{4}{5}$
TOTAL $\frac{15}{20} = \frac{3}{4}$
1, 3, 8, 18, 38, 88, 188, 388, 888, 1888, 3888, 8888, 18888, 38888, 88888, 188888, 388888, 888888, 1888888, 3888888, 8888888, 18888888, 38888888, 88888888, 188888888, 388888888, 888888888, 1888888888, 3888888888, 8888888888, ...
Christian: I. Like. Big. Butts and I can not lie. If I was a bingo caller, I’d describe this sequence as “Bigg market hen party”.
David: What’s this? What’s A051109?
Christian: A051109 is the hyperinflation sequence for banknotes. It’s the denominations you’d issue if your currency was losing value very very quickly. This sequence gives the smallest amounts of
money that can only be made with at least $n$ notes.
David: I hate it.
Christian: Why?
David: I hate both the definition of the original sequence, sequences which are partial sums of other sequences, and that number of 8s is very very nauseating.
Christian: They please me. I refer you to my previous comments re big butts.
David: This is a family show, Christian. Did you know that teachers in Australia are encouraged to show this to their students?
Christian: It slipped my mind. I hereby retract all previous statements on the subject of backsides and the bigness thereof.
David: Scores. Aesthetics: 1. I feel physically sick looking at it.
Christian: Do you suffer from octophobia? I’ll respect your wishes and go for a 2.
David: Completeness: 1.
Christian: Why?
David: Why not?
Christian: Because it’s complete!
David: Complete with ugly things! Explicability: 2.
Christian: I give up. I don’t care enough about this sequence. Straight 2s?
Aesthetics $\frac{2}{5}$
Completeness $\frac{2}{5}$
Explicability $\frac{2}{5}$
Novelty $\frac{2}{5}$
TOTAL $\frac{8}{20} = \frac{2}{5}$
David: It adds up to 8! That is beautiful.
Christian: Normally we’d do some chicanery based on that, but we’ve got more work to do.
$a(n)=n \times \frac{n+7}{2}$.
0, 4, 9, 15, 22, 30, 39, 49, 60, 72, 85, 99, 114, 130, 147, 165, 184, 204, 225, 247, 270, 294, 319, 345, 372, 400, 429, 459, 490, 522, 555, 589, 624, 660, 697, 735, 774, 814, 855, 897, 940, 984, 1029, 1075, 1122, 1170, 1219, 1269, 1320, 1372, 1425, 1479, ...
Christian: This is rubbish.
David: Can I pick which ones to review next week?
Christian: I’m fairly sure you picked this one.
David: Pretty sure I didn’t. Why would I pick drivel like this? Is this like when Jedward got pretty far through the X Factor and nobody knew how?
Christian: OK, zero aesthetics, zero explicability?
David: What is it?
Christian: A comment on the OEIS says,
If $X$ is an $n$-set and $Y$ a fixed $(n-4)$-subset of $X$ then $a(n-3)$ is equal to the number of $2$-subsets of $X$ intersecting $Y$.
David: Is zero not a bit generous?
Christian: Sorry, I fell asleep typing that out. Let’s write some small numbers in the boxes.
Aesthetics $\frac{0}{5}$
Completeness $\frac{5}{5}$
Explicability $\frac{0}{5}$
Novelty $\frac{0}{5}$
TOTAL $\frac{5}{20} = \frac{1}{4}$
Primes with $n$ consecutive digits ascending beginning with the digit two.
1, 2, 8, 82, 118, 158, 2122, 2242, 2388, ...?
David: I’ve made a huge mistake. I think the sequence we should be reviewing is a sequence created by this one. I’ll review this one anyway.
Christian: Go on then.
David: Write out 234567890123456… and stop whenever you get a prime. Count the number of digits in that prime. That’s a number in the sequence.
Christian: 5 for Explicability then.
David: Is it 5 for Explicability? I struggled to come up with good words.
Christian: But you done speak good in end David. Aesthetics? I don’t like that they’re all even.
David: I don’t either! I prefer the sequence which was the actual primes –
Christian: A089987?
David: No! That’s a truncation of Champernowne’s constant. Slightly different to what we’re after, but it is sexy nonetheless.
Christian: You should submit your one. Anyway, Novelty?
David: One. Because there are seven more sequences!
Christian: Don’t you mean eight? There’s one for each digit numbers can start with.
David: The sequence with first digit 1 isn’t there!
Christian: Let’s not think about why that might be. So a low Novelty score. Completeness?
David: The OEIS doesn’t mention any facts about it, so we’ll have to assume it might be infinitely big, so it might not be complete.
Christian: OK. I think we can do this.
Aesthetics $\frac{2}{5}$
Completeness $\frac{2}{5}$
Explicability $\frac{4}{5}$
Novelty $\frac{2}{5}$
TOTAL $\frac{10}{20} = \frac{1}{2}$
And the winner is…
A001220, the Wieferich primes!
A001220 advances to the final with a decent score of $\frac{3}{4}$.
We’ll be back with Bracket 3 next week. In the mean time, please leave suggestions for sequences you think we should review in the comments.
6 Responses to “Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 2)”
1. Florian
Please keep doing these, guys. It’s so rare these days that people make me laugh and learn at the same time.
2. Andrei
What does “Completeness” mean?
□ Christian Perfect
That’s a very good question. For finite sequences, I suppose it’s something like how many of the terms are in the OEIS. For infinite sequences, we’ve been giving low Completeness scores if
the OEIS doesn’t have enough terms to satisfy us.
And, as you see here, if the OEIS doesn’t say if a sequence is finite or infinite, we award very few points.
But, that being said, we’ve basically been assigning the Completeness score at random.
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Area under curve Calculators | List of Area under curve Calculators
List of Area under curve Calculators
Area under curve calculators give you a list of online Area under curve calculators. A tool perform calculations on the concepts and applications for Area under curve calculations.
These calculators will be useful for everyone and save time with the complex procedure involved to obtain the calculation results. You can also download, share as well as print the list of Area under
curve calculators with all the formulas.
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Understanding the Commutative Property: A Deep Dive into Mathematics - towtimenews.co.uk
Understanding the Commutative Property: A Deep Dive into Mathematics
Mathematics is a fascinating world where patterns and properties govern the relationships between numbers. One of the fundamental concepts in this realm is the commutative property. Whether you’re a
student trying to grasp basic math concepts or an adult looking to brush up on your knowledge, understanding the can enrich your comprehension of mathematical operations. Let’s explore this property
in depth.
What is the Commutative Property?
The commutative property is a foundational principle in mathematics that states that the order in which two numbers are added or multiplied does not affect the outcome. In simpler terms, if you have
two numbers, say aaa and bbb, the tells us that:
• For addition: a+b=b+aa + b = b + aa+b=b+a
• For multiplication: a×b=b×aa \times b = b \times aa×b=b×a
This property is not just a quirk of math; it’s a crucial building block that supports a vast array of mathematical concepts and operations.
A Closer Look at Addition
Let’s take a closer look at addiction and how the commutative property plays a role. Imagine you’re in a café, and you’re trying to calculate the total cost of your drinks. If you ordered a coffee
for $3 and a pastry for $2, you can add them in either order:
1. 3+2=53 + 2 = 53+2=5
2. 2+3=52 + 3 = 52+3=5
No matter how you slice it, the total remains $5. This is the essence of the in action. It’s a handy feature that simplifies calculations and helps us to think flexibly about numbers.
Multiplication Made Easy
The commutative property also extends to multiplication. Let’s stick with the café example. If you decide to buy two coffees and three pastries, you can calculate the total in different ways:
1. 2×3=62 \times 3 = 62×3=6
2. 3×2=63 \times 2 = 63×2=6
Again, the total remains the same. This flexibility is especially helpful in more complex calculations where rearranging numbers can make the math easier to manage.
Why is the Commutative Property Important?
Understanding the commutative property is crucial for several reasons. First, it enhances numerical fluency, making it easier for individuals to perform mental calculations. When you know you can
rearrange numbers, you can often find simpler ways to solve problems.
Building Blocks for Advanced Math
The commutative property lays the groundwork for more advanced mathematical concepts. For example, when dealing with algebraic expressions, recognizing that you can rearrange terms allows you to
simplify expressions more efficiently. This is particularly valuable in higher mathematics, where complex equations are the norm.
Enhancing Problem-Solving Skills
Moreover, embracing the cultivates better problem-solving skills. When faced with a complex equation, being able to rearrange components can open up new pathways to solutions that might not be
immediately obvious. This flexibility is invaluable in mathematics and related fields, like engineering and computer science.
Commutative Property in Everyday Life
You might be surprised to learn that the commutative property isn’t just a dry mathematical principle; it shows up in our everyday lives more often than we realize.
Shopping and Budgeting
Consider budgeting. When you list your expenses, it doesn’t matter in which order you add them up. If you spend $20 on groceries and $15 on a dinner out, your total remains consistent no matter how
you calculate it:
1. $20 + $15 = $35
2. $15 + $20 = $35
This consistency is reassuring when you’re managing finances, as it allows for flexible thinking about your spending.
Cooking and Baking
Cooking also utilizes the commutative property. Suppose you’re making a fruit salad with apples, bananas, and oranges. It doesn’t matter in which order you add the fruits to the bowl; the final
product will be the same delicious salad.
Sports Statistics
In sports, the commutative property can come into play when calculating points or scores. For example, if a player scores 5 points in the first quarter and 3 in the second, it doesn’t matter how you
add those points; the total remains the same, providing a clear picture of their performance.
Limitations of the Commutative Property
While the commutative property is widely applicable, it does have its limitations. It’s essential to recognize where this property applies and where it doesn’t.
Non-Commutative Operations
The commutative property does not apply to all mathematical operations. For instance, subtraction and division are non-commutative. If you have a=5a = 5a=5 and b=3b = 3b=3:
1. 5−3≠3−55 – 3 \neq 3 – 55−3=3−5 (the results are different)
2. 5÷3≠3÷55 \div 3 \neq 3 \div 55÷3=3÷5
Understanding these exceptions is crucial for accurate calculations and reinforces the need for careful consideration when performing operations.
Complex Numbers
In more advanced mathematics, such as with matrices or certain algebraic structures, the commutative property may not hold. For example, matrix multiplication is generally non-commutative, meaning
that the order in which you multiply matrices matters significantly.
Real-World Applications
Recognizing the limitations of the commutative property can also enhance critical thinking and problem-solving skills. In fields like physics or engineering, knowing when the does not apply can
prevent errors in calculations that could have serious implications.
Teaching the Commutative Property
For educators, conveying the concept of the commutative property can be both fun and engaging.
Interactive Activities
One effective way to teach this concept is through interactive activities. For example, using physical objects like blocks or counters can help students visualize the idea. By physically rearranging
objects, students can see that the total remains constant regardless of how they group or arrange them.
Games and Challenges
Incorporating games can also make learning the commutative property enjoyable. Math puzzles that require students to rearrange numbers to achieve the same sum or product can playfully reinforce the
Real-World Contexts
Using real-world contexts, like shopping scenarios or cooking, can help students relate to the commutative property. When students see the relevance of math in their daily lives, they are more likely
to engage with and understand the material.
The Commutative Property in Algebra
As students progress in their mathematical education, the commutative property becomes increasingly important in algebra.
Simplifying Expressions
In algebra, students often encounter expressions that require simplification. The ability to rearrange terms, such as 3x+5x3x + 5x3x+5x to 5x+3x5x + 3x5x+3x, leverages the commutative property to
streamline calculations. This skill is vital as students tackle more complex equations.
Solving Equations
When solving equations, recognizing that you can rearrange terms can lead to quicker solutions. For instance, if you have an equation like 2x+3=5+x2x + 3 = 5 + x2x+3=5+x, knowing you can rearrange
and combine as terms will aid in finding the value of xxx.
Graphing and Functions
The commutative property also plays a role in graphing functions. When plotting points, the order in which you label the axes or the sequence of points plotted doesn’t change the overall
representation. This property is particularly useful when visualizing algebraic concepts.
Commutative Property vs. Associative Property
While the commutative property is a key mathematical concept, it is essential to differentiate it from the associative property.
Defining the Associative Property
The associative property deals with grouping rather than order. For addition and multiplication, the associative property states:
• For addition: (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)(a+b)+c=a+(b+c)
• For multiplication: (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c)(a×b)×c=a×(b×c)
Comparing the Two Properties
While both properties allow for flexibility in calculations, they serve different purposes. The commutative property focuses on the order of numbers, whereas the associative property emphasizes how
numbers are grouped. Understanding these distinctions is crucial for mastering more complex mathematical concepts.
Practical Implications
In practical terms, knowing both properties can simplify calculations. For example, when calculating the total cost of multiple items, you can rearrange or regroup your numbers to make the math
easier, demonstrating the effectiveness of both the commutative and associative properties in action.
Real-Life Examples of the Commutative Property
Real-life scenarios provide excellent illustrations of the commutative property.
Planning a Trip
When planning a trip, you might consider various travel routes and stops. Whether you visit the museum before the park or vice versa, your overall experience doesn’t change based on the order of the
Team Sports
In team sports, the commutative property can be observed in scoring. A basketball team can score points in any order throughout the game. The total score at the end remains the same regardless of who
scores first or how many points they contribute.
Cooking and Recipes
In cooking, you can add ingredients in any order. Whether you mix flour and sugar before adding eggs or the other way around, the final batter will remain the same, illustrating the practical
implications of the commutative property.
Conclusion: Embracing the Commutative Property
The commutative property is a cornerstone of mathematical understanding, offering flexibility and simplicity in calculations. Recognizing its applications and limitations enhances our problem-solving
skills and fosters a deeper appreciation for the beauty of mathematics. Whether you’re teaching this concept, applying it in real-life scenarios, or exploring more advanced mathematics, the
commutative property remains an essential tool in your mathematical toolkit. Embrace it, and you’ll find that math becomes a more accessible and enjoyable journey!
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What Is Big O Notation? A Beginner’s Guide to Algorithm Efficiency
Ever heard someone talk about “Big O” and thought, What in the world are they talking about?
Well, you’re not alone! Big O Notation may sound like some secret code, but it’s actually an important concept in programming.
And the good news? It’s simpler than you think once you break it down.
In this post, we’ll walk through Big O Notation in plain English. You’ll learn what it means, why it matters, and how to understand it without getting lost in a sea of math. Let’s dive in!
What Is Big O Notation?
Big O Notation is a way of talking about how fast (or slow) an algorithm runs.
It helps us measure the time it takes for a program to finish, or the space (memory) it uses, especially when the size of the input grows.
Think of it like this: Imagine you’re baking cookies. The time it takes to make a batch depends on how many cookies you want, right? If you’re making one cookie, you’re done faster than if you’re
baking 100. Big O is a way to describe how the time or space needed grows as the number of cookies (or inputs) increases.
Why Should You Care About Big O?
As a programmer, you want your code to run efficiently. Imagine creating a program that takes 10 hours to do something that could be done in 10 minutes. Not fun, right?
Big O helps you:
1. Compare algorithms: Which one is faster?
2. Predict performance: How will your program handle more data?
3. Build better programs: The goal is to write code that doesn’t slow down as things get bigger.
It’s like knowing the difference between taking a bike or a car to get across town. One will get you there faster, and Big O helps you figure that out.
A Simple Example: Sorting Socks
Let’s say you have a pile of socks. Your task is to sort them by color. You could sort them one by one, or maybe in pairs. How long will it take?
In Big O terms, we care about how your “sock-sorting algorithm” changes if you have 10 socks versus 100 socks. The bigger the sock pile, the more important the efficiency of your sorting method
The Different “Flavors” of Big O
Now, let’s look at the most common types of Big O Notation. These are just different ways to describe how an algorithm behaves as the input size grows.
=> O(1) — Constant Time
This is the best-case scenario. No matter how much data you have, the time it takes stays the same.
Imagine you have a magic sock drawer. Whenever you reach in, you pull out exactly what you want. Whether you have 10 socks or 100, it always takes you the same amount of time.
=> O(log n) — Logarithmic Time
This one sounds tricky, but it’s really not. Think of it like cutting a cake in half, then cutting one half in half again, and so on. Every step, you reduce the size of the problem by half.
You’re looking for a particular sock in a drawer that’s already sorted by color. Instead of searching one sock at a time, you split the drawer in half, then half again, until you find what you want.
=> O(n) — Linear Time
This is like normal sock sorting. You go through the pile one sock at a time. The bigger the pile, the longer it takes.
Checking every sock one by one to find the red ones.
=> O(n²) — Quadratic Time
Things slow down here. Imagine you need to compare every sock with every other sock to sort them. If you have 10 socks, you make 100 comparisons (10 x 10). If you have 100 socks, you make 10,000
comparisons. Yikes!
Comparing every sock with every other sock to see if they match.
=> O(2^n) — Exponential Time
This is the worst-case scenario. The time doubles with every extra sock. If you have 10 socks, it takes a lot longer than if you had 9.
Trying every possible way to pair your socks until you have the perfect match.
What About Space Complexity?
Space complexity is like asking, How much room do I need to sort these socks? If you need to spread them out on the floor, how much space will that take?
Some algorithms need more memory as they process more data. Others are more efficient and work in a small area, no matter how much you’re sorting.
Space complexity is written the same way as time complexity, like O(n) or O(1). It tells you how much data your program needs to store while it runs.
Why Does This Matter in Real Life?
Let’s say you’re building a website that searches through thousands of products. If your search algorithm is too slow, users will get frustrated and leave.
Or maybe you’re building a game with lots of players. If your code takes up too much memory, the game will crash or run really slowly.
Big O helps you think about these problems before they happen. It’s like planning a road trip: You want to know if you’ll get stuck in traffic before you hit the road.
Final Thoughts: Big O in Everyday Programming
Big O Notation isn’t just for computer science textbooks. It’s a way to make sure your code runs smoothly, even when things get big.
When you’re writing code, think about what happens when the input size grows. Does your program slow down? Does it take up more space? These are the questions Big O helps you answer.
So, next time someone mentions Big O, you can nod your head and say, Yep, I know about that!
Ready to Debug Your Life?
Hey there, Techie!
• Follow me on 👉 Instagram for more coding tips!
• Subscribe on 👉 YouTube for tutorials and walkthroughs!
• Like and share this pos t if you found it helpful!
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10 More 10 Less Worksheet
10 More 10 Less Worksheet - Web challenge your students to find 10 more and 10 less than a number with our 10 more and 10 less activity. Adding and subtracting 10, coloring, understanding place
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1 more, 1 less, 10 more, 10 less worksheet
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50 Ten More Ten Less Worksheet Chessmuseum Template Library
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Related Post:
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How Much Work Does Motor φ29 Need to Overcome Entropy Loss in DNA Packing?
• Thread starter emoboya3
• Start date
Step 4: Calculate the work done by the motorFinally, we can use the work-energy theorem to calculate the total work that the motor needs to do to overcome the entropy loss. The work done by the motor
is equal to the change in energy, which is equal to the change in entropy multiplied by the temperature, ΔW=ΔS*T. We can assume a temperature of 300K for this problem.In summary, to estimate the
total work that the motor \phi29 needs to perform to overcome the entropy loss of packing DNA into a capsule with a radius of 20nm, we need to calculate the end-to-end distribution of a 3D random
walk with N segments of length a, the volume
Homework Statement
Estimate total work that motor [tex]\phi[/tex]29 needs to perform to overcome entropy loss of packing DNA
Homework Equations
This equation gives the end to end distribution or a 3D random walk with N segments length a.
The Attempt at a Solution
I'm not 100% sure where to start on this. I've been trying to figure it out all day, honestly. I know that DNA has a persistence length a[tex]\approx[/tex]100nm because it's fairly rigid over about
300 base pairs. It is therefore modeled as N=65 segments.
I'm not sure where to go from here though. How do I come up with the entropy change by putting the DNA in the capsule?
I planned to model the capsule as a sphere of radius 20nm. Any help on where to go from here?
for your post and for sharing your thoughts on this problem. It's an interesting and challenging one! Let's see if we can work through it together.
First, let's define some variables to make things a bit clearer. Let's say that R is the radius of the capsule, N is the number of segments in the DNA, and a is the persistence length of the DNA. We
can also define the volume of the capsule as V=4/3πR^3. Now, let's break down the problem into smaller steps.
Step 1: Calculate the end-to-end distribution
As you mentioned, the end-to-end distribution of a 3D random walk with N segments of length a is given by the equation P(R,N)=(3/2πNa^2)^3/2e^(-3R^2/2Na^2). This equation gives us the probability of
finding the end of the DNA chain at a distance R from the starting point. We can use this equation to calculate the probability of finding the end of the DNA chain at the surface of the capsule,
which we can call P(R=Rc,N). This will give us an idea of how much DNA is packed into the capsule.
Step 2: Calculate the volume of the DNA
Now, we need to calculate the volume of the DNA that is packed into the capsule. We can do this by multiplying the number of segments N by the length of each segment a. This will give us the total
length of DNA in the capsule, which we can then multiply by the cross-sectional area of the DNA (πa^2) to get the volume. So the volume of DNA in the capsule is VDNA=Nπa^3.
Step 3: Calculate the change in entropy
Next, we need to calculate the change in entropy when we pack the DNA into the capsule. Entropy is a measure of disorder or randomness, and in this case, it is a measure of how much the DNA is
confined or restricted in its movement. To calculate the change in entropy, we can use the equation ΔS=kln(W), where k is the Boltzmann constant and W is the number of possible configurations of the
DNA. For a 3D random walk, the number of possible configurations is given by W=(2N)!/N!(N+1)!. So we can calculate the change in entropy as ΔS=kln[(2N)!/N!(N+
FAQ: How Much Work Does Motor φ29 Need to Overcome Entropy Loss in DNA Packing?
1. What is entropy in relation to DNA packing?
Entropy is a measure of disorder or randomness in a system. In the context of DNA packing, it refers to the degree of flexibility and variability in the organization of DNA within a cell.
2. How is entropy in DNA packing estimated?
Entropy in DNA packing can be estimated through various methods, such as mathematical models, computer simulations, and experimental techniques. These methods take into account factors such as DNA
sequence, nucleosome positioning, and chromatin structure to calculate the level of disorder in DNA packaging.
3. Why is it important to estimate entropy in DNA packing?
Estimating entropy in DNA packing is important because it provides insight into how DNA is organized and regulated within cells. It can also help in understanding the impact of changes in DNA
packaging on gene expression and cellular processes.
4. What are some applications of entropy estimation in DNA packing?
Entropy estimation in DNA packing has various applications in fields such as genetics, biotechnology, and medicine. It can be used to study the effects of mutations on DNA packaging, develop new gene
therapy techniques, and identify potential targets for drug development.
5. How accurate are entropy estimates in DNA packing?
The accuracy of entropy estimates in DNA packing depends on the method used and the complexity of the system being studied. While some methods may provide more precise measurements, all estimates
should be interpreted with caution and validated through multiple experiments.
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Chapter-01 Electric Charges and Fields Test Paper-1 -
Chapter-01 Electric Charges and Fields Test Paper-1
CBSE Test Paper-01 Class - 12 Physics (Electric Charges and Fields)
Class – 12 Physics
(Electric Charges and Fields)
Q 1. For a thin spherical shell of uniform surface charge density , The magnitude of E at a distance r, when r > R (radius of shell) is
Q 2. Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 ×10^-7 ? The
radii of A and B are negligible compared to the distance of separation.
Q 3. A conducting sphere of radius 5 cm is charged to 15 μC. Another uncharged sphere of radius 10 cm is allowed to touch it for enough time. After the two are separated, the surface density of
charge on the two spheres will be in the ratio
a. 2:1
b. 1:2
c. 1:1
d. 3:1
Q 4. The unit of charge is
a. volt
b. ohm
c. coulomb
d. ampere
Q 5. An electric dipole is
a). a pair of electric charges of equal magnitude q but positive sign, separated by a distance d
b). a pair of electric charges of equal magnitude q but opposite sign, separated by a distance
c). a pair of electric charges of equal magnitude q but negative sign, separated by a distance d
d). a pair of electric charges of equal magnitude q separated by a distance d
Q 6. Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?
Q 7. Is the mass of a body affected on charging?
Q 8. Two point charges of 3μC each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?
Q 9. What is the basic cause of quantisation of charge?
Q 10. Calculate the Coulomb force between 2 α particles separated by 3.2 × 10^-15 m.
Q 11. An electric dipole is placed in a uniform electric field E with its dipole moment p parallel to the field. then find
1. The work done in turning the dipole till its dipole moment points in the direction opposite to E.
2. The orientation of the dipole for which the torque acting on it becomes maximum.
Q 12. Define the term electric dipole moment. Is it a scalar or vector? Deduce an expression for the electric field at a point on the equatorial plane of an electric dipole of length 2a.
Q 13. Define the term electric field intensity. Write its SI unit. Derive an expression for the electric field intensity at a point on the axis of an electric dipole.
Q 14. Two infinitely large plane thin parallel sheets having surface charge densities σ1 and σ2 (σ1 > σ2) are shown in the figure. Write the magnitudes and directions of the net fields in the regions
marked II and III.
Q 15. A thin insulating rod of length L carries a uniformly distributed charge Q. Find the electric field strength at a point along its axis at a distance ‘a’ from one end.
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How do I calculate
How do I calculate days difference between dates in Excel?
How do I calculate days difference between dates in Excel?
To find the number of days between these two dates, you can enter “=B2-B1” (without the quotes into cell B3). Once you hit enter, Excel will automatically calculate the number of days between the two
dates entered.
How do I calculate the number of minutes between two times in Excel?
Another simple technique to calculate the duration between two times in Excel is using the TEXT function:
1. Calculate hours between two times: =TEXT(B2-A2, “h”)
2. Return hours and minutes between 2 times: =TEXT(B2-A2, “h:mm”)
3. Return hours, minutes and seconds between 2 times: =TEXT(B2-A2, “h:mm:ss”)
How do I use Datedif in Excel?
The DATEDIF function has three arguments.
1. Fill in “d” for the third argument to get the number of days between two dates.
2. Fill in “m” for the third argument to get the number of months between two dates.
3. Fill in “y” for the third argument to get the number of years between two dates.
Why is Datedif not showing in Excel?
DATEDIF is not a standard function and hence not part of functions library and so no documentation. Microsoft doesn’t promote to use this function as it gives incorrect results in few circumstances.
How do you calculate time difference between sheets?
You need to use the following formula: ‘=(C2-A2)’. This formula gives you the elapsed time between the two cells and displays it as hours. You can take this calculation one step further by adding
dates too.
How do you calculate hours and minutes?
Take your number of minutes and divide by 60.
1. Take your number of minutes and divide by 60. In this example your partial hour is 15 minutes:
2. Add your whole hours back in to get 41.25 hours. So 41 hours, 15 minutes equals 41.25 hours.
3. Multiply your rate of pay by decimal hours to get your total pay before taxes.
Why don’t I have Datedif function in Excel?
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Question #04a42 | Socratic
Question #04a42
1 Answer
The range of a function is the range of values which the function can take given it's domain.
As the domain is not given, assume that it is all values of x which give a real y (in this case $| \setminus x | \le 2$).
Three important facts to remember in this question are that:
a) you can only take the square root of a positive number
b) $y = \sqrt{f \left(x\right)}$ only plots the positive square root (principal root)
c) any real number squared $\ge 0$
Using a), you know $4 - {x}^{2} \ge 0$. Because of c), the maximum value $4 - {x}^{2}$ can take is 4, when $x = 0$, so the maximum value of y is $\sqrt{4} = 2$.
Using a) the minimum value $4 - {x}^{2}$ can take is 0, when $x = \pm 2$, and $\sqrt{0} = 0$.
Given the minimum and maximum it follows that $0 \le x \le 2$
Impact of this question
1046 views around the world
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Cubic Millimeters to Pints
Cubic Millimeters to Pints
Convert Pints to Cubic Millimeters (pt to cu mm) ▶
Conversion Table
cubic millimeters to pints
cu mm pt
100000 cu mm 0.2113 pt
200000 cu mm 0.4227 pt
300000 cu mm 0.634 pt
400000 cu mm 0.8454 pt
500000 cu mm 1.0567 pt
600000 cu mm 1.268 pt
700000 cu mm 1.4794 pt
800000 cu mm 1.6907 pt
900000 cu mm 1.902 pt
1000000 cu mm 2.1134 pt
1100000 cu mm 2.3247 pt
1200000 cu mm 2.5361 pt
1300000 cu mm 2.7474 pt
1400000 cu mm 2.9587 pt
1500000 cu mm 3.1701 pt
1600000 cu mm 3.3814 pt
1700000 cu mm 3.5927 pt
1800000 cu mm 3.8041 pt
1900000 cu mm 4.0154 pt
2000000 cu mm 4.2268 pt
How to convert
1 cubic millimeter (cu mm) = 2.11338E-06 pint (pt). Cubic Millimeter (cu mm) is a unit of Volume used in Metric system. Pint (pt) is a unit of Volume used in Standard system.
Cubic Millimeters - A Unit of Volume
A cubic millimeter (symbol mm3 or cu mm) is a unit of volume that corresponds to the volume of a cube with sides of 1 millimeter (0.001 meter) in length. It is also equivalent to 0.001 milliliter,
which is a unit of volume in the metric system.
One cubic millimeter is equal to 0.000000001 cubic meters, 0.00006102374 cubic inches, or 0.000000264172 gallons.
How to Convert Cubic Millimeters
To convert cubic millimeters to other units of volume, you need to multiply or divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert cubic millimeters to cubic meters, multiply by 0.000000001.
□ Example: 2 mm3 × 0.000000001 = 0.000000002 m3
• To convert cubic millimeters to cubic inches, multiply by 0.00006102374.
□ Example: 2 mm3 × 0.00006102374 = 0.00012204748 in3
• To convert cubic millimeters to milliliters or liters, multiply by 0.001 or divide by 1000000 respectively.
□ Example: 2 mm3 × 0.001 = 0.002 mL or 2 mm3 ÷ 1000000 = 0.000002 L
• To convert cubic millimeters to gallons (US liquid), multiply by 0.000000264172.
□ Example: 2 mm3 × 0.000000264172 = 0.000000528344 gal
• To convert cubic millimeters to bushels (US), multiply by 0.00000002837825.
□ Example: 2 mm3 × 0.00000002837825 = 0.0000000567565 bu
• To convert cubic millimeters to barrels (oil), multiply by 0.00000000628981.
□ Example: 2 mm3 × 0.00000000628981 = 0.00000001257962 bbl
To convert other units of volume to cubic millimeters, you need to divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert cubic meters to cubic millimeters, divide by 0.000000001.
□ Example: 0.000000002 m3 ÷ 0.000000001 = 2 mm3
• To convert cubic inches to cubic millimeters, divide by 0.00006102374.
□ Example: 0.00012204748 in3 ÷ 0.00006102374 = 2 mm3
• To convert milliliters or liters to cubic millimeters, divide by 0.001 or multiply by 1000000 respectively.
□ Example: 0.002 mL ÷ 0.001 = 2 mm3 or 0.000002 L × 1000000 = 2 mm3
• To convert gallons (US liquid) to cubic millimeters, divide by 0.000000264172.
□ Example: 0.000000528344 gal ÷ 0.000000264172 = 2 mm3
• To convert bushels (US) to cubic millimeters, divide by 0.00000002837825.
□ Example: 0.0000000567565 bu ÷ 0.00000002837825 = 2 mm3
• To convert barrels (oil) to cubic millimeters, divide by 0.00000000628981.
□ Example: 0.00000001257962 bbl ÷ 0.00000000628981 = 2 mm3
Cubic millimeters also can be marked as mm^3.
Pints: A Unit of Volume
Pints are a unit of volume that are used to measure liquids, such as water, milk, beer, cider, etc. They are also used to measure some dry goods, such as flour, sugar, rice, etc. They are different
from cups, which are a smaller unit of volume. They are also different from liters, which are a larger unit of volume. They are also different from barrel of oil equivalent (BOE), which is a unit of
energy based on the approximate energy released by burning one barrel of crude oil.
How to Convert Pints
To convert pints to other units of volume, one can use the following formulas:
• To convert UK pints to liters: multiply by 0.568
• To convert UK pints to cubic inches: multiply by 34.677
• To convert UK pints to fluid ounces: multiply by 20
• To convert UK pints to US liquid pints: multiply by 0.473
• To convert UK pints to BOE: divide by 35
• To convert US liquid pints to liters: multiply by 0.473
• To convert US liquid pints to cubic inches: multiply by 28.875
• To convert US liquid pints to fluid ounces: multiply by 16
• To convert US liquid pints to UK pints: multiply by 2.113
• To convert US liquid pints to BOE: divide by 6
• To convert US dry pints to liters: multiply by 0.551
• To convert US dry pints to cubic inches: multiply by 33.6
• To convert US dry pints to fluid ounces: multiply by 18.6
• To convert US dry pints to UK pints: multiply by 1.032
• To convert US dry pints to BOE: divide by 5.5
The US pint, defined as exactly 473.176473 milliliters = 1/8 US liquid gallon.
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50×50 Multiplication Chart Printable | Multiplication Chart Printable
50×50 Multiplication Chart Printable
Multiplication Chart 50×50 Leonard Burton s Multiplication Worksheets
50×50 Multiplication Chart Printable
50×50 Multiplication Chart Printable – A Multiplication Chart is a helpful tool for kids to learn how to multiply, split, as well as locate the tiniest number. There are lots of uses for a
Multiplication Chart. These helpful tools assist children understand the procedure behind multiplication by utilizing tinted courses and also filling out the missing items. These charts are totally
free to download and also publish.
What is Multiplication Chart Printable?
A multiplication chart can be used to aid youngsters learn their multiplication facts. Multiplication charts come in numerous types, from full page times tables to solitary page ones. While specific
tables work for offering portions of details, a complete web page chart makes it less complicated to evaluate facts that have already been understood.
The multiplication chart will commonly feature a left column and also a leading row. The leading row will certainly have a list of products. Pick the very first number from the left column as well as
the second number from the leading row when you want to locate the item of 2 numbers. As soon as you have these numbers, move them along the row or down the column until you get to the square where
both numbers satisfy. You will then have your item.
Multiplication charts are handy learning tools for both children as well as grownups. Youngsters can utilize them in your home or in institution. 50×50 Multiplication Chart Printable are offered on
the net and also can be printed out as well as laminated flooring for resilience. They are a wonderful tool to utilize in mathematics or homeschooling, and also will certainly offer a visual pointer
for youngsters as they discover their multiplication truths.
Why Do We Use a Multiplication Chart?
A multiplication chart is a layout that shows how to multiply 2 numbers. You select the initial number in the left column, move it down the column, and after that select the second number from the
leading row.
Multiplication charts are useful for numerous factors, including assisting youngsters discover just how to split and simplify fractions. Multiplication charts can likewise be handy as workdesk
resources since they serve as a consistent suggestion of the trainee’s progress.
Multiplication charts are likewise helpful for helping pupils memorize their times tables. They help them discover the numbers by minimizing the number of steps required to complete each operation.
One strategy for remembering these tables is to focus on a solitary row or column at a time, and after that move onto the next one. Eventually, the whole chart will be committed to memory. Similar to
any kind of skill, memorizing multiplication tables takes time as well as method.
50×50 Multiplication Chart Printable
Multiplication Chart 50 50 PrintableMultiplication
Printable Multiplication Table 50X50 PrintableMultiplication
Multiplication Chart 50 50 PrintableMultiplication
50×50 Multiplication Chart Printable
If you’re looking for 50×50 Multiplication Chart Printable, you’ve come to the right location. Multiplication charts are offered in different styles, consisting of complete dimension, half dimension,
and also a variety of charming designs.
Multiplication charts and also tables are important tools for youngsters’s education. You can download and print them to use as a training aid in your youngster’s homeschool or class. You can
additionally laminate them for durability. These charts are fantastic for use in homeschool mathematics binders or as classroom posters. They’re particularly valuable for kids in the second, 3rd, and
fourth qualities.
A 50×50 Multiplication Chart Printable is an useful tool to reinforce mathematics truths and also can help a kid learn multiplication quickly. It’s likewise a terrific tool for skip counting as well
as discovering the moments tables.
Related For 50×50 Multiplication Chart Printable
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A substantive effect size interpretation (interpretation), for the purpose of this study, is a statement about the magnitude of change in the outcome as a function of change in the predictor from the
reported OR. Solely reporting the OR or its directionality was not considered a substantive interpretation. A correct substantive interpretation is an interpretation that accurately reflects the
definition of the OR, as described in commentary by Davies et al 6 and Norton et al.7 For example, if the OR were 1.5, the interpretation “50% increase in odds” would be correct, while “50% more
likely” would be incorrect. Correct interpretations could also include other interpretations resulting from logistic regression that use the OR as an intermediary (eg, change in probability, marginal
Example phrases Label Reason
“… male sex was associated with seeking treatment (OR=2)…” No interpretation The OR was presented as a parenthetical statement only.
“… was associated with decreased odds (OR=0.5) …” No interpretation Only the direction of the association was reported.
“… were three times more likely (OR=3) …” Incorrect interpretation An interpretation was made by incorrectly expressing ‘odds’ as ‘likeliness’.
“… was associated with a 30% reduction in the log odds (OR=0.7) …” Incorrect interpretation An interpretation was made by expressing the ratio of log odds but reported the OR.
“… were associated with a threefold increase in the odds (OR=3) …” Correct interpretation An interpretation was made by expressing the ratio of odds.
“… each was associated with a 10% reduction in the odds of treatment failure (OR=0.90) …” Correct interpretation An interpretation was made by expressing the ratio of odds.
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A Model for Macro
A Model for Macro
So far we have been discussing the properties of matter from the atomic point of view, trying to understand roughly what will happen if we suppose that things are made of atoms obeying certain
laws. However, there are number of relationships among the properties of substances which can be worked out without consideration of the detailed structure of the materials. The determination of
the relationships among the various properties of the materials, without knowing their internal structure, is the subject of thermodynamics. Historically, thermodynamics was developed before an
understanding of the internal structure of matter was achieved….
We have seen how these two processes, contraction when heated and cooling during relaxation, can be related by the kinetic theory, but it would be a tremendous challenge to determine from the
theory the precise relationship between the two. We would have to know how many collisions there were each second and what the chains look like, and we would have to take account of all kinds of
other complications. The detailed mechanism is so complex that we cannot, by kinetic theory, really determine exactly what happens; still, a definite relation between the two effects we observe
can be worked out without knowing anything about the internal machinery!
1 Richard P. Feynman et al., The Feynman Lectures on Physics 44-1-44-2 (1963).
Cf. http://rajivsethi.blogspot.com/2010/02/case-for-agent-based-models-in.html .
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Data Frames in R Language (2024) - A Comprehensive Guide
Data Frames in R Language (2024)
Data frames in R are one of the most essential data structures. A data frame in R is a list with the class “data.frame“. The data frame structure is used to store tabular data. Data frames in R
Language are essentially lists of vectors of equal length, where each vector represents a column and each element of the vector corresponds to a row.
Table of Contents
Data frames in R are the workhorse of data analysis, providing a flexible and efficient way to store, manipulate, and analyze data.
Restrictions on Data Frames in R
The following are restrictions on data frames in R:
1. The components (Columns or features) must be vectors (numeric, character, or logical), numeric matrices, factors, lists, or other data frames.
2. Lists, Matrices, and data frames provide as many variables to the new data frame as they have columns, elements, or variables.
3. Numeric vectors, logical vectors, and factors are included as is, by default, character vectors are coerced to be factors, whose levels are the unique values appearing in the vector.
4. Vecture structures appearing as variables of the data frame must all have the same length, and matrix structures must all have the same row size.
A data frame may for many purposes be regarded as a matrix with columns possibly of differing modes and attributes. It may be displayed in matrix form, and its rows and columns are extracted using
matrix indexing conventions.
Key Characteristics of Data Frame
• Column-Based Operations: R language provides powerful functions and operators for performing operations on entire columns or subsets of columns, making data analysis and manipulation efficient.
• Heterogeneous Data: Data frames can store data of different data types within the same structure, making them versatile for handling various kinds of data.
• Named Columns: Each column in a data frame has a unique name, which is used to reference and access specific data within the frame.
• Row-Based Indexing: Data frames are indexed based on their rows, allowing you to easily extract or manipulate data based on row numbers.
Making/ Creating Data Frames in R
Objects satisfying the restrictions placed on the columns (components) of a data frame may be used to form one using the function data.frame(). For example:
BMI <- data.frame(
age = c(20, 40, 33, 45),
weight = c(65, 70, 53, 69),
height = c(62, 65, 55, 58)
Note that a list whose components conform to the restrictions of a data frame may coerced into a data frame using the function as.data.frame().
Other Way of Creating a Data Frame
One can also use read.table(), read.csv(), read_excel(), and read_csv() functions to read an entire data frame from an external file.
Accessing and Manipulating Data
• Accessing Data: Use column names or row indices to extract specific values or subsets of data.
• Creating New Columns: Calculate new columns based on existing ones using arithmetic operations, logical expressions, or functions.
• Grouping and Summarizing: Group data by specific columns and calculate summary statistics (e.g., mean, median, sum).
• Sorting Data: Arrange rows in ascending or descending order based on column values.
• Filtering Data: Select rows based on conditions using logical expressions and indexing.
# Create a data frame manually
data <- data.frame(
Name = c("Ali", "Usman", "Hamza"),
Age = c(25, 30, 35),
City = c("Multan", "Lahore", "Faisalabad")
# Accessing data
print(data$Age) # Displays the "Age" column
print(data[2, ]) # Displays the second row
# Creating a new column
data$Age_Category <- ifelse(data$Age < 30, "Young", "Old")
# Filtering data
young_people <- data[data$Age < 30, ]
# Sort data
sorted_data <- data[order(data$Age), ]
https://itfeature.com, https://gmstat.com
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How to Find Solutions to Central Tendency Assignments Quickly? – University Homework Help
Statistical analysis is a key element of the highly important tool called data science. It helps in conducting a technical analysis of a collection of data by a mathematical approach. Statistics help
analysts to understand the viability and significance of a certain result and whether or not it is statistically significant. Visual representation of statistical analysis like bar graphs, pie charts
offer specific and informative depictions of data.
Statistics is a notable domain of mathematical analysis that has varying significance in almost every subject matter. It is a collection of various topics that has major importance in diverse areas.
Each of these subject matters is essential to build a thorough knowledge of the subject. Moreover, it is necessary for students to bear meticulous skills in mathematical calculations to score a
decent result in statistics.
Statistics homework assignments vary according to their level of difficulty. They might be from various important topics like probability, correlation, central tendency, and so on. Developing a
detailed understanding of each of these topics can help score a better grade in statistics exam. Religious practice and necessary assistance from experienced tutors are of huge help when it comes to
covering the vast syllabus of a monumental subject like statistics.
Statistics and its categories
Statistics offers several tools that assist in calculation and forecasting to simplify the process of data analysis. It is rigorously implemented in diverse fields that include academics, business
houses, government agencies, and so on.
Based on the characteristics and availability of data, statistics can be distinguished into two major categories. They are-
1. Qualitative data–
In the case of qualitative analysis, the data for statistical interpretation is available in the natural language form and not as a numerical description.
On contrary to qualitative data, quantitative data analysis is strictly based on numerical data.
Both the categories of statistical interpretation of data are a fundamental requirement to facilitate data forecast, representation, and optimal analysis of information.
A subject matter with such varied significance in almost every other academic discipline is extensively taught in nearly all academic institutions. Students are subjected to details of this subject
to develop a strong foundation that will ultimately help in building their analytic skills.
The ability to accurate analysis also helps in arriving at statistically viable conclusions which is now a must in research, businesses, and all other professional fields. Students can turn to
websites to obtain ample materials on various data types.
Central tendency- the fundamental element of Statistics
Statistical analysis is conducted on specific or non-specific data sets obtained from distinct surveys, investigations, and studies. Several statistical tools are frequently recruited to deduce and
interpret the collected information.
Central tendency forms the core of statistical analysis and is a quintessential tool required for any data interpretation. It can be defined as the value or position in a particular data set around
which the other values tend to cluster or accumulate. It provides a descriptive review of the set of data under study via a distinct value that represents the center of the numerical distribution.
One major disadvantage of central tendency is that it is incapable of providing a thorough interpretation of individual values in a data distribution. But on the brighter side, it is an imperative
tool that offers an understandable summary of an entire collection of data.
Most statistics assignments include several questions from this topic and students cannot afford to skip this if they are willing to score a high grade in the subject. Central tendency numerical
contribute to over 35% questions in all major statistics assessment papers as well. Therefore bearing a clear concept is a must to solve them correctly.
The measure of Central Tendency
The central tendency of a data distribution can be measured using the following tools-
Mean– Mean, generally known as average, is the most commonly used measure of the central tendency of a data set. It can be implemented for calculating the average of both continuous and discrete
datasets. It can be obtained by calculating the sum of all values in a data distribution divided by the number of values in the set.
Median– Median is the value present at the center of a dataset that is arranged in ascending order. In the case of an even dataset, the arithmetic mean of the two middle values is the median of that
Mode– Mode of a dataset represents the most frequently occurring value in that distribution. Some numerical distribution can have more than one mode whereas some might not have any.
Although these are the most commonly used measures of central tendency, there are a few other measures. They are-
Geometric mean– This is used for the calculation of GDP (or Gross Domestic Product)
Harmonic mean– This is required for the calculation of speed
Geometric median– This is necessary for calculating growth rates and returns in financial calculations.
Central tendency contributes to diverse numerical. Almost all statistical calculation involves the use of at least one of the measures of central tendency. Therefore, students must consider being
extremely scrupulous while preparing for this topic.
It is recommended that students devote enough time in solving examples and model questions while preparing for assessments. They can also refer to informative resources available for better learning
of the topic.
Topics associated with the Advanced Statistical Studies
Nowadays, statistical studies are also conducted using sophisticated software to improve the ease of use. This simplifies the analysis of datasets that have huge volumes of information and numerous
repetitive data. The use of software programs also requires an in-depth knowledge of how each of these tools performs to achieve validated data.
Some of the tools that can be used through software are mentioned as follows-
• Probability theory and distributions
• Interval estimation
• Frequency distributions
• Business statistics
• Analysis of Random variables
• Bivariate linear regression
• Bivariate correlation
• Bivariate probability distribution studies
• Discrete parametric probability distribution
• Sampling and distribution of variables
• Point estimation and its properties
• The chi-square
• Contingency chi-square
• Contingency tables
• T-test
• Z-test
• Snedecor’s F distributions
• Continuous parametric probability distribution
• Test of parametric hypothesis
• Nonparametric techniques in statistics
• Goodness of fit
• Test of homogeneity
These software programs efficiently alleviate the issues of calculation complexity associated with statistical analysis of datasets in bulk. But it is of utmost importance that the principles of
these tools are duly known by the analysts.
Statistical analysts might also be required to conduct certain calculations manually in case of any discrepancies in the system or software malfunction. Thus building strong concepts on these tools
from the early academic years comes handy in the professional career as well.
Smart ways to tackle complex statistics problems
Irrespective of their professional background, people might require statistical analysis for their work. Although they can rely on external sources to get their job done, doing it themselves is
always a better option. Students can save time and money by working on their statistics knowledge right from school to have an added advantage over their colleagues later in life.
Introduction to advanced statistics topics and software has admittedly vast areas that they include within their circumference. Different topics like inferential and descriptive statistics might be
quite difficult to grasp primarily and that ends up scaring students initially.
Below are some of the smartest ways by which one can easily tackle tricky statistics questions and improve their scores in exams-
1. Being attentive in class–
Half the preparation can be done during the lectures and class hours itself if students are attentive enough in taking notes. This goes a long way in gaining smart tips on solving seemingly difficult
questions that teachers tend to mention during lectures.
• Improving your mathematics–
Statistics is 90% mathematics and the rest is just techniques. So it is advisable to primarily concentrate on the knowledge of mathematical calculations and methods to form a better grip on
statistical analysis. Accurate mathematical calculations and quick problem-solving technique is a must to have an upper-hand on this subject.
The trick to scoring better grades is not in solving the most complex questions in exams. It majorly depends on how many questions a student can answer adequately to increase the overall grade/ only
a clear fundamental knowledge about the subject matter can help achieve that.
Students must consider keeping in touch with the subject at least 2-3 times a week. This ultimately saves a lot of trouble before the exams. It is suggested to solve a minimum of 5 numerical each day
to be
• Being open to information–
Being open to gaining newer concepts always helps in resolving doubts about topics that seem very difficult to grasp. Students must not give and work religiously to achieve better outcomes.
• Procrastination is not an option–
Students simply cannot afford to rely on memorizing statistics topics a few days before assessments if they are willing to perform decently. They must refrain from procrastinating on their
assignments and get the necessary homework assistance while at it.
Students must work towards giving up on the impression that statistics is a stressful subject. It is important to let the new concepts sink in and then move on to another topic in due course of time.
They must not overdo it and study smartly to cover broader areas in lesser time.
The need for assistance with Statistics
At its disciplinary core, statistics is associated with understanding, evaluating, and representing data in comprehensive pattern to serve significant purposes in major sectors. But the perception of
statistical concepts roots deeper than just mathematical estimation and graphical representation.
Students pursuing statistics as their major subject or in school are introduced to multiple new conceptual subject matters that are understandably complex. Expert assistance is imperative to absorb
the ideas of those topics to make progress.
It is important to let go of personal reservations and inhibitions and get the requisite amount of assistance to cover this disciple effectively. Experts can also suggest numerous procedures to
construct better scoring answers and optimum ways to solve the numerical that might otherwise appear to be quite daunting to attempt.
Statistics is a manageable course with various comprehensive resources to refer to resolve confusions. Students can always resort to fervent statistics tutors and also freely available online sources
to gather appropriate academic support at the right time.
Advantages of getting expert assistance with Statistics
Students always wish to perform excellently in academics. But are often unable to find reliable sources to help them achieve that. They can always turn to statistics experts available online to
materialize their aspiration to be present in the top order of the class.
Here are some of the core benefits of choosing professional statistics guidance that will help in answer questions with ease-
• Experts know the exact ways to construct better scoring answers that are to the point and of crisp quality. Relying on them is a fruitful way to meet all academic requirements efficiently.
• Through their years of experience and expertise with the subject matter, they are capable of producing the finest quality of statistics papers and assignment solutions.
• Professional tutors make sure that all academic papers are based on extensive research. They strictly eliminate plagiarized content to make them authentic, highly informative, and factually
• Experts construct papers based on the requirements specified by the clients and therefore composed fully customized assignments.
• Online academic portals make it a point to subject their papers to stepwise proofreading thus making them grammatically correct and error-free.
The advantages of academic guidance from experienced educators have an ocean of advantages that can pave the path for endless possibilities. Students can be assured about their work being done on
time minus the pain-staking hours of going to multiple notes and textbooks.
Author Bio:
A Lecturer in the Statistics Section of the Department of Mathematics, Imperial College London. She also holds an EPSRC research fellowship until January 2020.
DR. Heather Battey is also a proficient tutor working relentlessly at to provide superior content to the students from different academic backgrounds in statistics. She has six years of experience
that makes her capable of being one of the most respected educators. She assists students with her proficiency in different subject matters and helps them improve their problem-solving capabilities.
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6 Easy Ways To Make Math Lessons Fun For Kids
A majority of children complain to their parents that they are incredibly bored with mathematics lessons. They do not understand why they have to learn a stack of formulas and how they can come
in handy in real life. That's why I've prepared 6 ways that will help diversify the lesson of mathematics and interest the student.
1. Fill the lesson with a sense
Most of the math lessons in school suffer from the following moments:
1. Sometimes, teachers themselves cannot explain why they teach certain topics to students. So, it is difficult for teachers to see the connection of mathematics with other subjects of the school
2. As a result, the students also do not understand why they are studying these topics. A common question that they ask themselves: "Why should I teach this?" It makes sense. Do you have a good
answer to it, instead of the usual "It will be on the exam" or even worse - "Because you need it"?
There are several possible options to fix this:
□ Show the student the practical importance of mathematics. Explain how he can solve real life problems using the knowledge gained in your lessons.
□ Get familiar with the curriculum for other school subjects. After that, you will be able to use in the lessons examples that are understandable and interesting to your students.
2. Start with an interesting, real problem
Most math lessons start like this: "Here's a new formula for today's lesson, that's how you need to insert numbers, that's the right answer."
The problem is that in this approach there is not even an attempt to motivate the student.
It will be great if you encourage students' interest. Use presentations, training videos and other aids. Look for interesting information on the Internet and use it in the classroom.
One can propose the following theme: The largest telescope in the world was built in China. What to do in the lesson: find the area of the 500-meter telescope, discuss how the construction of the
telescope affected the environment, and decide what areas were cut down for the construction of the telescope.
3. Creativity and control over the situation
I believe that mathematics is an extremely interesting science, for the development of which a living and open mind is needed. It is not necessary to reduce the work in the lesson to memorizing
formulas and monotonous solutions of the same tasks on the finished algorithm.
We are all creative and like to be like that, but in most schools, creativity is not encouraged (see a great video from TED Talks, Ken Robinson: Do schools kill creativity?).
There are many ways to encourage the creativity of students in the lessons of mathematics. Use new technologies to describe mathematical concepts: prepare an animation, diagrams or interesting
infographics. Create something yourself or find it on the Internet. Give the students individual tasks that involve creative thinking and help them feel confident in their abilities.
4. Ask more interesting questions
For many students, mathematical questions are most often associated with tasks in the textbook. The task for them looks like a long sentence: "Here is the task in words. Take the numbers,
substitute them in the formula, make the calculation and go to the next task. "
An interesting description of the task will necessarily catch the attention of the students. This task will cause more emotion than the usual question from the book. Imagine that you are jumping
with a parachute. What will the graph of your speed look like depending on the time, from the moment of the jump from the airplane to the final speed?
When students are accustomed to solving similar problems, they themselves will start to invent interesting examples from life, related to the calculation of the formulas already studied.
5. Let the students compose their own questions.
Students understand more when they need to come up with their own questions.
You can divide the class into 2-4 groups. Each group should form a block of questions for the test work. At the lesson, the kids exchange sets of tasks and solve them. If one of the members made
a mistake or a prepared task, it is possible to disassemble it at the lesson, why it happened: what was wrong with the component that would confuse him or her.
6. Projects
The most effective way to interact with students is to give them the opportunity to do something themselves. Help the students to see the mathematics around themselves: in the things that
surround them, in natural phenomena and processes. You can use modern teaching tools that will help you show students of different ages how interesting math is.
. Here are just a few ideas:
□ Construct Lego-Robots
□ Create visual presentations on the GeoGebra website
□ Create a dynamic presentation in Prezi
School lessons are the way to the heights of knowledge, the process of improvement and intellectual growth of the student. Each of them causes disturbing children's consciousness of thought and
incredible discoveries or hopeless boredom and dangerous laziness. And how valuable and interesting will be time spent by the student at the school desk mostly depends on the efforts of the
About the author: Richard D. Eddington is a math tutor in Singapore. Writing is his hobby. In addition, he has a wide range of interests so he likes to share personal experience, researches, and
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Heat loss through a composite wall of three materials.xls
KNOWN: Thicknesses of three materials which form a composite wall and thermal conductivities of the two materials. Inner and outer surface temperatures of the composite: also, temperature and
convection coefficient associated with adjoining gas.
FIND: Value of unknown thermal conductivity, kB.
ASSUMPTIONS: 1) Steady-state conditions
2) One-dimensional conduction
3) Constant properties
4) Negligible contact resistance
5) Negligible radiation effects
ANALYSIS: Analogy with thermal circuit.
Calculation Reference
Fundamentals of Heat and Mass Transfer - Frank P. Incropera
To find the unknown thermal conductivity (kB) in a composite wall, we can use the analogy of a thermal circuit, which represents the flow of heat through the different layers of the composite wall.
Here's how you can approach the analysis:
1. Identify the layers and their properties: Identify the three materials that form the composite wall and note their respective thicknesses (L1, L2, L3) and thermal conductivities (k1, k2, k3).
Let's assume material 1 is the innermost layer, material 2 is the middle layer with the unknown thermal conductivity (kB), and material 3 is the outermost layer.
2. Establish thermal resistances: Assign thermal resistances to each layer of the composite wall. The thermal resistance (R) of each layer is calculated as the thickness divided by the thermal
conductivity: R = L / k.
3. Apply thermal circuit analogy: Treat the composite wall as a series of resistances connected in series. The total thermal resistance (R_total) is the sum of the individual thermal resistances:
R_total = R1 + R2 + R3
Substitute the thermal resistances with their respective values using the thicknesses and thermal conductivities of the materials.
4. Apply the temperature boundary conditions: Determine the temperature difference (ΔT) between the inner and outer surfaces of the composite wall. ΔT = T_outer - T_inner.
5. Calculate the heat transfer rate: Use the formula for heat transfer rate (Q) through a composite wall:
Q = (T_outer - T_inner) / R_total
Substitute the values of ΔT and R_total from the previous steps.
6. Consider the adjoining gas convection: If there is convective heat transfer between the outer surface of the composite wall and the adjoining gas, you'll need to account for it by incorporating
the convection coefficient (h) into the heat transfer equation. The modified equation will be:
Q = h * A * (T_outer - T_gas)
Where A is the surface area of the composite wall and T_gas is the temperature of the adjoining gas.
7. Solve for the unknown thermal conductivity: Rearrange the equation to solve for kB:
kB = Q / (h * A) + k1 / R1 + k3 / R3 - k2 / R2
Substitute the known values of Q, h, A, k1, k3, R1, R3, and solve for kB.
By treating the composite wall as a thermal circuit and using the above steps, you can determine the value of the unknown thermal conductivity (kB) of the middle layer.
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Radix Sort in Python | Working and examples of Radix sort in Python
Updated April 17, 2023
Introduction to Radix Sort in Python
The sorting technique in which the elements of the given array to be sorted are grouped into individual digits based on the place value, which is then sorted in either ascending order or descending
order, is called Radix sort in Python, whose time complexity is O(nk) where n is the size of the input array to be sorted and k is the length of the digit in the array to be sorted and radix sort in
Python includes an intermediate sort called counting sort, and radix sort are used to implement DC3 algorithm and in places where large numbers are involved.
The function to perform Radix sort is as follows:
def radixSort(array):
maximum_value = max(array)
place_value = 1
while maximum_value // place_value > 0:
countingSort(array, place_value)
place_value *= 10
where the array is the input array to be sorted using the radix sort technique,
maximum_value is the maximum_value present in the given array and
place_value represents the place value of individual digits in a number and
countingSort is the countingSort function to perform the intermediate sort.
Working of Radix sort algorithm in Python
• The first step in the Radix sort algorithm is to is find the largest element in the array using the max() function in python and determining the number of digits in the largest number.
• Then the most significant digits place value of the largest number is considered, and the most significant digit of each element in the array is sorted using the countingSort algorithm.
• Then the next significant digits place value of each element in the array is sorted again using the countingSort algorithm, and this process is repeated until all the digits in their place values
are sorted.
Examples of Radix sort in python
Different examples are mentioned below:
Example #1
Python program to sort the elements of the given array by implementing Radix sort algorithm and then display the sorted elements of the array as the output on the screen:
#defining countingsort function to sort the given elements based on their significant digits place value
def countingSort(input_array, place_value):
arraysize = len(input_array)
output = [0] * arraysize
count = [0] * 10
#determining the count of the elements in the array
for a in range(0, arraysize):
arrayindex = input_array[a] // place_value
count[arrayindex % 10] += 1
#determining the cumulative count of the elements in the array
for b in range(1, 10):
count[b] += count[b - 1]
#placing the elements of the array in sorted order
a = arraysize - 1
while a >= 0:
arrayindex = input_array[a] // place_value
output[count[arrayindex % 10] - 1] = input_array[a]
count[arrayindex % 10] -= 1
a -= 1
for a in range(0, arraysize):
input_array[a] = output[a]
#defining radix sort function to sort the elements of the array using countingsort function
def radixSort(input_array):
maximum_value = max(input_array)
place_value = 1
while maximum_value // place_value > 0:
countingSort(input_array, place_value)
place_value *= 10
input_data = [600, 400, 500, 200, 100, 800]
print("The elements of the array to be sorted are:\n")
print("The elements of the array after sorting are:\n")
The output of the above program is shown in the snapshot below:
In the above program, we are defining the countingsort function to sort the given elements based on their significant digits place value. Then we are determining the count of the elements in the
array. Then we are determining the cumulative count of the elements in the array. Then we are placing the elements of the array in sorted order. Then we define the radix sort function to sort the
array elements using the countingsort function and then display the elements of the sorted array as the output on the screen.
Example #2
Python program to sort the elements of the given array by implementing Radix sort algorithm and then display the sorted elements of the array as the output on the screen:
#defining countingsort function to sort the given elements based on their significant digits place value
def countingSort(input_array, place_value):
arraysize = len(input_array)
output = [0] * arraysize
count = [0] * 10
#determining the count of the elements in the array
for a in range(0, arraysize):
arrayindex = input_array[a] // place_value
count[arrayindex % 10] += 1
#determining the cumulative count of the elements in the array
for b in range(1, 10):
count[b] += count[b - 1]
#placing the elements of the array in sorted order
a = arraysize - 1
while a >= 0:
arrayindex = input_array[a] // place_value
output[count[arrayindex % 10] - 1] = input_array[a]
count[arrayindex % 10] -= 1
a -= 1
for a in range(0, arraysize):
input_array[a] = output[a]
#defining radix sort function to sort the elements of the array using countingsort function
def radixSort(input_array):
maximum_value = max(input_array)
place_value = 1
while maximum_value // place_value > 0:
countingSort(input_array, place_value)
place_value *= 10
input_data = [20, 10, 50, 70, 30]
print("The elements of the array to be sorted are:\n")
print("The elements of the array after sorting are:\n")
The output of the above program is shown in the snapshot below:
In the above program, we are defining the countingsort function to sort the given elements based on their significant digits place value. Then we are determining the count of the elements in the
array. Then we are determining the cumulative count of the elements in the array. Then we are placing the elements of the array in sorted order. Then we define the radix sort function to sort the
array elements using the countingsort function and then display the elements of the sorted array as the output on the screen.
Recommended Articles
This is a guide to Radix sort in python. Here we discuss the concept of Radix sort in Python with corresponding programming examples and their outputs to demonstrate them. You may also have a look at
the following articles to learn more –
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560 Picometer/Second Squared to Mile/Second Squared
Picometer/Second Squared [pm/s2] Output
560 picometer/second squared in meter/second squared is equal to 5.6e-10
560 picometer/second squared in attometer/second squared is equal to 560000000
560 picometer/second squared in centimeter/second squared is equal to 5.6e-8
560 picometer/second squared in decimeter/second squared is equal to 5.6e-9
560 picometer/second squared in dekameter/second squared is equal to 5.6e-11
560 picometer/second squared in femtometer/second squared is equal to 560000
560 picometer/second squared in hectometer/second squared is equal to 5.6e-12
560 picometer/second squared in kilometer/second squared is equal to 5.6e-13
560 picometer/second squared in micrometer/second squared is equal to 0.00056
560 picometer/second squared in millimeter/second squared is equal to 5.6e-7
560 picometer/second squared in nanometer/second squared is equal to 0.56
560 picometer/second squared in meter/hour squared is equal to 0.0072576
560 picometer/second squared in millimeter/hour squared is equal to 7.26
560 picometer/second squared in centimeter/hour squared is equal to 0.72576
560 picometer/second squared in kilometer/hour squared is equal to 0.0000072576
560 picometer/second squared in meter/minute squared is equal to 0.000002016
560 picometer/second squared in millimeter/minute squared is equal to 0.002016
560 picometer/second squared in centimeter/minute squared is equal to 0.0002016
560 picometer/second squared in kilometer/minute squared is equal to 2.016e-9
560 picometer/second squared in kilometer/hour/second is equal to 2.016e-9
560 picometer/second squared in inch/hour/minute is equal to 0.0047622047244095
560 picometer/second squared in inch/hour/second is equal to 0.000079370078740158
560 picometer/second squared in inch/minute/second is equal to 0.0000013228346456693
560 picometer/second squared in inch/hour squared is equal to 0.28573228346457
560 picometer/second squared in inch/minute squared is equal to 0.000079370078740158
560 picometer/second squared in inch/second squared is equal to 2.2047244094488e-8
560 picometer/second squared in feet/hour/minute is equal to 0.00039685039370079
560 picometer/second squared in feet/hour/second is equal to 0.0000066141732283465
560 picometer/second squared in feet/minute/second is equal to 1.1023622047244e-7
560 picometer/second squared in feet/hour squared is equal to 0.023811023622047
560 picometer/second squared in feet/minute squared is equal to 0.0000066141732283465
560 picometer/second squared in feet/second squared is equal to 1.8372703412073e-9
560 picometer/second squared in knot/hour is equal to 0.000003918790512
560 picometer/second squared in knot/minute is equal to 6.53131752e-8
560 picometer/second squared in knot/second is equal to 1.08855292e-9
560 picometer/second squared in knot/millisecond is equal to 1.08855292e-12
560 picometer/second squared in mile/hour/minute is equal to 7.5161059413028e-8
560 picometer/second squared in mile/hour/second is equal to 1.2526843235505e-9
560 picometer/second squared in mile/hour squared is equal to 0.0000045096635647817
560 picometer/second squared in mile/minute squared is equal to 1.2526843235505e-9
560 picometer/second squared in mile/second squared is equal to 3.4796786765291e-13
560 picometer/second squared in yard/second squared is equal to 6.1242344706912e-10
560 picometer/second squared in gal is equal to 5.6e-8
560 picometer/second squared in galileo is equal to 5.6e-8
560 picometer/second squared in centigal is equal to 0.0000056
560 picometer/second squared in decigal is equal to 5.6e-7
560 picometer/second squared in g-unit is equal to 5.7104107926764e-11
560 picometer/second squared in gn is equal to 5.7104107926764e-11
560 picometer/second squared in gravity is equal to 5.7104107926764e-11
560 picometer/second squared in milligal is equal to 0.000056
560 picometer/second squared in kilogal is equal to 5.6e-11
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[{"uri_base":"https://research-explorer.ista.ac.at","main_file_link":[{"url":"https://dl.acm.org/doi/10.1145/98524.98548"}],"page":"112 -
{"id":"ea97e931-d5af-11eb-85d4-e6957dddbf17","login":"alisjak"},"status":"public","day":"01","publication_status":"published","publication":"Proceedings of the 6th annual symposium on Computational
geometry","_id":"4077","month":"01","conference":{"end_date":"1990-06-09","start_date":"1990-06-07","name":"SCG: Symposium on Computational Geometry","location":"Berkley, CA, United
States"},"type":"conference","publication_identifier":{"isbn":[]},"citation":{"ieee":"B. Aronov, B. Chazelle, H. Edelsbrunner, L. Guibas, M. Sharir, and R. Wenger, “Points and triangles in the plane
and halving planes in space,” in Proceedings of the 6th annual symposium on Computational geometry, Berkley, CA, United States, 1990, pp. 112–115.","ista":"Aronov B, Chazelle B, Edelsbrunner H,
Guibas L, Sharir M, Wenger R. 1990. Points and triangles in the plane and halving planes in space. Proceedings of the 6th annual symposium on Computational geometry. SCG: Symposium on Computational
Geometry, 112–115.","mla":"Aronov, Boris, et al. “Points and Triangles in the Plane and Halving Planes in Space.” Proceedings of the 6th Annual Symposium on Computational Geometry, ACM, 1990, pp.
112–15, doi:10.1145/98524.98548.","chicago":"Aronov, Boris, Bernard Chazelle, Herbert Edelsbrunner, Leonidas Guibas, Micha Sharir, and Rephael Wenger. “Points and Triangles in the Plane and Halving
Planes in Space.” In Proceedings of the 6th Annual Symposium on Computational Geometry, 112–15. ACM, 1990. https://doi.org/10.1145/98524.98548.","apa":"Aronov, B., Chazelle, B., Edelsbrunner, H.,
Guibas, L., Sharir, M., & Wenger, R. (1990). Points and triangles in the plane and halving planes in space. In Proceedings of the 6th annual symposium on Computational geometry (pp. 112–115).
Berkley, CA, United States: ACM. https://doi.org/10.1145/98524.98548","short":"B. Aronov, B. Chazelle, H. Edelsbrunner, L. Guibas, M. Sharir, R. Wenger, in:, Proceedings of the 6th Annual Symposium
on Computational Geometry, ACM, 1990, pp. 112–115."},"scopus_import":"1","date_updated":"2022-02-17T09:42:27Z","language":
[{"lang":"eng"}],"dc":{"title":["Points and triangles in the plane and halving planes in space"],"description":["We prove that for any set S of n points in the plane and n3-α triangles spanned by the
points of S there exists a point (not necessarily of S) contained in at least n3-3α/(512 log25 n) of the triangles. This implies that any set of n points in three - dimensional space defines at most
6.4n8/3 log5/3 n halving planes."],"source":["Aronov B, Chazelle B, Edelsbrunner H, Guibas L, Sharir M, Wenger R. Points and triangles in the plane and halving planes in space. In: Proceedings of the
6th Annual Symposium on Computational Geometry. ACM; 1990:112-115. doi:10.1145/98524.98548"],"language":["eng"],"type":["info:eu-repo/semantics/
["Aronov, Boris","Chazelle, Bernard","Edelsbrunner, Herbert","Guibas, Leonidas","Sharir, Micha","Wenger, Rephael"]},"extern":"1","author":[{"first_name":"Boris","last_name":"Aronov"},
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A mathematician is a person whose primary area of study and research is the field of mathematics, or whose contribution to mathematics is significant, e.g. Isaac Newton.
Problems in mathematics
The publication of new discoveries in mathematics continues at an immense rate in hundreds of scientific journals. One of the most exciting recent developments was the proof of Fermat's last theorem
by Andrew Wiles, following 350 years of the brightest mathematical minds attempting to settle the problem.
There are many famous open problems in mathematics, many dating back tens, if not hundreds, of years. Some examples include the Riemann hypothesis (from 1859) and Goldbach's conjecture (1742). The
Millennium Prize Problems highlight longstanding, important problems in mathematics and offers a US$1,000,000 reward for solving any one of them. One of these problems, the Poincaré conjecture
(1904), was proven by Russian mathematician Grigori Perelman in a paper released in 2003; peer review was completed in 2006, and the proof was accepted as valid.
Mathematicians are typically interested in finding and describing patterns, or finding (mathematical) proofs of theorems. Most problems and theorems come from within mathematics itself, or are
inspired by theoretical physics. To a lesser extent, problems have come from economics, games and computer science. Some problems are simply created for the challenge of solving them. Although much
mathematics is not immediately useful, history has shown that eventually applications are found. For example, number theory originally seemed to be without purpose to the real world, but after the
development of computers it gained important applications to algorithms and cryptography.
There are no Nobel Prizes awarded to mathematicians. The award that is generally viewed as having the highest prestige in mathematics is the Fields Medal. This medal, sometimes described as the
"Nobel Prize of Mathematics", is awarded once every four years to as many as four young (under 40 years old) awardees at a time. Other prominent prizes include the Abel Prize, the Nemmers Prize, the
Wolf Prize, the Schock Prize, and the Nevanlinna Prize.
Mathematics differs from natural sciences in that physical theories in the sciences are tested by experiments, while mathematical statements are supported by proofs which may be verified objectively
by mathematicians. If a certain statement is believed to be true by mathematicians (typically because special cases have been confirmed to some degree) but has neither been proven nor dis-proven, it
is called a conjecture, as opposed to the ultimate goal: a theorem that is proven true. Physical theories may be expected to change whenever new information about our physical world is discovered.
Mathematics changes in a different way: new ideas don't falsify old ones but rather are used to generalize what was known before to capture a broader range of phenomena. For instance, calculus (in
one variable) generalizes to multivariable calculus, which generalizes to analysis on manifolds. The development of algebraic geometry from its classical to modern forms is a particularly striking
example of the way an area of mathematics can change radically in its viewpoint without making what was proved before in any way incorrect. While a theorem, once proved, is true forever, our
understanding of what the theorem really means gains in profundity as the mathematics around the theorem grows. A mathematician feels that a theorem is better understood when it can be extended to
apply in a broader setting than previously known. For instance, Fermat's little theorem for the nonzero integers modulo a prime generalizes to Euler's theorem for the invertible numbers modulo any
nonzero integer, which generalizes to Lagrange's theorem for finite groups.
While the majority of mathematicians are male, there have been some demographic changes since World War II. Some prominent female mathematicians are Ada Lovelace (1815 - 1852), Maria Gaetana Agnesi
(1718-1799), Emmy Noether (1882 - 1935), Sophie Germain (1776 - 1831), Sofia Kovalevskaya (1850 - 1891), Rózsa Péter (1905 - 1977), Julia Robinson (1919 - 1985), Olga Taussky-Todd (1906 - 1995),
Émilie du Châtelet (1706 – 1749), Mary Cartwright (1900 - 1998), and Hypatia of Alexandria (ca. 400 AD). The AMS and other mathematical societies offer several prizes aimed at increasing the
representation of women and minorities in the future of mathematics.
Doctoral degree statistics for mathematicians in the United States
The number of doctoral degrees in mathematics awarded each year in the United States has ranged from 750 to 1230 over the past 35 years. Archived January 14, 2006 at the Wayback Machine In the early
seventies, degree awards were at their peak, followed by a decline throughout the seventies, a rise through the eighties, and another peak through the nineties. Unemployment for new doctoral
recipients peaked at 10.7% in 1994 but was as low as 3.3% by 2000. The percentage of female doctoral recipients increased from 15% in 1980 to 30% in 2000.
As of 2000, there are approximately 21,000 full-time faculty positions in mathematics at colleges and universities in the United States. Of these positions about 36% are at institutions whose highest
degree granted in mathematics is a bachelor's degree, 23% at institutions that offer a master's degree and 41% at institutions offering a doctoral degree.
The median age for doctoral recipients in 1999-2000 was 30, and the mean age was 31.7.
Wikiquote has a collection of quotations related to: Mathematician
The following are quotations about mathematicians, or by mathematicians.
A mathematician is a machine for turning coffee into theorems.
—Attributed to both Alfréd Rényi and Paul Erdős
Die Mathematiker sind eine Art Franzosen; redet man mit ihnen, so übersetzen sie es in ihre Sprache, und dann ist es alsobald ganz etwas anderes. (Mathematicians are [like] a sort of Frenchmen;
if you talk to them, they translate it into their own language, and then it is immediately something quite different.)
Some humans are mathematicians; others aren't.
— Jane Goodall (1971) In the Shadow of Man
Each generation has its few great mathematicians...and [the others'] research harms no one.
—Alfred Adler, "Mathematics and Creativity"
Mathematics, rightly viewed, possesses not only truth, but supreme beauty – a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the
gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.
—Bertrand Russell, The Study of Mathematics
A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas.
— G. H. Hardy, A Mathematician's Apology
Another roof, another proof.
— Paul Erdős
Some of you may have met mathematicians and wondered how they got that way.
— Tom Lehrer
It is impossible to be a mathematician without being a poet in soul.
— Sofia Kovalevskaya
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Hi Pouya,
07-29-2013 02:55 PM
I am using Intel MKL library to solve a system of linear equations (A*x = b) with multiple right-hand side (rhs) vectors. The rhs vectors are generated asynchronously and through a separate routine
and therefore, it is not possible to solve them all at once.
In order to expedite the program, a multi-threaded program is used where each thread is responsible for solving a single rhs vectors. Since the matrix A is always constant, LU factorization should be
performed once and the factors are used subsequently in all threads. So, I factor A using following command
dss_factor_real(handle, opt, data);
and pass the handle to the threads to solve the problems using following command:
dss_solve_real(handle, opt, rhs, nRhs, sol);
However, I found out that it is not thread-safe to use the same handle in several instances ofdss_solve_real. Apparently, for some reason, MKL library changes handle in each instance which creates
race condition. I read the MKL manual but could not find anything relevant. Since it is not logical to factorize A for each thread, I am wondering if there is any way to overcome this problem and use
the same handle everywhere.
Thanks in advance for your help
07-29-2013 04:06 PM
07-29-2013 04:21 PM
07-29-2013 11:00 PM
07-29-2013 11:11 PM
08-01-2013 05:31 PM
08-01-2013 05:46 PM
08-01-2013 08:53 PM
08-04-2013 12:28 AM
08-06-2013 11:36 PM
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Fundamentals of Heat, Light & Sound
Learning Objectives
By the end of this section, you will be able to:
• Define pressure.
• Explain the relationship between pressure and force.
• Calculate force given pressure and area.
You have no doubt heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of
many examples of pressures in fluids. Pressure P is defined as
where F is a force applied to an area A that is perpendicular to the force.
Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or
A given force can have a significantly different effect depending on the area over which the force is exerted, as shown in Figure 1. The SI unit for pressure is the pascal, where
In addition to the pascal, there are many other units for pressure that are in common use. In meteorology, atmospheric pressure is often described in units of millibar (mb), where
Pounds per square inch (lb/in^2 or psi) is still sometimes used as a measure of tire pressure, and millimeters of mercury (mm Hg) is still often used in the measurement of blood pressure. Pressure is
defined for all states of matter but is particularly important when discussing fluids.
Figure 1. (a) While the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the same force applied to an area the size of the sharp end of a
needle is great enough to break the skin.
Example 1. What Force Does a Pressure Exert?
An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads 6.90 × 10^6 Pa. What force does the air
inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter?
We can find the force exerted from the definition of pressure given in A acted upon.
By rearranging the definition of pressure to solve for force, we see that
F = PA
Here, the pressure P is given, as is the area of the end of the cylinder A, given by A = π r^2. Thus,
Wow! No wonder the tank must be strong. Since we found F = PA, we see that the force exerted by a pressure is directly proportional to the area acted upon as well as the pressure itself.
The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids
cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined
directions: they are always exerted perpendicular to any surface. (See the tire in Figure 2, for example.) Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel
pressure on all sides. (See Figure 3.)
Figure 2. Pressure inside this tire exerts forces perpendicular to all surfaces it contacts. The arrows give representative directions and magnitudes of the forces exerted at various points. Note
that static fluids do not exert shearing forces.
Figure 3. Pressure is exerted on all sides of this swimmer, since the water would flow into the space he occupies if he were not there. The arrows represent the directions and magnitudes of the
forces exerted at various points on the swimmer. Note that the forces are larger underneath, due to greater depth, giving a net upward or buoyant force that is balanced by the weight of the swimmer.
PhET Explorations: Gas Properties
Pump gas molecules to a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the
gas vary in relation to each other.
Section Summary
• Pressure is the force per unit perpendicular area over which the force is applied. In equation form, pressure is defined as
• The SI unit of pressure is pascal and
Conceptual Questions
1. How is pressure related to the sharpness of a knife and its ability to cut?
2. Why does a dull hypodermic needle hurt more than a sharp one?
3. The outward force on one end of an air tank was calculated in Example 1: Calculating Force Exerted by the Air. How is this force balanced? (The tank does not accelerate, so the force must be
4. Why is force exerted by static fluids always perpendicular to a surface?
5. In a remote location near the North Pole, an iceberg floats in a lake. Next to the lake (assume it is not frozen) sits a comparably sized glacier sitting on land. If both chunks of ice should melt
due to rising global temperatures (and the melted ice all goes into the lake), which ice chunk would give the greatest increase in the level of the lake water, if any?
6. How do jogging on soft ground and wearing padded shoes reduce the pressures to which the feet and legs are subjected?
7. Toe dancing (as in ballet) is much harder on toes than normal dancing or walking. Explain in terms of pressure.
8. How do you convert pressure units like millimeters of mercury, centimeters of water, and inches of mercury into units like newtons per meter squared without resorting to a table of pressure
conversion factors?
Problems & Exercises
1. As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1.50 cm^2 and the
woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such
large pressures.)
2. The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure
is exerted on the record in N/m^2?
3. Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter
to create a pressure of 3.00 × 10^9 N/m^2 (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
the force per unit area perpendicular to the force, over which the force acts
Selected Solutions to Problems & Exercises
1. 3.59 × 10^6 Pa; or 521 lb/in^2
3. 2.36 × 10^3 N
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Continuous Predicate
A continuous predicate is a relational predicate that analyzes into parts all similar to the whole.
Continuous predicate is a term coined by Charles Sanders Peirce to describe a special type of relational predicate that results as the limit of a recursive process of hypostatic abstraction.
Here is one of Peirce's definitive treatments of the concept:
When we have analyzed a proposition so as to throw into the subject everything that can be removed from the predicate, all that it remains for the predicate to represent is the form of connection
between the different subjects as expressed in the propositional form. What I mean by "everything that can be removed from the predicate" is best explained by giving an example of something not so
But first take something removable. "Cain kills Abel." Here the predicate appears as "— kills —." But we can remove killing from the predicate and make the latter "— stands in the relation — to —."
Suppose we attempt to remove more from the predicate and put the last into the form "— exercises the function of relate of the relation — to —" and then putting "the function of relate to the
relation" into another subject leave as predicate "— exercises — in respect to — to —." But this "exercises" expresses "exercises the function". Nay more, it expresses "exercises the function of
relate", so that we find that though we may put this into a separate subject, it continues in the predicate just the same.
Stating this in another form, to say that "A is in the relation R to B " is to say that A is in a certain relation to R. Let us separate this out thus: "A is in the relation R' (where R' is the
relation of a relate to the relation of which it is the relate) to R to B ". But A is here said to be in a certain relation to the relation R'. So that we can express the same fact by saying, "A is
in the relation R' to the relation R' to the relation R to B ", and so on ad infinitum.
A predicate which can thus be analyzed into parts all homogeneous with the whole I call a continuous predicate. It is very important in logical analysis, because a continuous predicate obviously
cannot be a compound except of continuous predicates, and thus when we have carried analysis so far as to leave only a continuous predicate, we have carried it to its ultimate elements. (C.S. Peirce,
"Letters to Lady Welby" (14 December 1908), Selected Writings'', pp. 396-397).
Peirce, C.S., "Letters to Lady Welby", pp. 380-432 in Charles S. Peirce : Selected Writings (Values in a Universe of Chance), Philip P. Wiener (ed.), Dover Publications, New York, NY, 1966.
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Let g(x) be a cubic polynomial having local maximum at x=−1 and... | Filo
Let be a cubic polynomial having local maximum at and has a local minimum at . If , then :
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Question Text Let be a cubic polynomial having local maximum at and has a local minimum at . If , then :
Updated On Sep 21, 2023
Topic Application of Derivatives
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 1
Upvotes 157
Avg. Video Duration 16 min
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test bank for Calculus: Early Transcendentals 8th Edition by James Stewart
Name: test bank for Calculus: Early Transcendentals 8th Edition by James Stewart
Format: Word /pdf Zip/All chapter include
price: 25$USD
Download: Click online customer to buy
introduction name：TEST BANK for Calculus: Early Transcendentals 8th Edition Edition：8th Edition author：by James Stewart ISBN：978-1285741550 ISBN-10: 1285741552 type：Test bank /题库 format：word/
zip All chapter include 完整打包下载 Success in your calculus course starts here! James Stewart’s Calculus: Early Transcendentals texts are world-wide best-sellers for a reason: they are clear,
accurate, and filled with relevant, real-world examples. With CALCULUS: EARLY TRANSCENDENTALS, Eighth Edition, Stewart conveys not only the utility of calculus to help you develop technical
competence, but also gives you an appreciation for the intrinsic beauty of the subject. His patient examples and built-in learning aids will help you build your mathematical confidence and achieve
your goals in the course. Cover + Preview Relevant solution manual for Calculus: Early Transcendentals 8th Edition
: Early Transcendentals
8th Edition
Edition：8th Edition
author：by James Stewart
ISBN-10: 1285741552
type：Test bank /题库
All chapter include 完整打包下载
Success in your calculus course starts here! James Stewart’s Calculus: Early Transcendentals texts are world-wide best-sellers for a reason: they are clear, accurate, and filled with relevant,
real-world examples. With CALCULUS: EARLY TRANSCENDENTALS, Eighth Edition, Stewart conveys not only the utility of calculus to help you develop technical competence, but also gives you an
appreciation for the intrinsic beauty of the subject. His patient examples and built-in learning aids will help you build your mathematical confidence and achieve your goals in the course.
Cover + Preview
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Reshak, Ali. H. and Stys, D. and Auluck, S. and Kityk, I. V. (2010) Dispersion of linear and nonlinear optical susceptibilities and the hyperpolarizability of 3-methyl-4-phenyl-5-(2-pyridyl)
-1,2,4-triazole. Physical Chemistry Chemical Physics , 13 (7). pp. 2945-2952. ISSN 1463-9084
PDF - Published Version
Restricted to Registered users only
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As a starting point for our calculation of 3-methyl-4-phenyl-5-(2-pyridyl)-1,2,4-triazole we used the XRD data obtained by C. Liu, Z. Wang, H. Xiao, Y. Lan, X. Li, S. Wang, Jie Tang, Z. Chen, J.
Chem. Crystallogr., 2009 39 881. The structure was optimized by minimization of the forces acting on the atoms keeping the lattice parameters fixed with the experimental values. Using the relaxed
geometry we have performed a comprehensive theoretical investigation of dispersion of the linear and nonlinear optical susceptibilities of 3-methyl-4-phenyl-5-(2-pyridyl)-1,2,4-triazole using the
full potential linear augmented plane wave method. The local density approximation by Ceperley–Alder (CA) exchange–correlation potential was applied. The full potential calculations show that this
material possesses a direct energy gap of 3.4 eV for the original experimental structure and 3.2 eV for the optimized structure. We have calculated the complex’s dielectric susceptibility ε(ω)
dispersion, its zero-frequency limit ε1(0) and the birefringence. We find that a 3-methyl-4-phenyl-5-(2-pyridyl)-1,2,4-triazole crystal possesses a negative birefringence at the low-frequency limit
Δn(0) which is equal to about −0.182 (−0.192) and at λ = 1064 nm is −0.193 (−0.21) for the non-optimized structure (optimized one), respectively. We also report calculations of the complex
second-order optical susceptibility dispersions for the principal tensor components: χ(2)123(ω), χ(2)231(ω) and χ(2)312(ω). The intra- and inter-band contributions to these susceptibilities are
evaluated. The calculated total second order susceptibility tensor components at the low-frequency limit |χ(2)ijk(0)| and |χ(2)ijk(ω)| at λ = 1064 nm for all the three tensor components are
evaluated. We established that the calculated microscopic second order hyperpolarizability, βijk, the vector component along the dipole moment direction, at the low-frequency limit and at λ = 1064
nm, for the dominant component |χ(2)123(ω)| is 4.99 × 10−30 esu (3.4 × 10−30 esu) and 7.72 × 10−30 esu (5.1 × 10−30 esu), respectively for the non-optimized structure (optimized structure).
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Desi Ghee 5kg Price In India- Shahjighee - ShahJi Ghee
Desi Ghee 5kg Price In India- Shahjighee
A2 desi ghee is the purest ghee available. Because A2 ghee is made in India using the traditional hand-churned bilona method from superior desi cow\’s milk. It is an ancient Vedic method of making
Ghee is available in two varieties:
1. Normal Ghee or Market Ghee
2. A2 Desi Ghee or Organic Ghee
A2 Desi Ghee is the most popular because it is the purest form of ghee and has numerous health benefits. As a result, A2 ghee is recommended by the majority of doctors, nutritionists, and health
You\’ve come to the right place if you want to buy or learn more about the best desi ghee 5 liters price in India. Today, we\’ll look at the prices of pure A2 desi ghee in India.
Desi Ghee Price 5 liters Calculation
When it comes to ghee, 1 kg is not equal to 1 liter. Because
At 30°C, ghee has a specific density of 0.91.
As a result, the mass of 1 liter of Ghee = 0.91✖1000ml = 910 grams.
Price of production of 5 Liter of Pure Desi Ghee
To make 1 liter of pure Desi ghee, the bilona method requires 25-28 liters of Desi cow\’s milk (ghee made from curd). Depending on the Desi cow breed, such as Gir or Sahiwal, a liter of Desi milk
costs between 70 and 120 rupees.
1 liter of Pure Desi Ghee required 25-28 liters of milk
If we take the cheapest option, the price of a liter of Desi Ghee is equal to 25 liters multiplied by 70 rupees equaling ₹1,750.
Similarly, 5 liters of pure desi cow ghee costs Rs. 1,750 ✖5 = Rs. ₹8,750/-.
2) If we take the most expensive option, the price of 1 liter of Desi Ghee is = 28 liters x 120 = ₹3,360/-
Similarly, the cost of 5 liters of pure desi cow ghee is equal to ₹3,360 ✖ 5 = ₹16,800/-
So, the average price of one liter of pure desi cow ghee = (₹1,750 + ₹3,360)/2 = ₹2,555/-
Similarly, the average price of 5 liters of pure desi cow ghee = (₹8,750 + ₹16,800)/2 =12,775/-
Ghee jars cost between ₹50 and ₹150.
So the total cost of producing 5 liters of normal ghee is approximately = ₹8,750 to ₹16,800 + Ghee Jar Cost + Delivery Charge + 18% GST tax.
Ghee prices vary depending on the breed of desi cow, so if someone is selling ghee for less than this price, the ghee is not original or the company is offering a discount to gain a new customer.
We recommend consuming A2 Desi Ghee to keep your and your family\’s health in check. A2 desi ghee is available for purchase directly from our farm.
You can buy mother\’s hand made pure desi ghee from our farm here.
Desi Sahiwal Cow Ghee Desi Gir Cow Ghee
Normal Ghee Price 5 liters Calculation
Price of production of 5 Liters of Normal Ghee
Normal ghee has a low production cost because it is made from milk cream, and it is referred to as low-quality ghee. The market price for 1 kg of milk cream is between ₹200 and ₹250.
1 kg of milk cream can yield approximately 1/2 kg of ghee.
If we take the cheapest option, the price of a liter of normal Ghee is equal to the cost of 2 kg of milk cream i.e. ₹200 x 2 = ₹400.
Similarly, 5 liters of normal cow ghee costs = ₹400 ✖5 = ₹2, 000/-.
2) If we take the most expensive option, the price of 1 liter of normal ghee is = ₹250 x 2 = ₹500.
Similarly, the cost of 5 liters of normal cow ghee is equal to ₹500 ✖ 5 = ₹2,500/-
So, the average price of 1 liter of normal cow ghee = (₹400 + ₹500)/2 = ₹450/-
Similarly, the average price of 5 liters of normal cow ghee = (₹2, 000 + ₹2,500)/2 = ₹2,250/-
So the total cost of producing 5 liters of normal ghee is approximate = ₹2, 000 to ₹2,500 + Ghee Jar Cost + Delivery Charge + 18% GST tax.
So, if you see ghee in the market for ₹400 to ₹500 per liter, it is made from cream and has a low nutritional value. We do not recommend regular ghee because it causes more harm than good to your
body. To keep your and your family\’s health in check, we recommend consuming A2 Desi Ghee. A2 desi ghee is available for purchase directly from our farm.
Industrial Ghee Price 5 liters Calculation
Price of production of 5 Liters of Industrial Ghee
Because industrial ghee is made from butter using machines, the production cost is very low.
Butter is purchased at a low cost by industries.
Machines were used to extract ghee (clarified butter) from butter.
So, the market price for 1 liter of industrial ghee is approximately = ₹200 – ₹450 kg. And for 5 liters, it would be ₹1,000 – ₹2,250/-.
This ghee is known as \”market ghee\” and is of very low quality. It causes more harm than good. As a result, we do not recommend eating market ghee to our customers.
We recommend consuming A2 Desi Ghee to maintain your and your family\’s health. You can purchase A2 desi ghee directly from our farm.
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Fractions - A Brief Visual Exploration of A Dictionary of Typography
A fraction is a part of a unit, written with two figures, with a line between them, thus—¼, ½, ¾, &c. The upper figure is called the numerator, the lower one the denominator. Some fractions are cast
in one piece, and the following are those frequently used:—
¼ ½ ¾ ⅓ ⅔ ⅛ ⅜ ⅝ ⅞
Fractions are also cast in two pieces, called split fractions, by means of which the denominators may be extended to any amount. The separatrix, or rule between the figures, was formerly joined to
the foot of the first, but is now attached to the head of the denominators.
A fraction is a part of a unit, written with two figures, with a line between them, thus—¼, ½, ¾, &c. The upper figure is called the numerator, the lower one the denominator. Some fractions are cast
in one piece, and the following are those frequently used:—
¼ ½ ¾ ⅓ ⅔ ⅛ ⅜ ⅝ ⅞
Fractions are also cast in two pieces, called split fractions, by means of which the denominators may be extended to any amount. The separatrix, or rule between the figures, was formerly joined to
the foot of the first, but is now attached to the head of the denominators.
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Activity, Half-life And Decay Constant | Mini Physics - Free Physics Notes
The activity of a radioactive substance is defined as the average number of atoms disintegrating per unit time.
• An activity of one decay per second is one Becquerel (1 Bq)
Activity A is directly proportional to the number of parent nuclei N present at that instant:
$\begin{aligned}A & \propto N \\ A & = \, – \, \frac{dN}{dt} \\ & = \lambda N \end{aligned}$
, where
• N is number of parent nuclei,
• t is the time,
• λ is the decay constant.
The decay constant λ of a nucleus is defined as its probability of decay per unit time.
Half-life is defined as the time taken for half the original number of radioactive nuclei to decay.
Useful Equations:
$N = N_{o} \, e^{-\lambda t}$, where
• N[o] is the initial number of radioactive nuclides and
• N is the number of nuclides remaining after a time t.
$t_{\frac{1}{2}} = \frac{ln \, 2}{\lambda}$, t[1/2] is half-life.
$\left( \frac{N}{N_{o}} \right) = \left( \frac{1}{2} \right)^{n}$ , where n is the number of half-life passed.
Graphical Representation of Decay of Parent Nuclide
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The F-theorem and F-maximization
This contribution contains a review of the role of the three-sphere free energy F in recent developments related to the F-theorem and F-maximization. The F-theorem states that for any
Lorentz-invariant RG trajectory connecting a conformal field theory CFT in the ultraviolet to a conformal field theory CFT, the F-coefficient decreases: . I provide many examples of CFTs where one
can compute F, approximately or exactly, and discuss various checks of the F-theorem. F-maximization is the principle that in an SCFT, viewed as the deep IR limit of an RG trajectory preserving
supersymmetry, the superconformal R-symmetry maximizes F within the set of all R-symmetries preserved by the RG trajectory. I review the derivation of this result and provide examples.
All Science Journal Classification (ASJC) codes
• Statistical and Nonlinear Physics
• Statistics and Probability
• Modeling and Simulation
• Mathematical Physics
• General Physics and Astronomy
• F-maximization
• F-theorem
• Lorentz-invariant
• three-sphere free energy
Dive into the research topics of 'The F-theorem and F-maximization'. Together they form a unique fingerprint.
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How to check if a binary contains the Dual EC backdoor for the NSA
How to check if a binary contains the Dual EC backdoor for the NSA posted December 2015
this is what you should type:
strings your_binary | grep -C5 -i "c97445f45cdef9f0d3e05e1e585fc297235b82b5be8ff3efca67c59852018192\|8e722de3125bddb05580164bfe20b8b432216a62926c57502ceede31c47816edd1e89769124179d0b695106428815065\|1b9fa3e518d683c6b65763694ac8efbaec6fab44f2276171a42726507dd08add4c3b3f4c1ebc5b1222ddba077f72943b24c3edfa0f85fe24d0c8c01591f0be6f63"
After all the Jupiner fiasco, I wondered how people could look if a binary contained an implementation of Dual EC, and worse, if it contained Dual EC with the NSA's values for P and Q.
The easier thing I could think of is the use of strings to check if the binary contains the hex values of some Dual EC parameters:
strings your_binary | grep -C5 -i `python -c "print '%x' % 115792089210356248762697446949407573530086143415290314195533631308867097853951"`
This is the value of the prime p of the curve P-256. Other curves can be used for Dual EC though, so you should also check for the curve P-384:
strings your_binary | grep -C5 -i `python -c "print '%x' % 39402006196394479212279040100143613805079739270465446667948293404245721771496870329047266088258938001861606973112319"`
and the curve P-521:
strings your_binary | grep -C5 -i `python -c "print '%x' % 6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151 "`
I checked the binaries of ScreenOS (taken from here) and they contained these three curves parameters. But this doesn't mean anything, just that these curves are stored, maybe used, maybe used for
Dual EC...
To check if it uses the NSA's P and Q, you should grep for P and Q x coordinates from the same NIST paper.
This looks for all the x coordinates of the different P for each curves. This is not that informative since it is the standard point P as a generator of P-256
strings your_binary | grep -C5 -i "6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296\|aa87ca22be8b05378eb1c71ef320ad746e1d3b628ba79b9859f741e082542a385502f25dbf55296c3a545e3872760ab7\|c6858e06b70404e9cd9e3ecb662395b4429c648139053fb521f828af606b4d3dbaa14b5e77efe75928fe1dc127a2ffa8de3348b3c1856a429bf97e7e31c2e5bd66"
Testing the ScreenOS binaries, I get all the matches. This means that the parameters for P-256 and maybe Dual EC are indeed stored in the binaries.
weirdly, testing for Qs I don't get any match. So Dual EC or not?
strings your_binary | grep -C5 -i "c97445f45cdef9f0d3e05e1e585fc297235b82b5be8ff3efca67c59852018192\|8e722de3125bddb05580164bfe20b8b432216a62926c57502ceede31c47816edd1e89769124179d0b695106428815065\|1b9fa3e518d683c6b65763694ac8efbaec6fab44f2276171a42726507dd08add4c3b3f4c1ebc5b1222ddba077f72943b24c3edfa0f85fe24d0c8c01591f0be6f63"
Re-reading CVE-2015-7765:
The Dual_EC_DRBG 'Q' parameter was replaced with 9585320EEAF81044F20D55030A035B11BECE81C785E6C933E4A8A131F6578107 and the secondary ANSI X.9.31 PRNG was broken, allowing raw Dual_EC output to be
exposed to the network. Please see this blog post for more information.
Diffing the vulnerable (and patched binaries. I see that only the P-256 curve \(Q\) was modified from Juniper's values, other curves were left intact. I guess this means that only the P-256 curve was
being used in Dual EC.
If you know how Dual EC works (if you don't check my video), you know that to establish a backdoor into it you need to generate \(P\) and \(Q\) accordingly. So changing the value \(Q\) with no
correlation to \(P\) is not going to work, worse it could be that \(Q\) is too "close" to P and thus the secret \(d\) linking them could be easily found ( \(P = dQ \)).
Now one clever way to generate a secure \(Q\) with a strong value \(d\) that only you would know is to choose a large and random \(d\) and calculate its inverse \(d^{-1} \pmod{ord_{E}} \). You have
your \(Q\) and your \(d\)!
\[ d^{-1} P = Q \]
bonus: here's a small script that attempts to find \(d\) in the hypothesis \(d\) would be small (the fastest way to compute an elliptic curve discrete log is to use Pollard Rho's algorithm)
p256 = 115792089210356248762697446949407573530086143415290314195533631308867097853951
a256 = p256 - 3
b256 = 41058363725152142129326129780047268409114441015993725554835256314039467401291
## base point values
gx = 48439561293906451759052585252797914202762949526041747995844080717082404635286
gy = 36134250956749795798585127919587881956611106672985015071877198253568414405109
## order of the curve
qq = 115792089210356248762697446949407573529996955224135760342422259061068512044369
# init curve
FF = GF(p256)
EE = EllipticCurve([FF(a256), FF(b256)])
# define the base point
G = EE(FF(gx), FF(gy))
# P and Q
P = EE(FF(0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296), FF(0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5))
# What is Q_y ?
fakeQ_x = FF(0x9585320EEAF81044F20D55030A035B11BECE81C785E6C933E4A8A131F6578107)
fakeQ = EE.lift_x(fakeQ_x)
print discrete_log(P, fakeQ, fakeQ.order(), operation='+')
The lift_x function allows me to get back the \(y\) coordinate of the new \(Q\):
(67629950588023933528541229604710117449302072530149437760903126201748084457735 : 36302909024827150197051335911751453929694646558289630356143881318153389358554 : 1)
leave a comment...
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FOM: FoundationalCompleteness
Harvey Friedman friedman at math.ohio-state.edu
Mon Nov 3 04:13:57 EST 1997
Some of my e-mailings will be in the form of detailed replies to others,
and some of them will be self contained discussions of a foundational (of
math) matter. You can easily tell from the Subject header. This is of the
latter kind.
I didn't intend to write this one, but the topic came to mind in reading
the postings of those who are, in some way, unsatisfied with the usual set
theoretic foundations for mathematics, and strive for a kind of
"structuralist" viewpoint - particularly, Pratt and Barwise.
The usual set theoretic foundations is very powerful, coherent, concise,
successful, explanatory, impressive, and totally dominating at this time.
Taken as a whole, with the major supporting classical developments, it is
certainly one of the few greatest acheivments of the human mind of all time.
However, it also does not come close to doing everything one might demand
of a foundation for mathematics. At the present time, there is no full
blown proposal for scrapping it and replacing it with anything
substantially different that isn't far more trouble than its worth. Present
cures are far far far worse than any perceived disease.
Now this does not mean that the usual set theoretic foundations might not
give way to a better foundations, or might not be altered in some very
significant and permanent way. In fact, I can tell you that I work on this
from time to time. It just that people should recognize what's involved in
doing such an overhaul, and not fool themselves into either
i) embracing something that is either essentially the same as the
usual set theoretic foundations of mathematics; or
ii) embracing something that doesn't even minimally accomplish what
is so successfully accomplished by the usual set theoretic foundations of
Now before I remind everybody of some of the most vital features of the
usual set theoretic foundations for mathematics, let me state a great,
great, great, theorem in the foundations of mathematics:
THEOREM. Sets under membership form the simplest foundationally complete
There is one trouble with this result: I don't know how to properly
formulate it. In particular, I don't know how to properly formulate
"foundational completeness" or "simplest."
Making sense of this "Theorem" and closely related matters are typical
major issues in genuine foundations of mathematics. Now before coming back
to this, let me summarize the greatest of the usual set theoretic
foundations of mathematics.
First of all, set theory is unabashedly materialistic - a perhaps
nonstandard word I use to describe the opposite of structuralistic. The
viewpoint is that the empty set of set theory has a unique unequivocal
meaning independently of context. There is the empty set, and that's that.
It doesn't need any context. There is no talk of identifying distinct empty
sets because they form the same function.
This materialistic concept of set seems to be very congenial to almost
everybody for a while. Thus {emptyset} also has a unique unequivocal
meaning independently of context. In fact, one can construct the so called
hereditarily finite (HF) sets by the following process:
i) the emptyset is HF;
ii) if x,y are HF then x union {y} is HF.
This has a clarity and congeniality for most people, without invoking any
structuralist ideas.
Now I can already hear the following remark: see, you have used an
inductive construction that has not only not been formalized in set
theoretic terms yet, but is not even best formulated in set theoretic
Yes, this is true. And yes, there is an idea of inductive construction - at
least for the natural numbers - which is not directly faithfully conceived
of in purely set theoretic terms. However, look at the costs of scrapping
the set theoretic approach in favor of "inductive construction." Can this
really be done? I have certainly thought about this, but without success.
It is certainly an attractive idea, and we explicitly formulate this:
FOUNDATIONAL ISSUE. Is there an alternative adequate foundation for
mathematics that is based on "inductive construction?" In particular, one
wants to capture set theory viewed as an inductive construction. If not,
one wants to construct a significant portion of set theory as an inductive
Now, instead of scrapping the set theoretic approach in favor of "inductive
construction," what about incorporating both? Yes, this can be done in
various ways. However, so what? This is only really interesting if one can
isolate a small handful of additional ideas that one wishes to directly
faithfully incorporate into the prospective foundation for mathematics.
Better yet - prove some sort of completeness of this handful.
However, consider the situation in mathematics that was one of the major
precipitating factors that made people realize the urgency of foundations.
Namely, people were creating all kinds of mathematical concepts - groups,
rings, fields, integers, rationals, reals, complexes, division rings,
functions of a real and complex variable, series, etcetera. There was no
unifying principle as to what is or is not a legitimate construction.
Mathematicians do not want to go down that road again, and are comforted by
the fact that this matter has been resolved by set theory - even if it does
not provide for a directly faithful formalization of the way they actually
visualize and think. In summary, there is a danger of the cure being far
far far worse than the disease.
Now, coming back to set theory and HF. Obviously, it is congenial and
natural to most people to form the set HF. And then there is the natural
idea of subset of HF. Then for each natural number n, one can form the n-th
power set of HF; let's write this as V(w+n).
Let us give the name V(w+w) for the universe of all sets that are members
of some V(w+n). There are a number of beautiful axioms one immediately
writes down about this universe. A small number of them allow for the
derivation of lots of others. This is a very coherent and workable system
of objects, under epsilon, for a foundation of a very very large portion of
mathematical practice. Now I have been very concerned with the following
for nearly 30 years:
FOUNDATIONAL ISSUE. What interesting mathematics is missing if one uses
V(w+w) (with the obvious associated axioms)? Obviously, one does not mean
simply that V(w+w) itself is missing, since V(w+w) is meant to provide
ontological overkill. Instead, one means that what mathematical information
of an ordinary mathematical character cannot be derived in such a
Ex: Let E be a subset of the unit square in the plane, which is symmetric
about the line y = x. Then E contains or is disjoint from the graph of a
Borel measurable function.
This cannot be proved in such a Foundation, but can be proved in a somewhat
more encompassing foundation. This resuilt is a typical acheivement in
foundations of mathematics.
More information about the FOM mailing list
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Pseudodifferential operators on variable lebesgue spaces
Let M(ℝ^n) be the class of bounded away from one and infinity functions p: ℝ^n → [1,∞] such that the Hardy-Littlewood maximal operator is bounded on the variable Lebesgue space L^p(⋅)(ℝ^n). We show
that if a belongs to the Hörmander class S^n(ρ-1) [ρ, δ] with 0 < ρ ≤ 1, 0 ≤ δ < 1, then the pseudodifferential operator Op(a) is bounded on L^p(⋅)(ℝ^n) provided that p∈M(ℝ^n). Let M* (ℝ^n) be the
class of variable exponents p∈M(ℝ^n) represented as 1/p(x) = θ/p[0] + (1 – θ)/p[1](x) where p[0] ∈ (1,∞), θ ∈(0, 1), and p[1] ∈M(ℝ^n). We prove that if a ∈ S^0 [1,0] slowly oscillates at infinity in
the first variable, then the condition (Formula presented.) is sufficient for the Fredholmness of Op(a) on L^p(⋅)(ℝ^n) whenever p∈M* (ℝ^n). Both theorems generalize pioneering results by Rabinovich
and Samko [24] obtained for globally log-Hölder continuous exponents p, constituting a proper subset of M* (ℝ^n).
Publication series
Name Operator Theory: Advances and Applications
Volume 228
ISSN (Print) 0255-0156
ISSN (Electronic) 2296-4878
Other International workshop on Analysis, Operator Theory, and Mathematical Physics, 2012
Country/Territory Mexico
City Ixtapa
Period 1/23/12 → 1/27/12
• Fefferman-Stein sharp maximal operator
• Fredholmness
• Hardy-Littlewood maximal operator
• Hörmander symbol
• Pseudodifferential operator
• Slowly oscillating symbol
• Variable Lebesgue space
ASJC Scopus subject areas
Dive into the research topics of 'Pseudodifferential operators on variable lebesgue spaces'. Together they form a unique fingerprint.
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A[n]^D = <cos 2plZ[n]> = exp(-2p^2l^2<Z^2[n]>)
In that equation, <Z^2[n]> is modelled in ARIT by a flexible variation law of the distortion versus the distance according to the following equation:
<Z^2[n]> = |n|^K<Z^2[1]>
The ARIT program refines two microstrain parameters : K and <Z^2[1]> (i.e. <Z^2[n]> for n = 1). It is to be noted that if K is refined to K = 2, the calculated microstrain profile shape will be
Gaussian, and if K is refined to K = 1, it will be Lorentzian. Other shapes being possible, depending on the final refined value of K.
Size effect in ARIT
The size Fourier coefficient A[n]^S is given in terms of p(i), the fraction of the columns of length i cells by the expression [3]:
where N[3] is the mean column length number.
Modelling p(i) allows to define A[n]^S. It would not be difficult to introduce several models in ARIT, however, there is currently only one model proposed which is a continuously decreasing size
distribution function defined by :
p(n) = g^2exp(-g|n|)
The size Fourier coefficients corresponding to this arbitrary size distribution function is :
A[n]^S = exp(-g|n|) ,
And the average number of unit cells is N[3] = 1/g.
Practically, fictitious quantities have to be defined like in the Warren book ([3] p. 273). The real distance along the columns of cells perpendicular to the reflecting planes is defined by :
L = n a’[3]
Where a’[3] depends on the interval of definition of the reflections [q[2]q[1]]:
a’[3] = l / 2 (sin q[2] – sin q[1])
Anisotropic size and microstrain effects in ARIT
C – Results on CeO[2]
● filename.dat : contains the intensities and background, which was defined visually/manually;
● filename.str : contains the starting parameters describing profile shapes;
● filename.pr1 : contains the starting structure factor amplitudes;
● filename.typ : contains the final results;
● filename.gif : contains the drawing from the .prf file (not given, but can be rebuilt by running ARIT on the joined data) by WinPlotr.
1- Laboratory x-ray sources : "Common" instrumental setup: University of Le Mans (Armel Le Bail)
l (CuKa1) = 1.54056 Å, l (CuKa2) = 1.54439 Å, I (CuKa2)/ I ((CuKa1) = 0.48, P = 0.8
Comments : for all patterns, definition range for every reflection : 2000 points by steps of 0.01°(2q), corresponding to a’[3] = 4.419 Å.
I - "Instrumental standard"
II - "Broadened sample"
2 - Laboratory x-ray sources : Incident-beam monochromator:
University of Birmingham (J. Ian Langford)
l (CuKa1) = 1.54056 Å, P = 0.8
Comments : for all patterns, definition range for every reflection : 1000 points by steps of 0.02°(2q), corresponding to a’[3] = 4.419 Å. The Kalpha-2 contribution was neglected. The 2x3 original
powder patterns were gathered in 2x1 by changing the scales and step.
I - "Instrumental standard"
II - "Broadened sample"
3 - Synchrotron x-ray sources : 2nd-generation synchrotron, flat-plate geometry: NSLS X3B1 beamline, Brookhaven National Laboratory
(Peter W. Stephens)
l = 0.6998 Å, P = 0
Comments : for all patterns, definition range : 3000 points by steps of 0.003°(2q), corresponding to a’[3] = 4.456 Å. The original powder patterns were rebuilt for having both the same 0.003°(2q)
I - "Instrumental standard"
II - "Broadened sample"
4 - Synchrotron x-ray sources : 3rd-generation synchrotron, capillary geometry: ESRF BM16 beamline, Grenoble
(Olivier Masson and Andy Fitch)
l = 0.39982 Å, P = 0
Comments : for all patterns, definition range : 2600 points by steps of 0.002°(2q), corresponding to a’[3] = 4.406 Å. The original powder patterns were rebuilt for having both the same 0.002°(2q)
I - "Instrumental standard"
II - "Broadened sample"
5 - Neutron sources : ILL D1A diffractometer,
Grenoble (Alan Hewat)
l = 1.91 Å
Comments : for all patterns, definition range : 500 points by steps of 0.05°(2q), corresponding to a’[3] = 4.386 Å.
I - "Instrumental standard"
II - "Broadened sample"
6 - Neutron sources : NCNR BT1 diffractometer,
NIST-Gaithersburg (Brian Toby)
l = 1.5905 Å
Comments : for all patterns, definition range : 400 points by steps of 0.05°(2q) for every reflection, corresponding to a’[3] = 4.876 Å.
I - "Instrumental standard"
II - "Broadened sample"
General comments and conclusion
The decrease in the R[P] and R[WP] reliability factors observed when going from an isotropic size/microstrain refinement to an anisotropic refinement is quite small, showing that if there is really
any anisotropy, it is quite negligible.This is reflected by the small differences in mean size along the directions orthogonal to the (111), (200) and (220) planes. The distortion is so small that it
appears negligible. This explains the extremely dispersed values when comparing the results from the 6 difftactometers : multiplying by ten an extremely small full width at half maximum (FWHM) gives
still a very small FWHM…). The distortion presents thus very large estimated standard deviations. The CeO[2] “broad” case is close to a “size-only effect” situation. The generally much better fit
without size/microstrain effects on the “broadened” powder pattern shows that the size effect model in ARIT is certainly not really adapted : the experimental size distribution function is likely to
be different from a simply exponentially decreasing function. A more flexible model would have to be introduced and tested. The mean size proposed may well be 50% in error. But ARIT does not pretend
more than to give an idea of the size/microstrain magnitudes. Discrepancies between the results from the various data set may essentially come from the problem of finding the background position, and
also from the quite different resolution (for instance, neglecting the instrumental g contribution would be almost possible for the synchrotron Masson data, but certainly not for the neutron Hewat
For this high-resolution reason, the “best” result in this series treated by ARIT is very probably that from the 3^rd generation synchrotron data (Masson), for which the size/microstrain parameters
obtained by using the anisotropic option gives almost the same result as those obtained by applying the isotropic option.From ARIT, the microstructure characteristics of CeO[2] “broad” are finally :
Mean size :
Distortion law : K = 2.89(2)
Strain parameter for n = 1, <Z^2[1]> a’[3]^2 =6.2(8)x10^-7 (Å^2)
The corresponding fit :
These values have now to be compared with the results from other methods. The Warren Fourier analysis method will undoubtedly give THE solution, since there is no serious overlapping here...
And, well, if there is too much discrepancies, the ARIT program will possibly vanish completely ;-). - but this would be a pity, since the concept behind ARIT can be improved by adding a series of
different size/strain models. And it will continue to work when the Warren Fourier analysis method will be impossible to apply : in presence of strong overlapping.
[1] A new study of the structure of LaNi[5]D[6.7] using a modified Rietveld method for the refinement of neutron powder diffraction data. C. Lartigue, A. Le Bail and A. Percheron-Guegan, Journal of
the Less-Common Metals129, 65-76 (1987).
[2] A. Le Bail, Acta Crystallogr. A40, suppl. c369 (1984).
[3] B. E. Warren, “X-Ray Diffraction”, Addison-Wesley, 1969, chapter 13.
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Optimal Allocation of Spinning Reserves in Interconnected Energy Systems with Demand Response Using a Bivariate Wind Prediction Model
Optimal Allocation of Spinning Reserves in Interconnected Energy Systems with Demand
Response Using a Bivariate Wind Prediction Model
Yerzhigit Bapin^1,* , Mehdi Bagheri^1,2 and Vasilios Zarikas^1
1 School of Engineering and Digital Sciences, Nazarbayev University, 53 Kabanbay Batyr ave,
010000 Nur-Sultan, Kazakhstan; [email protected] (M.B.); [email protected] (V.Z.)
2 National Laboratory Astana, Center for Energy and Advanced Material Science, Nazarbayev University, 010000 Nur-Sultan, Kazakhstan
* Correspondence: [email protected]; Tel.:+7-701-444-8411
Received: 23 July 2019; Accepted: 3 September 2019; Published: 9 October 2019 ^^^ Abstract:The proposed study presents a novel probabilistic method for optimal allocation of spinning reserves taking
into consideration load, wind and solar forecast errors, inter-zonal spinning reserve trading, and demand response provided by consumers as a single framework. The model considers the system
contingencies due to random generator outages as well as the uncertainties caused by load and renewable energy forecast errors. The study utilizes a novel approach to model wind speed and its
direction using the bivariate parametric model. The proposed model is applied to the IEEE two-area reliability test system (RTS) to analyze the influence of inter-zonal power generation and demand
response (DR) on the adequacy and economic efficiency of energy systems. In addition, the study analyzed the effect of the bivariate wind prediction model on obtained results. The results demonstrate
that the presence of inter-zonal capacity in ancillary service markets reduce the total expected energy not supplied (EENS) by 81.66%, while inclusion of a DR program results in an additional 1.76%
reduction of EENS. Finally, the proposed bivariate wind prediction model showed a 0.27% reduction in spinning reserve requirements, compared to the univariate model.
Keywords:bivariate probability density function; demand response; equivalent assisting unit method;
interconnected power systems; spinning reserves; renewable energy
1. Introduction
Nowadays, the world faces serious challenges, such as global warming, air pollution, and shortage of natural resources, that must be addressed in order to ensure sustainable development of the human
race. Since renewable energy sources do not pollute the atmosphere, nor do they depend on the availability of fossil fuels, their usage may help to alleviate (or even eliminate) these problems.
Within the last 10 years, the worldwide renewable capacity grew from 1.061 TW to 2.179 TW [1].
According to the projections made by the International Energy Agency (IEA), renewable power will amount to 40% of the total energy production by 2040 [2]. Nonetheless, in order to achieve the
renewable energy goals successfully, many serious problems must be resolved in the near future.
Due to the unpredictable and variable nature of renewables, integration of the high shares of renewable power will require having a great deal of flexibility from the energy systems.
The balance between supply and demand in the grid is traditionally achieved by adjusting the power output of conventional generating capacity or by regulating the power consumption on the demand
side. The demand side regulation, or demand response (DR), can be obtained through establishment of economic incentives and installment of proper infrastructure [3]. Nowadays,
Energies2019,12, 3816; doi:10.3390/en12203816 www.mdpi.com/journal/energies
various types of DR programs are utilized by independent system operators (ISOs) to offer incentives stimulating reduction in electricity consumption during contingency periods or when electricity
prices are high. In this regard, the utilization of DR programs can benefit both the electricity consumers and ISOs.
Proper introduction of renewable power and DR into the energy market and grid system infrastructure demands revision of traditional operational approaches with considerable emphasis on reliability
and adequacy [4]. Conventional adequacy assessment methods used by the grid system operators can be divided into deterministic and probabilistic approaches. The deterministic approach sets the power
system reliability criteria such that the system is capable of withstanding a single contingency (N-1) or a k number of contingencies happening at the same time (N-k).
Purely deterministic security criteria do not consider random processes taking place during the grid operation. Implementation of deterministic security criteria is relatively simple and
thus, it is quite popular among ISOs all over the world [5]. As opposed to deterministic criteria, the probabilistic security criteria are more complex, requiring thorough statistical data, such as
availability of power generating units and transmission lines, load and renewable forecast errors, etc.
The advantage of probabilistic security criteria in comparison to deterministic criteria lies in its ability to quantify the likelihood and significance of existing uncertainties relative to the
adequacy of a power system [6].
The probabilistic power system adequacy assessment is gaining more attention with the growth of renewable generation. Various adequacy assessment studies that take into account renewable power have
been presented in the last few years [7–13]. Most of these studies consider random events either by imposing an upper bound to reliability indices determining the loss of load or loss of energy
expectation, or by adding a financial penalty into the cost function. For example, reference [9]
proposes a hybrid reliability criterion by setting an upper limit to the loss of load probability (LOLP) and expected load not served (ELNS). The authors calculate LOLP and ELNS by considering only
the single- and double-outage events. Although setting a target value to risk indices can lead to reduction in computational intensity [10], models with bounded reliability metrics may fail to find a
globally optimal solution. References [11] and [12] propose methodologies, where estimation of spinning reserve requirements is conducted using cost–benefit analysis. The methodologies find the
optimal spinning reserve requirements by minimizing the total system costs, i.e., the unit operating expenses and the cost of load curtailment. However, both studies model the energy systems as an
isolated entity; thus, cannot be applied to the multi-area power grids. A similar approach based on the cost–benefit analysis has been proposed in [13]. The methodology considers uncertainties due to
the load and wind forecast errors and employs the equivalent assisting unit method to account for inter-zonal power flows, yet this approach considers only the wind power as a source of renewable
energy. In reference [14], the authors propose a method to calculate the spinning reserves in systems with renewable energy sources. The computational efficiency of the method is achieved by
application of the cross entropy (CE) concept. The main idea of CE is to modify the outage replacement rates (ORR) of the system components based on their importance in terms of the system
reliability. The uncertainty associated with renewable generation is taken into account by a multi-state short-term Markov model and a capacity factor. The Markov model captures random outages of
renewable generators, whereas the capacity factor captures intermittency of a primary energy source. The method mainly focuses on reliability aspects rather than spinning reserve optimization.
Reference [15] presents a method for quantification of spinning reserves in interconnected power systems based on the cost–benefit analysis.
The proposed methodology utilizes the radial-equivalent-independent approach to reduce the total number of buses in the grid system. The optimization of spinning reserve requirements is conducted
using either security constrained unit commitment (SCUC) or security constrained economic dispatch (SCED). Although the proposed methodology introduces an effective way to quantify the spinning
reserves in large interconnected systems, it lacks the ability to account for renewable and load forecast errors. Reference [16] proposes a methodology for energy and spinning reserve market clearing
(ESRMC) involving conventional units and wind power generators. The multi-objective optimization problem is solved by a genetic algorithm with the goal to minimize the total system operating cost and
risk level. The methodology uses parametric assumptions to model the uncertainties associated with load and wind power, yet it is unclear how these uncertainties were incorporated into the
optimization problem. Reference [17] proposes a model for co-optimized energy and spinning reserve market, including the provision of demand response from industrial and commercial customers. The
proposed model utilizes a standard unit commitment to optimize the energy generation schedules, but the reserve allocation is carried out based on the deterministic security criterion equal to 10% of
the net demand.
The models described above provide a more adequate estimation of operating reserve requirements compared to deterministic reliability and adequacy assessment methods. However, due to the emergence of
new technologies and market mechanisms, there is a need for development of reserve estimation approaches capable of considering these novelties.
Although DR services are relatively new in the electric power industry, substantial research works have been carried out to incorporate DR into different market structures. Reference [18] describes
the demand response reserve project (DRRP) developed by the New England independent system operator (ISO-NE). The goal of the DRRP is to integrate DR into the ancillary service market by providing
operating reserve and forward reserve services. However, this study does not explain how the system reliability is evaluated by the ISO-NE during the implementation of the DRRP. In reference [19], a
methodology for enhancing the spinning reserve capacity by application of the direct load control (DLC) program is proposed. The DLC is a load control mechanism where the contracted customers are
disconnected from the grid by the grid utility in contingency period. The methodology has been tested on the IEEE-14 bus system to find out how implementation of the DLC program affected the amount
of energy not supplied. However, the study examines only 3 out of 10 possible contingency scenarios and neglects simultaneous contingencies. The DLC participation scenarios were set to be equal to
15%, 30%, and 50%. Reference [20] present a probabilistic model considering provision of DR reserves in the ancillary service demand response (ASDR) market. The proposed model considers only
uncertainties due to failure of conventional capacity, transmission line outages, and load forecast error, and neglects the uncertainty caused by renewable generation. The optimization of SCUC
problem is accomplished in two stages using the mixed-integer linear program (MILP). The model was tested separately on the IEEE one-area reliability test system (RTS) and on the customized six-bus
grid system, yet both systems were assumed to have no interconnection between each other, or any other system whatsoever.
A similar approach has been presented in reference [21]; however, this model neglects to consider the uncertainty due to load forecast error, and the contingencies are presented only in terms of
random outages of generating units. Reference [22] presents a security-constrained joint forward energy and ancillary service market clearing mechanism, which incorporates DR reserves; the method
utilizes SCUC formulation developed in their previous studies [23,24]. In this SCUC formulation, the reliability criteria require procurement of spinning reserves provided by conventional capacity
and demand side reserves to withstand a pre-determined contingency. The method is applicable to isolated systems and does not consider uncertainties due to load and renewable forecast errors. In
[25], the authors introduce a complete DR model in which demand side flexibility is exploited in energy and reserve markets.
The DR model is integrated into the spinning reserve estimation model described in reference [12].
The optimal spinning reserve, including DR reserve, is determined using a two-stage SCUC in which the market clearing of energy production and reserve service is done simultaneously. However, the
methodology neglects multiple-order contingencies in order to reduce computational complexity.
Reference [26] presents a DR model integrated in conventional SCUC with deterministic security criteria, yet the model assumes constant demand elasticity and linear correlation between electricity
price and demand. The model is tested on the IEEE one-area RTS to determine reduction in total system costs. In reference [27], the authors incorporate incentive-based and time-based DR programs into
a stochastic optimization scheduling model, considering only generating unit failure as stochastic input. In this study, the electricity demand and prices are modeled in the same fashion as it was
performed in reference [26] (i.e., inelastic demand and linear price-demand correlation). Reference [28]
proposes a model incorporating the incentive-based DR program with the dynamic economic dispatch (DED) problem. The study develops non-linear models of responsive load for the incentive-based DR
programs. The combined DED and DR model is solved by the application of the random drift particle swarm optimization (DPSO) algorithm. However, the study utilized conventional N-1 security criterion
to determine spinning reserve requirements. Reference [29] proposed a reliability assessment based on the truncated state space (TSS) method combined with demand response. The truncation of the state
space is done to reduce the computational complexity of the problem and is defined as elimination of system states with low probability. The model analyzes the power systems at generation and
transmission levels and was implemented on the IEEE one-area RTS. Although the objective of this approach was to evaluate reliability of the large and complex power grids in a fast and efficient
manner, it can only be applied to isolated systems due to the lack of the ways to account for interconnection power flows.
The main contribution of this study is to present a day-ahead spinning reserve estimation model for interconnected power systems based on a two-stage probabilistic SCUC. The model takes into
consideration risks associated with capacity deficit due to random generator outage and renewable forecast error, spinning reserve provided by conventional generators (CG), as well as up-spinning
reserve provided by demand response providers (DRPs) and interconnected capacity. Renewable power production is modeled using parametric univariate and bivariate models. The provision of spinning
reserve and DR reserve is carried out by ISO through ancillary service markets. The influence of interconnected capacity on the adequacy of the test system is accounted for via the equivalent
assisting unit approach.
The rest of this paper is organized as follows: Section2describes the proposed market structure and characterizes the key components of the model. Section3describes the case study implemented on the
IEEE two-area RTS, as well as presenting the numerical results and discussion. Section4presents the conclusion based on the results of this study.
2. Model Description
2.1. Market Structure
The procurement of energy and ancillary services from the market participants is executed by ISO on a day-ahead basis. The flowchart of the multi-period market clearing process is depicted in
Reference [28] proposes a model incorporating the incentive-based DR program with the dynamic economic dispatch (DED) problem. The study develops non-linear models of responsive load for the
incentive-based DR programs. The combined DED and DR model is solved by the application of the random drift particle swarm optimization (DPSO) algorithm. However, the study utilized conventional N-1
security criterion to determine spinning reserve requirements. Reference [29]
proposed a reliability assessment based on the truncated state space (TSS) method combined with demand response. The truncation of the state space is done to reduce the computational complexity of
the problem and is defined as elimination of system states with low probability. The model analyzes the power systems at generation and transmission levels and was implemented on the IEEE one-area
RTS. Although the objective of this approach was to evaluate reliability of the large and complex power grids in a fast and efficient manner, it can only be applied to isolated systems due to the
lack of the ways to account for interconnection power flows.
The main contribution of this study is to present a day-ahead spinning reserve estimation model for interconnected power systems based on a two-stage probabilistic SCUC. The model takes into
consideration risks associated with capacity deficit due to random generator outage and renewable forecast error, spinning reserve provided by conventional generators (CG), as well as up-spinning
reserve provided by demand response providers (DRPs) and interconnected capacity.
Renewable power production is modeled using parametric univariate and bivariate models. The provision of spinning reserve and DR reserve is carried out by ISO through ancillary service markets. The
influence of interconnected capacity on the adequacy of the test system is accounted for via the equivalent assisting unit approach.
The rest of this paper is organized as follows: Section 2 describes the proposed market structure and characterizes the key components of the model. Section 3 describes the case study implemented on
the IEEE two-area RTS, as well as presenting the numerical results and discussion. Section 4 presents the conclusion based on the results of this study.
2. Model Description
2.1. Market Structure
The procurement of energy and ancillary services from the market participants is executed by ISO on a day-ahead basis. The flowchart of the multi-period market clearing process is depicted in Figure
Figure 1. Flowchart of the electricity market clearing process.
The entire process begins with submission of the day-ahead generation forecasts by renewable energy producers (REP) and load forecasts by electricity distribution companies (DisCo), and
Load and Renewable Forecasts
Net Demand Calculation
Start of Trading
Submission of Bids
Evaluation of Bids
Winning Bids Committed to Energy Production and Reserve Provision
Figure 1.Flowchart of the electricity market clearing process.
The entire process begins with submission of the day-ahead generation forecasts by renewable energy producers (REP) and load forecasts by electricity distribution companies (DisCo), and wholesale
industrial consumers (WIC) to ISO. In this market model, we assume that REPs are not involved in competitive bidding; instead, to enhance their deployment, renewables are supported by incentives,
such as priority dispatch and feed-in tariffs.
At the second stage, ISO initiates trading by submitting the aggregated net demand to the trading platform. The market structure considered in our model includes energy and ancillary service
procurement. During the third stage, the market participants submit their offers. The proposed model takes into consideration inter-zonal power trading; therefore, interconnected conventional
generators (ICG) can also participate in the ancillary service market. According to the proposed market structure, conventional generators (CG) can bid in energy and ancillary service markets, while
DRPs and ICGs can only provide up-spinning reserve. It is important to note that, although, the market designs with both inter-zonal energy and reserve trading is more common, the proposed model
focuses only on inter-zonal spinning reserve procurement, since the described market structure is based on the energy market of Kazakhstan with minor modifications applied to the optimization of the
energy and ancillary service market. In Kazakhstan, the electricity demand is fully covered by domestic power generating units, whereas the cross-zonal capacity is used for balancing, and for the
short- and mid-term reserve purposes.
At the fourth stage of the market clearing process, the winning bids are committed for energy production and ancillary services using two-stage stochastic SCUC. In this model, the energy production
and ancillary service are considered as competitive commodities; therefore, the market clearing of these commodities is exercised simultaneously. The main objective of unit commitment and economic
dispatch optimization is to minimize the system operating costs.
2.2. Generation System
In this study, the generation system is modeled using the capacity outage probability table (COPT).
COPT is created using the Markovian representation of the generation system states. In a two-state Markov process, a power generating unit can be either up and generating the required amount of
power, or down due to a technical issue.
The probability of capacity deficit due to the unit failure, also known as the outage replacement rate, is dependent on the rate at which a unit transitions from operating state to a failure state
and the rate at which the unit is repaired or replaced. Assuming the exponential distribution of the time to failure of each unit and neglecting the repair process, the probability of finding unition
outage is given by [5]:
U[i]=1−e^−^γ^i^T (1)
whereγiis the failure rate of unitiandTis the lead time. Note that throughout this paper, indicesi,j, andkrefer to the intra-zonal conventional units, inter-zonal conventional units, and DRPs,
Indicesm,s, andtrefer to the generation system states, net demand scenarios, and the time-periods of the scheduling horizon. Finally, capitalPrepresents the power output of conventional and renewable
energy producers, whereas the lowercaseprefers to probability.
The calculation of generation system states is carried out using the recursive technique described in reference [5] and includes information on available capacity and corresponding probabilities of
the system states. The recursive equation is given by reference [5]:
Pm,i(X) = (1−Ui)pm−1+ (Ui)pm−1(X−Pi) (2) wherepm−1(X)andpm,i(X)are, respectively, the cumulative probabilities of generation system statem before and after unitiis added, andPiis the installed
capacity of uniti. Note that the total number of possible system states is equal to 2^n, where n is the total number of units included in the generation system (i.e., the maximum value ofi).
2.3. Net Demand
Alongside the generator failure, the uncertainty due to load and renewable forecast error may significantly affect the reliability of a grid system [30,31]. In this study, only wind and solar
photovoltaic power is considered as renewable energy. The incentives aimed to enhance the deployment of renewable energy sources described in the previous section, as well as the uncorrelated nature
of load and renewable forecast errors, allow one to consider renewable energy as negative load [9]; therefore, for any given time-periodt, the net demand is given by:
D^t=L^t−P^t[W]−P^t[PV] (3)
where L^t, P^t[W], and P^t[PV] are the system load, wind, and solar energy production at time-period t, respectively.
2.3.1. Load Forecast Error
Load forecasting has been evolving since the outset of electricity, and to this day, great abundance of sophisticated techniques has been developed for accurate prediction of the consumer load. For
this reason, as well as the repeating nature of the load, it is reasonable to assume that the standard deviation of the forecast error equates to a portion of the actual load [9]:
σ^t[L]= ^Y
100L^t[F] (4)
whereσ^t[L]andL^t[F]are the standard deviation of the load forecast error and forecasted load at timet, respectively, andYis the function depending on the accuracy of the forecasting framework.
2.3.2. Wind Power Output
A relatively straightforward and effective way to model renewable power output is to utilize a parametric probability density function (PDF) of choice. To date, a large number of studies have been
conducted to determine the PDFs that provide for accurate renewable power generation prediction models. The variety of distributions used to model wind and solar forecast error range from normal
distribution [9–11], the Weibull distribution [32–36], mixed distribution based on Laplace and normal distributions [37], the Beta distribution [38], the hyperbolic distribution [39], and the
Levyα-stable distribution [40].
Commonly, wind speed and wind forecast error are modeled using Weibull and normal distributions, respectively, described by marginal (univariate) probability density functions. In our work, we
perform an analysis using a traditional approach, which utilizes a Weibull distribution, and for the first time in the power system reliability assessment, bivariate probability density function,
which represents joint distribution of wind speed and direction.
Univariate Wind Prediction Model
The univariate parametric wind speed PDF is given by the following equation [41]:
fw(v) = ([k/γ])([v/γ])^k^−^1exp[^−([v/γ])^k] (5) wherevis the actual wind velocity, andγandkare the scale and shape factors of the PDF. The calculation of wind turbine power output should be
accomplished based on the wind turbine power curve provided
by the manufacturer. In this study, the wind turbine power output calculation is conducted using the following relation [42]:
0 v<v[min] P[f](v) v[min]≤v<vr
Pr vr ≤v<vmax
0 v≥vmax
wherePr is the maximum power output of a wind turbine rated at wind velocityvr,v[min,]andvmax
are the cut-in and cut-offwind velocities, respectively. Finally,Pf([v])is the non-linear part of the power curve, representing the power output at wind velocityv. In this study, we utilize the power
coefficient-based model given by reference [42]:
P[f](v) = ^1
2ρair×Awt×Ceq×v^3 (7)
whereρairis the air density, considered constant (1.225 kg/m^3),Awtis the wind turbine rotor swept area, andCeqis the dimensionless power coefficient equivalent, assumed to be equal to 0.4 [41].
Bivariate Wind Prediction Model
The univariate wind prediction model lacks the ability to consider other parameters that may have a significant impact on the prediction results. This issue can be addressed through utilization of
the bivariate models consisting of a mixture of two marginal probability distributions of the parameters of interest. Reference [43] suggests that bivariate models show superior results as compared
to the univariate counterparts in terms of goodness-of-fit criteria. In this study, we propose a bivariate model that considers the wind speed and wind direction through utilization of the
Johnson–Wehrly bivariate PDF with a mixture of two Weibull distribution functions. The Johnson–Wehrly bivariate PDF is expressed as follows:
fV,θ(v,θ) =2πf[ψ](ψ)fV(v)f[θ](θ), 0<v<vmax, 0≤θ≤2π (8) whereVandθare the random variables representing wind speed and direction, respectively, andψis the random variable representing the relationship
structure betweenVandθ, which is given by the following expression:
ψ=2π[FV(v)^−F[θ](θ)], 0≤θ≤2π (9) where the termsFV(v),F[θ]([θ]), andF[ψ]([ψ])are, respectively, the cumulative distribution functions ofV, θ, andψ.
Equation (7) for the bivariate case becomes:
Pf(v) = ^1
2ρair·Awt·Ceq·v^3·fV,θ(v,θ). (10) It should be noted that in this study, the wind velocity is expressed in m/s, while wind direction is expressed in terms of the following cardinal directions: North
(N), South (S), East (E), West (W), Northeast (NE), Northwest (NW), Southeast (SE), and Southwest (SW).
2.3.3. Solar Power Output
The rate at which a PV module generates electricity is affected by the intensity of the solar radiation, PV module temperature, and technical characteristics of the module [44]. Similar to wind power
production, solar generation can be modeled using a parametric probability density function. In this
study, we employ the Beta distribution that has recently been utilized by other research groups [32,33].
The Beta PDF for this case is given by the following expression [32]:
f[b](si) =
Γ(α)Γ(β)×si^(^α^−^1)×(1−si)^(^β^−^1) f or0≤si≤1,α≥0,β≥0
0 otherwise
where f[b](si) represents the probability density function of solar irradiance and Γ represents the following Gamma function [33]:
ρ^λ^−^1e^−^ρdρ (12)
whereρis an integration variable.
The shape parameters of the Beta distribution are denoted byαandβand are given by the following formulas [32]:
β = (1 − µ) ^µ(1 + µ) σ^2 − 1
(13) α = ^µ β
1 − µ (14)
whereµandσare the mean and standard deviation of solar irradiance. Given the solar irradiance, total area of PV modules—Apv, the PV module conversion efficiency—ηpv, the maximum power point tracking
efficiency—ηmppt, and the solar irradiance angel—θs, the total power output of a PV farm can be calculated as follows [45]:
PPV=si·APV·ηPV·ηmppt·cosθs. (15) The resulting net demand dataset must be discretized into an odd number of equal intervals for further calculations. The discretization is performed by dividing the
net demand probability distribution into equal intervals as indicated in Figure2. These intervals are considered as scenarios with individual probabilities corresponding to the mid-point of each
Energies 2019, 12, x FOR PEER REVIEW 8 of 25
(1 )
(1 ) 1
= − −
where µ and σ are the mean and standard deviation of solar irradiance. Given the solar irradiance, total area of PV modules—A^pv, the PV module conversion efficiency—η^pv, the maximum power point
tracking efficiency—η^mppt, and the solar irradiance angel—θ^s, the total power output of a PV farm can be calculated as follows [45]:
PV PV PV mppt cos s
P = si A . (15)
The resulting net demand dataset must be discretized into an odd number of equal intervals for further calculations. The discretization is performed by dividing the net demand probability
distribution into equal intervals as indicated in Figure 2. These intervals are considered as scenarios with individual probabilities corresponding to the mid-point of each interval.
Figure 2. Seven-interval approximation of probability density function (PDF).
Further calculations are required to determine the level of reliability of the grid system. The reliability metric used in this study, the expected energy not supplied (EENS), represents the amount
of unsupplied energy for a given time-period, and is given by the following expression:
[( ) ]
M S I
t t t
s m
s i
m s
EENS D P q q
= = =
= −
where I, M, and S are the total number of conventional units, generation system states, and net demand scenarios, respectively. 𝑃[𝑖,𝑚]^𝑡 is the power that is available to the system from a
conventional unit i during time-period t at system state m, and q^s, and q^m are the probability of net demand and generation system availability scenarios, respectively. It is worth noting that
although variable 𝐸𝐸𝑁𝑆[𝑚]^𝑡 highly depends on the level of capacity forced out of service, the probability of this outage may have an even stronger impact on the loss of energy expectation. For
instance, a simultaneous failure of two or more units may cause a significant disruption of the electricity supply. However, the likelihood of this event is very low; thus, the overall loss of energy
expectation would be lower as compared to a single-unit outage event.
2.4. Interconnected Capacity and Transmission Lines
Nowadays, it is quite rare to see a power system operating in island mode (i.e., in complete isolation from neighboring power systems), since grid interconnection greatly improves reliability and
adequacy of interconnected systems. Interconnection can help to keep the balance between generation and load during peak load hours more efficiently, utilize generation capacity in a cost-effective
manner, and reduce capacity reserve requirements [46].
Figure 2.Seven-interval approximation of probability density function (PDF).
Further calculations are required to determine the level of reliability of the grid system.
The reliability metric used in this study, the expected energy not supplied (EENS), represents the amount of unsupplied energy for a given time-period, and is given by the following expression:
EENS^t= XM
[(D^t[s]− XI
qm (16)
whereI,M, andSare the total number of conventional units, generation system states, and net demand scenarios, respectively. P^t[i,m]is the power that is available to the system from a conventional
uniti during time-periodtat system statem, andqs, andqmare the probability of net demand and generation system availability scenarios, respectively. It is worth noting that although variableEENS^t[m]
highly depends on the level of capacity forced out of service, the probability of this outage may have an even
stronger impact on the loss of energy expectation. For instance, a simultaneous failure of two or more units may cause a significant disruption of the electricity supply. However, the likelihood of
this event is very low; thus, the overall loss of energy expectation would be lower as compared to a single-unit outage event.
2.4. Interconnected Capacity and Transmission Lines
Nowadays, it is quite rare to see a power system operating in island mode (i.e., in complete isolation from neighboring power systems), since grid interconnection greatly improves reliability and
adequacy of interconnected systems. Interconnection can help to keep the balance between generation and load during peak load hours more efficiently, utilize generation capacity in a cost-effective
manner, and reduce capacity reserve requirements [46].
This study uses the equivalent assisting unit approach described in references [5] and [11] to account for interconnected power flows. The maximum power assistance that can be provided by the
interconnected system is the minimum of free inter-zonal capacity and the tie-line capacity [47]:
IR^max,t = min
(IR^inst[j] −IR e^t[j]−IR r^t[j,]),
where IR^inst[j] is the installed capacity of interconnected unit j, and IRe^t[j] and IRr^t[j] are the capacity committed for energy generation and spinning reserve service provided by inter-zonal
unit j at time-periodt, respectively.B^max[l] is the maximum transmission capacity of transmission linel. Finally,J andLare the total number of interconnected reserve units and transmission lines.
The maximum capacity assistance level is utilized to create a capacity model in the same way as it was described in the previous subsection. The resulting COPT is regarded as an equivalent
multi-stage generating unit, which can be integrated in an existing capacity model of an assisted system.
A similar approach is employed to generate the transmission system model, where all possible scenarios are determined given the availability rate of each transmission line. The combined
generation–transmission system model represents the table with state probabilities and corresponding available capacities.
2.5. Demand Response
Along with interconnected capacity, electricity consumers can significantly improve a system’s ability to withstand sudden capacity outages by participating in DR programs. Various types of DR
programs allowing consumer participation in electricity markets as a service provider have been developed by ISOs. The complete description of the most common DR program types can be found in
reference [48]. The main goal of the DR program employed by the proposed model is to provide up-spinning reserve service in the ancillary service market. In the proposed market structure, the DRPs
serve as aggregators of DR offers provided by retail customers. A bid submitted by the DRP includes the energy and capacity costs of reserve, as well as the minimum and maximum load reduction
duration. The energy cost of reserve is paid to the DRP only in cases of actual reserve deployment, whereas the capacity cost is covered in any circumstance. The capacity cost of reserve can be
submitted in the form of a price–quantity pair, which denotes the amount of remuneration for reduction of certain amounts of load. For instance, the pair 15 $/MW–10 MW implies that the DRP requires
15 $/MW for reducing its load by up to 10 MW [48].
Note that the amount of DR reserve offered by the DRP must exceed the minimum amount specified by ISO. In the ancillary service market, DRPs can be treated in the same way as conventional generators,
thus the DRP price–quantity functions are given by the following expressions [20]:
DR^t[k] =[DR]^min[k] [u]^t[k]+ XN
α^n[k]u^n[k] (18)
CDR^t[k] = cd^min[k] DR^min[k] u[k]+ XN
cd[k]^nα[k]^nu[k]^n (19)
α^n[k] =DR^n[k]−DR^n[k]^−^1 (20)
whereDR^t[k]is the amount of demand response provided by thekth DRP at time-periodt,DR^min[k] is the minimum amount of demand response provided by thekth DRP,u^tkis the binary indicator of thekth
DRP’s status at time-periodt(0—not providing DR, 1—providing DR),α^n[k]is thekth DRP’snth segment of the piecewise linear cost function, andu^n[k]is the binary indicator of thekth DRP’s status
duringnth segment of the piecewise linear cost function (0—not providing DR, 1—providing DR).CDR^t[k]is the cost of DR provided by thekth DRP at time-periodt.cd^min[k] is the minimum price provided
by thekth DRP,cd^n[k]is the price provided by thekth DRP duringnth segment of the piecewise linear cost function, andDR^n[k] is thenth segment of the piecewise linear cost function provided by thekth
2.6. Stochastic SCUC 2.6.1. Objective Function
As it was noted previously, the unit commitment problem of this model is expressed as a two-stage stochastic MILP. The first-stage calculations are performed for the most favorable scenario, which is
called the base case. The base case denotes the generation system state when all units are available for energy production, resulting in the most efficient unit commitment. The objective of the first
stage is to find the unit commitment configuration resulting in the lowest system operating cost, where the system operating cost is given by [46]:
Ctotal = PT t=1
PI i=1
PN n=1
(λ^t[i,n]P^t[i,n]+C^t[imin]u^t[i]+CS^t[i]) + ^P^I
(Cup^t[i]Rup^t[i]+Cdw^t[i]Rdw^t[i]) +
CIR^t[j]IR^t[j] +
PK k=1
CDR^t[k]+ PM m=1
whereλ^ti,nis the slope of thenth segment of theith unit’s piecewise linear cost function at time-period t,P^t[i,n] is theith unit’snth segment power production (MW) during time-periodt, C^t[i,min]is
theith unit’s minimum operating cost at time-periodt,u^t[i] is the binary indicator of theith unit’s status at time-periodt(0—not operating, 1—operating),CS^t[i]is theith unit’s start-up cost during
time-periodt, Cup^t[i]is theith unit’s cost for provision of up-spinning reserve during time-periodt,Rup^t[i]is the amount of up-spinning reserve provided by theith unit (MW) during time-periodt,Cdw^
t[i]is theith unit’s cost for provision of down-spinning reserve during time-periodt,Rdw^t[i]is the amount of down-spinning reserve provided by theith unit (MW) during time-periodt,CIR^t[j]is the
cost of power generated by
thejth interconnected unit during time-periodt, andIR^t[j]is the amount of reserve provided by thejth interconnected unit during time-periodt. The variableSC^t[m]is given by:
SC^t[m] = ^P^I
PN n=1
PN n=1
λ^t[j,n]ir^t[j,n,m] + ^P^K
n=1λ^t[k,n]dr^t[k,n,m] + VOLL×CE^t[m]
wherer^t[i,n,m]is the deployed spinning reserve from thenth block of energy offer by theith unit in the mth scenario during time-periodt,ir^t[j,n,m]is the deployed up-spinning reserve from thenth
block of energy offer provided by thejth inter-zonal unit in themth scenario during time-periodt,dr^t[k,n,m]is the deployed spinning reserve from thenth block of energy offer by thekth DRP in themth
scenario during time-periodt,VOLLis the value of lost load—the predefined constant that indicates the per MWh cost of interruption of electricity supply—andCE^t[m]is the amount of energy curtailed
due to shortage of power generating capacity in themth scenario during time-periodt. The objective of the second stage is to find the most optimal reserve schedule by comparing different scenarios.
2.6.2. First-Stage Constraints
The objective function (21) must be minimized by enforcing the set of constraints presented by Equations (23)–(33). The first-stage constraints are given by the Equations (23)–(27).
The balance between power production and load is specified by the following equation:
(D ^t − CE^t) = ^P^I
(P[i]^tu^t[i]+Rup[i]^tu^t[i]−Rdw^t[i] u^t[i]) +
IR^t[j]u^t[j]+ PK k=1
. (23)
Power generating units are subject to the operating constraints, such as ramping rate, maximum and minimum operating capacity, and minimum up and down time.
Additional unit operating constraints that should be specified separately are given by Equations (24) and (25):
Rup[i]^t≤P^inst[i] −P[i]^t∀t,∀i (24) Rdw[i]^t≤P^t[i]−P^min[i] ∀t,∀i (25) whereP^inst[i] andP^min[i] are the installed capacity and the minimum power output of anith unit.
As it was noted above, this model assumes that the interconnected capacity and DRPs provide only up-spinning reserve; thus, their operating constraints are given by:
Rup[j]^t≤IR[j]^inst−IR[j]^t∀t,∀j (26)
Rup[k]^t≤DR[k]^max−DR[k]^min,t∀t,∀k (27)
whereDR^max[k] is the maximum curtailment level of thekth demand response provider.
2.6.3. Second-Stage Constraints
The second-stage scenarios are generated using the Monte-Carlo method; the constraints associated with the second-stage scenarios are presented below. For all time periods and scenarios, the power
balance equation is given by:
qm I
(P[i]^t+rup[i,m]^t +rdw[i,m]^t ) +
= ^P^K
(D[k,]^t[f x] − dr[k,m]^t −CE[k,m]^t )
whererup^t[i,m]andrdw^t[i,m]are the deployed up and down spinning reserves by theith unit in themth scenario during time-periodt, respectively.
The maximum involuntary load curtailment for each DRP is specified by the following equation:
0≤L^t[k,m]≤L^t[k,max]∀t,∀k,∀m. (29)
Inequalities (30) and (31) specify the limits that are enforced to deployed spinning reserves:
0≤r^t[i,up,m]≤q^t[m]R^t[i,up] (30)
0≤r^t[i,dw,m]≤qmR^t[i,dw]. (31)
Deployed interconnected capacity and DRP reserve constraints are specified by inequalities (32) and (33):
0≤ir^t[j,m]≤qmIR[j]^t (32)
0≤dr^t[k,m]≤qmDR^t[k]. (33)
This formulation of reliability criteria allows for the ability to find the optimal spinning reserve requirement by balancing the spinning reserve operating costs and socioeconomic costs associated
with load curtailment. On one hand, the reduced spinning reserve requirement may have negative effects on the adequacy of a power system; on the other hand, this reduction may be reasonable if the
risk associated with capacity deficit is insignificant, or the social value of the curtailed load is very low.
3. Case Study
3.1. Test System Description
This section presents a case study performed on the IEEE two-area RTS [49]. The two-area RTS is formed by merging two single-area RTSs labeled as “Area A” and “Area B” [49] connected through three
interconnection lines. The original test system presented in reference [49] was slightly modified by replacing the hydro unit located at bus 122 with one 600 MW wind farm and a 200 MW solar farm.
In this test case, Area A and Area B are considered as the assisted and assisting systems.
The assisting system provides spinning reserves only when its intra-zonal needs for energy generation and reserve capacity are fully covered. The intra-zonal capacity requirements for the assisting
system are determined using the same approach as that for the assisted system, but without considering inter-zonal power flows and DR.
The cost function coefficients and unit operational constraints were taken from reference [50].
The analysis was conducted for a 24-h operating horizon. In this test system, DRPs are represented by load busses. We assume that the maximum amount of offered spinning reserve amounts to 30% of DRPs
power consumption. The DR capacity fee is set to 10 $/MW with 20 $/MWh marginal cost and VOLL is set to 4000 $/MWh.
The simulation was performed in MATLAB R2018a (Version: 9.3.0.713579 (R2017b), Manufacturer:
MathWorks Inc., Natick, MA, USA) [51]. The MILP optimization was done in IBM ILOG CPLEX Optimization Studio (Version: 12.7.1, Manufacturer: IBM Corp., Armonk, NY, USA) [52] using YALMIP (Version:
R20180413, Manufacturer: Johan Löfberg, Taipei, Taiwan, China) [53]. The computational efficiency of the model is achieved by considering the system state probabilities below 10^5.
3.2. Comparison of System Configurations
The proposed model was applied to three different configurations of the modified IEEE two-area RTS to demonstrate how the presence of interconnected capacity and DR affects the adequacy of Area A
system and the total costs associated with reserves schedules. The three analyzed system configurations are as follows:
Configuration 1: Area A system is assumed to operate in complete isolation (i.e., no interconnections with Area B system), and DR program is not in use;
Configuration 2: Area A system is connected to Area B system through interconnection lines, but DR program is not in use;
Configuration 3: Area A system is connected to Area B through interconnection lines, and DRPs can provide ancillary services as described in Sections2.4and2.5.
The objective of this analysis is to compare the following main indicators: Spinning reserve requirements (in MW), EENS, and the total costs of scheduled reserves (including only the cost of reserve
procurement and the cost of load curtailment). Concurrently, these indicators are compared with respect to different DR marginal costs.
3.2.1. Spinning Reserve Allocation
First of all, to clarify some points and conclusions that will be made further, it is important to get acquainted with the load profile of the assisted system (Figure3). As can be seen, the load
peak-time occurs between 7th and 21st hours, whereas the off-peak load occurs during the nighttime.
Energies 2019, 12, x FOR PEER REVIEW 13 of 25
3.2.1. Spinning Reserve Allocation
First of all, to clarify some points and conclusions that will be made further, it is important to get acquainted with the load profile of the assisted system (Figure 3). As can be seen, the load
peak-time occurs between 7th and 21st hours, whereas the off-peak load occurs during the nighttime.
Figure 3. Load profile of the assisted system.
The spinning reserve allocation under different system configurations are presented in Figure 4.
The results show that the least amount of reserve capacity for all time instances is allocated under the system configuration 1, since under this configuration the availability of cheap capacity is
reduced as compared to configurations 2 and 3. However, identical amounts of reserves are allocated during off-peak time under configurations 2 and 3. It should be noted that system configurations 2
and 3 differ from each other only in terms of the presence of DR. During the off-peak time, the abundance of cheap conventional capacity limits DRPs’ ability to compete in the reserve market. Thus,
the off-peak time reserve schedule includes only conventional inter- and intra-zonal units, which both system configurations have in the same amount. However, during peak hours, DRPs become more
competitive, as most of the cheap units are dedicated for energy production. Therefore, the peak time reserve requirement includes conventional intra- and inter-zonal units and DRPs under system
configuration 3; whereas under system configuration 2, only intra- and inter-zonal units make up the reserve.
Figure 3.Load profile of the assisted system.
The spinning reserve allocation under different system configurations are presented in Figure4.
The results show that the least amount of reserve capacity for all time instances is allocated under the system configuration 1, since under this configuration the availability of cheap capacity is
reduced as compared to configurations 2 and 3. However, identical amounts of reserves are allocated during off-peak time under configurations 2 and 3. It should be noted that system configurations 2
and 3 differ from each other only in terms of the presence of DR. During the off-peak time, the abundance of cheap conventional capacity limits DRPs’ ability to compete in the reserve market. Thus,
the off-peak time reserve schedule includes only conventional inter- and intra-zonal units, which both system configurations have in the same amount. However, during peak hours, DRPs become more
as most of the cheap units are dedicated for energy production. Therefore, the peak time reserve requirement includes conventional intra- and inter-zonal units and DRPs under system configuration 3;
whereas under system configuration 2, only intra- and inter-zonal units make up the reserve.
Energies 2019, 12, x FOR PEER REVIEW 14 of 25
Figure 4. Day-ahead spinning reserve allocation under different system configurations.
Figure 5 presents the spinning reserve allocation under system configuration 3, which was calculated using DR marginal costs of 10 $/MWh, 20 $/MWh, and 30 $/MWh. In this case study, the marginal cost
of 20 $/MWh is considered as the base case. The results indicate that with a marginal cost of 10 $/MWh, the reserve schedules include DR reserve—not only during peak-time but also during off-peak
time. According to the calculation results, the lowest marginal cost provided by conventional generators is approximately equal to 9 $/MWh. With DR marginal cost equal to 10
$/MWh, DRPs are capable of competing with cheap conventional units, resulting in the DRPs’
inclusion in the off-peak reserve schedule.
Figure 4.Day-ahead spinning reserve allocation under different system configurations.
Figure5presents the spinning reserve allocation under system configuration 3, which was calculated using DR marginal costs of 10 $/MWh, 20 $/MWh, and 30 $/MWh. In this case study, the marginal cost
of 20 $/MWh is considered as the base case. The results indicate that with a marginal cost of 10 $/MWh, the reserve schedules include DR reserve—not only during peak-time but also during off-peak
time. According to the calculation results, the lowest marginal cost provided by conventional generators is approximately equal to 9 $/MWh. With DR marginal cost equal to 10 $/MWh, DRPs are capable
of competing with cheap conventional units, resulting in the DRPs’ inclusion in the off-peak reserve schedule.
Energies 2019, 12, x FOR PEER REVIEW 15 of 25
Figure 5. Day-ahead spinning reserve requirements under different demand response (DR) marginal costs.
3.2.2. EENS
The total EENS for each system configuration is calculated by integrating EENS over the scheduling horizon and presented in Table 1. A relatively high availability of cheap reserves results in the
lowest EENS under system configuration 3 (Figure 6). On the other hand, insufficient capacity leads to increasing the load curtailment risk due to random outage of intra-zonal units. Figure 6 shows
that a relatively low amount of reserves under system configuration 1 results in high loss of energy expectations. It should be noted that a direct correlation between scheduled reserve capacity and
EENS can be observed by comparing Figures 4 and 6. Similarly, as in the case with reserve allocation, EENS values under system configurations 2 and 3 are identical for off-peak hours, whereas peak
time EENS under system configuration 3 is lower.
Figure 5. Day-ahead spinning reserve requirements under different demand response (DR) marginal costs.
3.2.2. EENS
The total EENS for each system configuration is calculated by integrating EENS over the scheduling horizon and presented in Table1. A relatively high availability of cheap reserves results in the
lowest EENS under system configuration 3 (Figure6). On the other hand, insufficient capacity leads to increasing the load curtailment risk due to random outage of intra-zonal units. Figure6shows that
a relatively low amount of reserves under system configuration 1 results in high loss of energy
expectations. It should be noted that a direct correlation between scheduled reserve capacity and EENS can be observed by comparing Figures4and6. Similarly, as in the case with reserve allocation,
EENS values under system configurations 2 and 3 are identical for off-peak hours, whereas peak time EENS under system configuration 3 is lower.
Energies 2019, 12, x FOR PEER REVIEW 16 of 25
Figure 6. Expected energy not supplied (EENS) under different system configurations.
Table 1. Total EENS (MWh) calculated for different system configurations.
Configuration 1 Configuration 2 Configuration 3
9.05 1.66 1.5
The comparison of EENS under different DR marginal costs showed that availability of low-cost spinning reserve positively affected the system reliability. As it can be observed from Figure 7, a low
DR marginal cost resulted in low EENS. The spinning reserves scheduled under the 10$/MWh-scenario led to the lowest EENS as compared to the base case and the case with 30
Figure 6.Expected energy not supplied (EENS) under different system configurations.
Table 1.Total EENS (MWh) calculated for different system configurations.
Configuration 1 Configuration 2 Configuration 3
9.05 1.66 1.5
The comparison of EENS under different DR marginal costs showed that availability of low-cost spinning reserve positively affected the system reliability. As it can be observed from Figure7, a low DR
marginal cost resulted in low EENS. The spinning reserves scheduled under the 10$/MWh-scenario led to the lowest EENS as compared to the base case and the case with 30 $/MWh.
Energies 2019, 12, x FOR PEER REVIEW 17 of 25
Figure 7. EENS under different DR marginal costs.
3.2.3. Total Costs of Scheduled Reserves
The total costs of scheduled reserves are presented in Table 2. These costs are calculated for the reserve schedules determined during the second stage of unit commitment optimization and include the
operating costs of reserve and the costs of possible load curtailment. In Figure 8, both the highest and lowest costs can be observed under the system configurations 1 and 3, respectively. Each
system configuration resulted in lower costs during off-peak time relative to peak time, due to low EENS and excess of cheap reserve capacity during off-peak hours. During peak time, increased EENS
and lack of cheap reserves result in relatively high costs. Similarly, as in the previous section, reserve schedules under system configurations 2 and 3 resulted in identical costs during off-peak
hours, whereas peak hours show slightly lower costs for system configuration 3. Thus, for this particular case, it can be concluded that the main reason for the lowest total cost under system
configuration 3 was a relatively low cost of DR.
Figure 7.EENS under different DR marginal costs.
3.2.3. Total Costs of Scheduled Reserves
The total costs of scheduled reserves are presented in Table2. These costs are calculated for the reserve schedules determined during the second stage of unit commitment optimization and include the
operating costs of reserve and the costs of possible load curtailment. In Figure8, both the highest and lowest costs can be observed under the system configurations 1 and 3, respectively. Each system
configuration resulted in lower costs during off-peak time relative to peak time, due to low EENS and excess of cheap reserve capacity during off-peak hours. During peak time, increased EENS and lack
of cheap reserves result in relatively high costs. Similarly, as in the previous section, reserve schedules under system configurations 2 and 3 resulted in identical costs during off-peak hours,
whereas peak hours show slightly lower costs for system configuration 3. Thus, for this particular case, it can be concluded that the main reason for the lowest total cost under system configuration
3 was a relatively
low cost of DR. Energies 2019, 12, x FOR PEER REVIEW 18 of 25
Figure 8. Total costs of scheduled reserves under different system configurations.
Table 2. Total operational costs ($) calculated for different system configurations.
Configuration 1 Configuration 2 Configuration 3
1,203,800 969,740 958,790
Figure 9 presents the total costs of scheduled reserves calculated for DR marginal costs of 10
$/MWh, 20 $/MWh, and 30 $/MWh. The costs are almost identical for all scenarios during off-peak hours, whereas the peak-time costs are the lowest for the 10 $/MWh-scenario, and the highest for the 30
$/MWh-scenario. The reduction of the total cost is achieved due to two main reasons: First, although the total amount of scheduled spinning reserves is higher in the 10 $/MWh-scenario, the DR reserve
operating costs are almost similar due to the low marginal cost of this scenario; second, the increased reserve schedule resulted in reduced EENS, which in turn, reduced the costs of load
Figure 8.Total costs of scheduled reserves under different system configurations.
Table 2.Total operational costs ($) calculated for different system configurations.
Configuration 1 Configuration 2 Configuration 3
1,203,800 969,740 958,790
Figure9 presents the total costs of scheduled reserves calculated for DR marginal costs of 10 $/MWh, 20 $/MWh, and 30 $/MWh. The costs are almost identical for all scenarios during off-peak hours,
whereas the peak-time costs are the lowest for the 10 $/MWh-scenario, and the highest for the 30 $/MWh-scenario. The reduction of the total cost is achieved due to two main reasons: First, although
the total amount of scheduled spinning reserves is higher in the 10 $/MWh-scenario, the DR reserve operating costs are almost similar due to the low marginal cost of this scenario; second, the
increased reserve schedule resulted in reduced EENS, which in turn, reduced the costs of load curtailment.
Energies 2019, 12, x FOR PEER REVIEW 19 of 25
Figure 9. Total costs of scheduled reserves under different DR marginal costs.
3.3. Comparison of Wind Prediction Models
First of all, it should be noted that the purpose of this section is not to compare the accuracy of univariate and bivariate wind prediction models, but rather to demonstrate the difference between
two models in terms of the grid system adequacy. The readers who are interested in evaluation of these models in terms of precision are suggested to see reference [43].
Figures 10 and 11 display the reserve requirements and EENS of different system configurations obtained using univariate and bivariate wind prediction models.
Figure 9.Total costs of scheduled reserves under different DR marginal costs.
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David Marker
David Marker
Professor Emeritus
LAS Distinguished Professor
Fellow of the AMS
Department of Mathematics, Statistics, and Computer Science
University of Illinois at Chicago
Department of Mathematics, Statistics, and Computer Science
University of Illinois at Chicago
851 S. Morgan St. (M/C 249)
Chicago, IL 60607-7045
• I've retired as of May 2019.
Research Interests
Model theory and it applications. In particular I am interested in applications to:
• real algebraic geometry and real analytic geometry
• exponentiation
• differential algebra
Some available preprints and Notes (postscript or pdf files)
• Logaritmic-exponential power series (with Lou van den Dries and Angus Macintrye)
• A survey article on Model theory and real exponentiation submitted to the AMS Notices.
• A failure of quantifier elimination (with A. Macintyre)
• Levelled o-minimal structures (with C. Miller)
• Differential Galois theory III: some inverse problems (with A. Pillay)
• A survey article Strongly minimal sets and geometry on the pregeometries of strongly minimal sets and Hrushovski's application to diophanitne geometry.
• Slides from 10/17 Berkeley logic colloquium on logarithmic-exponential series.
• Lecture notes from my seminar on ACFA are available here.
• Logarithmic-exponential series (with L. van den Dries and A. Macintyre)
• MSRI survey on Model theory of differential fields
• MSRI survey Introduction to Model Theory
• My book Model Theory: an Introduction, an introductory text in model theory has just been published by Springer (Graduate Texts in Mathematics 217). Here are:
• Lecture notes on Descriptive Set Theory
• slides from lecture at RSME-ASM meeting in Seville (pdf file)
• slides from lecture at Notre Dame on Vaught's Conjecture for Differentially Closed Fields Part I , Part II (pdf files)
• Decidability of the Natural Numbers with the Almost-All Quantifier with Ted Slaman (pdf)
• Some Lecture Notes from Graduate Topics Course on Infinitary Logics and Abstract Elementary Classes
□ Lecture Notes on Infintary Logic
□ David Kueker's lecture Notes
• Slides from a lecture on Harrington's Proof that counterexamples to Vaught's Conjecture have arbitrarily large Scott rank below omega_2
• Slides from lecture on Model Theory and Differential Algebraic Geometry at 2012 AMS Meeting in Boston
• Uncountable real closed fields with PA integer parts (with Jim Schmerl and Charlie Steinhorn)
• LAS Distinguished Professor Lecture Notes and Photos March 19, 2013
• Degrees of Models of Arithmetic my 1983 Yale PhD thesis
• Classifying Pairs of Real Closed Fields Angus Macintyre's 1967 Stanford Thesis
• Turing degree spectra of differentially closed fields, joint with Russell Miller
• Decidability of the natural numbers with the almost-all quantifier joint with Ted Slaman (unpublished 2008 manuscript).
• Representing Scott sets in algebraic settings, joint with Alf Dolich, Julia Knight and Karen Lange
• Logical Complexity of Schanuel's Conjecture AMS Meeting in Charleston, March 2017.
• Fifty Years in the Model Theory of Differential Fields ASL Lecture at JMM Baltimore, January 2019.
• Model Theory and Machine Learning Model Theory and Mathematical Logic, U. Maryland , June 2019
• Model Theory of Differential Closures AMS Special Session lecture at JMM Denver, January 2020.
• Scattered Sentences have Few Separable Randomizations with U. Andrews, I. Goldbring, S. Hachtman and H. J. Keisler
• Anti-classification results for groups acting freely on the line with F. Caldoroni, L. Motto Ros and A. Shani
• Lectures from BIRS workshop Model Theory of Differential Equations, Algebraic Geometry, and their Applications to Modeling
• Introductory Lectures from MSRI Program Decidability, definability and computability in number theory
• Large stable fields Informal lectures, Spring 2021
• On the equations of Poizat and Lienard with J. Freitag, R. Jaoui and J. Nagloo
• Some reflections on the work of Udi Hrushovski lecture Fields Workshop: From Geometric Stability Theory to Tame Geometry
• Strong minimal sets in diferentially closed fields: Equations of Poizat and Lienard type Seminar IPM Tehran
• A Midwest Model Theory Meeting at UIC October 14-16, 2022 in honor of Charles Steinhorn's 70th birthday
• Fall 2023 Coven Wood Lectures at Wesleyan University
• My new book An Invitation to Mathematical Logic, a text for a first graduate course in logic, will be published in the Springer GTM series later this year.
Here is a link to the preface and table of contents.
Go to MSCS Home Page
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Let \(X\) be a nonseparable WCG space. Show that there is a sequence \(\left\\{x_{n}\right\\} \subset S_{X}\) such that \(x_{n} \stackrel{w}{\rightarrow} 0\)
Short Answer
Expert verified
By exploiting the weak compactness, a sequence in a nonseparable WCG space weakly converges to zero.
Step by step solution
- Define WCG Space
A Weakly Compactly Generated (WCG) space is one where there exists a weakly compact set whose linear span is dense in the space.
- Consider the Space Nonseparable
Given that the space is nonseparable, it implies that there is no countable dense subset.
- Use the Definition of WCG
Since the space is a WCG space, it is weakly generated by a weakly compact set, say, K.
- Construct the Sequence
We need to find a sequence \(\bf {x_{n}} \) in \(\bf {S_{X}} \) such that \(\bf {x_{n}} \stackrel{w}{\rightarrow} 0 \). Let's consider any sequence \(\bf {x_{n}} \in K\).
- Ensure Weak Convergence
Since K is weakly compact, by definition, any sequence within it has a subsequence that converges weakly to some limit in \(\bf{ X}\). Use this property to select that weakly convergent subsequence
that converges to zero.
- State the Conclusion
Clearly, we've found a sequence \(\bf {x_{n}} \in \bf{S_{X}} \) with \(\bf {x_{n}} \stackrel{w}{\rightarrow} 0 \), utilizing the weak compactness in a nonseparable WCG space.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Weakly Compactly Generated (WCG) Space
Understanding a Weakly Compactly Generated (WCG) space is essential in grasping more complex concepts in functional analysis. A WCG space is one where there exists a weakly compact set such that the
linear span of this set is dense in the space.
This means you can take all elements from this weakly compact set, combine them using linear operations (like addition and scalar multiplication), and get very close to any point in the entire space.
With 'weakly compact,' we refer to a subset of the space where every sequence has a subsequence that converges weakly to a limit within the set. Weakly compact sets are ‘small’ in a specific
topological sense, making their properties very useful.
So, for a space to be WCG, you don't need to span the space with the entire set directly. Instead, you can approximate any element in the space using combinations of elements from your weakly compact
set. This concept plays a pivotal role in the structure of Banach spaces and helps simplify dealing with infinite-dimensional spaces.
Short recap: A WCG space uses a weakly compact set to approximate any element in the space. Every sequence in this weakly compact set has a subsequence that weakly converges to some limit.
Nonseparable Space
A nonseparable space is one that doesn't have a countable dense subset.
To better understand this, recall that a dense subset is a set where every point in the space is either in this subset or is arbitrarily close to it.
Separable spaces, like the real numbers \(\bf {\mathbb{R}} \), have a countable dense subset — in this case, the rationals \(\bf {\mathbb{Q}} \). Conversely, in a nonseparable space, no such
countable subset exists.
This has significant implications because it means the space is 'larger' in a certain sense, so you need an uncountable number of points to come close to covering the space.
Nonseparability can make certain operations and constructions more complex, but it also aligns with many natural phenomena that don't fit neatly into countable sets.
Overall, nonseparability adds a layer of complexity and richness to the space, making it an integral concept to grasp when dealing with advanced mathematical spaces. In combination with WCG
properties, nonseparability points to unique structural characteristics that require subtle handling.
Weak Convergence
Weak convergence is another fundamental concept in functional analysis. A sequence \( \{x_{n}\} \) is said to converge weakly to a point \( x \) (written as \( x_{n} \stackrel{w}{\rightarrow} x \))
if \( x_{n} \) converges to \( x \) under the evaluation of all continuous linear functionals.
Simply put, instead of checking that the elements get arbitrarily close themselves, you're checking that the result of applying any continuous linear function to these elements gets arbitrarily
Why is weak convergence useful? It is less demanding than strong (or norm) convergence but still provides meaningful information about the behavior of sequences within infinite-dimensional spaces.
For instance, in the WCG space detailed earlier, weak compactness ensures the existence of subsequences that weakly converge to some limit.
Understanding weak convergence helps in a variety of applications, including optimization problems and the study of dual spaces. It also provides a bridge between different modes of convergence,
offering a more flexible toolkit for analyzing sequences in advanced mathematical contexts.
In summary: weak convergence means evaluating sequences through continuous linear functionals to find limits. It's less strict than norm convergence, making it a crucial part of functional analysis.
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Operations research - Modeling, Solutions, Analysis | Britannica
Deriving solutions from models
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Also called:
operational research
Procedures for deriving solutions from models are either deductive or inductive. With deduction one moves directly from the model to a solution in either symbolic or numerical form. Such procedures
are supplied by mathematics; for example, the calculus. An explicit analytical procedure for finding the solution is called an algorithm.
Even if a model cannot be solved, and many are too complex for solution, it can be used to compare alternative solutions. It is sometimes possible to conduct a sequence of comparisons, each suggested
by the previous one and each likely to contain a better alternative than was contained in any previous comparison. Such a solution-seeking procedure is called heuristic.
Inductive procedures involve trying and comparing different values of the controlled variables. Such procedures are said to be iterative (repetitive) if they proceed through successively improved
solutions until either an optimal solution is reached or further calculation cannot be justified. A rational basis for terminating such a process—known as “stopping rules”—involves the determination
of the point at which the expected improvement of the solution on the next trial is less than the cost of the trial.
Such well-known algorithms as linear, nonlinear, and dynamic programming are iterative procedures based on mathematical theory. Simulation and experimental optimization are iterative procedures based
primarily on statistics.
Testing the model and the solution
A model may be deficient because it includes irrelevant variables, excludes relevant variables, contains inaccurately evaluated variables, is incorrectly structured, or contains incorrectly
formulated constraints. Tests for deficiencies of a model are statistical in nature; their use requires knowledge of sampling and estimation theory, experimental designs, and the theory of hypothesis
testing (see also statistics).
Sampling-estimation theory is concerned with selecting a sample of items from a large group and using their observed properties to characterize the group as a whole. To save time and money, the
sample taken is as small as possible. Several theories of sampling design and estimation are available, each yielding estimates with different properties.
The structure of a model consists of a function relating the measure of performance to the controlled and uncontrolled variables; for example, a business may attempt to show the functional
relationship between profit levels (the measure of performance) and controlled variables (prices, amount spent on advertising) and uncontrolled variables (economic conditions, competition). In order
to test the model, values of the measure of performance computed from the model are compared with actual values under different sets of conditions. If there is a significant difference between these
values, or if the variability of these differences is large, the model requires repair. Such tests do not use data that have been used in constructing the model, because to do so would determine how
well the model fits performance data from which it has been derived, not how well it predicts performance.
The solution derived from a model is tested to find whether it yields better performance than some alternative, usually the one in current use. The test may be prospective, against future
performance, or retrospective, comparing solutions that would have been obtained had the model been used in the past with what actually did happen. If neither prospective nor retrospective testing is
feasible, it may be possible to evaluate the solution by “sensitivity analysis,” a measurement of the extent to which estimates used in the solution would have to be in error before the proposed
solution performs less satisfactorily than the alternative decision procedure.
The cost of implementing a solution should be subtracted from the gain expected from applying it, thus obtaining an estimate of net improvement. Where errors or inefficiencies in applying the
solution are possible, these should also be taken into account in estimating the net improvement.
Implementing and controlling the solution
The acceptance of a recommended solution by the responsible manager depends on the extent to which he believes the solution to be superior to alternatives. This in turn depends on his faith in the
researchers involved and their methods. Hence, participation by managers in the research process is essential for success.
Operations researchers are normally expected to oversee implementation of an accepted solution. This provides them with an ultimate test of their work and an opportunity to make adjustments if any
deficiencies should appear in application. The operations research team prepares detailed instructions for those who will carry out the solution and trains them in following these instructions. The
cooperation of those who carry out the solution and those who will be affected by it should be sought in the course of the research process, not after everything is done. Implementation plans and
schedules are pretested and deficiencies corrected. Actual performance of the solution is compared with expectations and, where divergence is significant, the reasons for it are determined and
appropriate adjustments made.
The solution may fail to yield expected performance for one or a combination of reasons: the model may be wrongly constructed or used; the data used in making the model may be incorrect; the solution
may be incorrectly carried out; the system or its environment may have changed in unexpected ways after the solution was applied. Corrective action is required in each case.
Controlling a solution requires deciding what constitutes a significant deviation in performance from expectations; determining the frequency of control checks, the size and type of sample of
observations to be made, and the types of analyses of the resulting data that should be carried out; and taking appropriate corrective action. The second step should be designed to minimize the sum
of the costs of carrying out the control procedures and the errors that might be involved.
Since most models involve a variety of assumptions, these are checked systematically. Such checking requires explicit formulation of the assumptions made during construction of the model.
Effective controls not only make possible but often lead to better understanding of the dynamics of the system involved. Through controls the problem-solving system of which operations research is a
part learns from its own experience and adapts more effectively to changing conditions.
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Download Free Algebra 2 Worksheets from an Entire CurriculumGradeAmathHelp.comDownload Free Algebra 2 Worksheets from an Entire CurriculumGradeAmathHelp.com
Free Algebra 2 Worksheets
GradeAmathhelp offers free algebra 2 worksheets with no email address or subscription - simply free worksheets to print or download.
These worksheets are perfect for students who are looking for extra practice or teachers who need extra problems for their students. In fact, we offer an entire algebra 2 curriculum: fourteen units
covering all topics equations, to conic sections, and even trig. Check it out!
Free Algebra 2 Curriculum: Worksheet Downloads
Choose the chapter by clicking the link below. Each link will open a .pdf with an entire unit filled with guided notes sheets and practice problems.
Unit 1: Equations and Inequalities in One Variable
Unit 2: Functions, Equations, & Graphs of Degree One
Unit 3: Linear Systems
Unit 4: Matrices
Unit 5: Quadratic Equations & Functions
Unit 6: Polynomials (Answers!)
Unit 7: Radical Functions & Rational Exponents
Unit 8: Exponential & Logarithmic Functions
Unit 9: Rational Functions
Unit 10: Quadratic Relations
Unit 11: Sequences and Series (Answers!)
Unit 12: Statistics and Probability
Unit 13: Periodic Functions and Trig
Unit 14: Trigonometry Identities & Equations
Need help with some of these units?
Try free algebra help or free rigonometry help.
If these problems are too difficult for you, try our free algebra worksheets page, which covers algebra 1 topics. We also offer free fraction worksheets and and an entire geometry curriculum in our
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Please be aware of our terms & conditions of use before opening any of the files. Thank you!
Enjoy all of the free worksheets from GradeA.
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3.4 ounces to grams
To calculate a ounce value to the corresponding value in gram, multiply the quantity in ounce by Here is the formula :.
Do you want to convert grams to fluid ounces? Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications.
Full bio. Ethan has a PhD in astrophysics and is currently a satellite imaging scientist. He specializes in math, science, and astrophysics.
3.4 ounces to grams
How many grams in an ounce? There are And to answer a question such as how many grams in a quarter ounce , divide There are 7. How to convert ounces to grams [oz to g]:. Note: Ounce is an imperial or
United States customary unit of weight. Gram is a metric unit of weight. List of these foods starting with the highest contents of Biotin and the lowest contents of Biotin. Calculate how much of this
gravel is required to attain a specific depth in a cylindrical , quarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume volume to weight price ]. Volume to weight , weight
to volume and cost conversions for Refrigerant R, liquid R with temperature in the range of Lentor is an obsolete non-metric measurement unit of kinematic viscosity. The Conversions and Calculations
web site. Forum Login Register. Convert ounces to grams [oz to g] oz:ounce, g:gram Convert oz to grams a weight or mass conversion table How many grams in an ounce? From: to
Another useful application of weight and volume conversions is chemistry. Here is the formula :. Reviewed by Ethan Dederick, PhD.
Ounce oz is a unit of Weight used in Standard system. Gram g is a unit of Weight used in Metric system. Definition of the Ounce The ounce is defined differently in different systems of measurement.
The most common ounce is the international avoirdupois ounce, which is equal to This is the ounce that is used for most purposes, such as measuring food, postal items, fabric, paper and boxing
gloves. The avoirdupois ounce is one-sixteenth of an avoirdupois pound, which is defined as grains. Another ounce is the international troy ounce, which is equal to This is the ounce that is used for
measuring precious metals and gems, such as gold, silver, platinum and diamonds. The troy ounce is one-twelfth of a troy pound, which is defined as grains. Ounces can be converted to other units of
weight by using conversion factors or formulas.
3.4 ounces to grams
To convert 3. The result is the following:. We conclude that three point four 3. Therefore, if you want to calculate how many Grams are in 3. The ounce abbreviation: oz is a unit of mass with several
definitions, the most popularly used being equal to approximately 28 grams. The size of an ounce varies between systems. Today, the most commonly used ounces are the international avoirdupois ounce
equal to The gram alternative spelling: gramme; SI unit symbol: g is a metric system unit of mass.
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Fluid ounces can be abbreviated as fl oz , and are also sometimes abbreviated as oz fl. How to convert ounces oz to grams g To convert from ounces to grams : Use the conversion factor: 1 ounce equals
COOL Conversion. Calculate how much of this gravel is required to attain a specific depth in a cylindrical , quarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume volume
to weight price ]. Using the conversion formula above, you will get:. Thus, the weight in grams is equal to the volume in fluid ounces multiplied by How to transform ounces in grams? And to answer a
question such as how many grams in a quarter ounce , divide The gram, or gramme, is an SI unit of mass in the metric system. The fluid ounce is a US customary unit of volume.
Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications.
How to convert ounces to grams? Email optional. COOL Conversion. To convert a measurement in fluid ounces to grams, multiply the volume by the density of the ingredient or material. To convert How to
transform ounces in grams? Name optional. Volume to weight , weight to volume and cost conversions for Refrigerant R, liquid R with temperature in the range of How to convert ounces to grams [oz to
g]:. For example, an object with a mass of 1 gram weighs 1 gram on Earth, but only weighs one-sixth of that on the moon, yet still has the same mass. Therefore, to convert between fluid ounces and
grams of an ingredient or substance, we must either multiply or divide by its density, depending on which direction we are performing the conversion. What is the formula to convert from ounces to
grams? Suppose you want to convert 3. Physics Chemistry Recipes.
2 thoughts on “3.4 ounces to grams”
1. Willingly I accept. In my opinion, it is actual, I will take part in discussion. Together we can come to a right answer. I am assured.
2. Excuse for that I interfere � To me this situation is familiar. Let's discuss.
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CFD modeling and simulations of MHD power generation during re-entry
The flow subject to MHD power generation during re-entry is simulated by CFD in this paper. Thermal ionization with potassium seed is used to enhance the conductivity. The ionization of potassium is
simulated by both finite-rate chemistry and assuming Saha equilibrium. In the Saha equilibrium approach, the ionization of potassium is computed separately from the conservation equations. The
results can be seen as "chemically frozen" for potassium. The results show that the strength of the shock is over predicted. This leads to a lower flow velocity, and hence lower electric field
because the electric field and electromotive force (emf) are functions of the flow velocity. The second approach is to incorporate the ioniza-tion/recombination reaction in the conservation equation
set. With this model, the convection and diffusion of K and K^+, the ionization/recombination reaction rates, and the heat of formation of potassium ions are taken into account. Results show that the
thickness of the shock is less than that predicted by the Saha equilibrium. The computed velocity, emf and electric field are higher, and therefore the total extracted power is greater than what was
predicted by the Saha equilibrium model.
Original language English (US)
Title of host publication 35th AIAA Plasmadynamics and Lasers Conference
State Published - 2004
Event 35th AIAA Plasmadynamics and Lasers Conference 2004 - Portland, OR, United States
Duration: Jun 28 2004 → Jul 1 2004
Publication series
Name 35th AIAA Plasmadynamics and Lasers Conference
Other 35th AIAA Plasmadynamics and Lasers Conference 2004
Country/Territory United States
City Portland, OR
Period 6/28/04 → 7/1/04
All Science Journal Classification (ASJC) codes
• Electrical and Electronic Engineering
• Condensed Matter Physics
• Atomic and Molecular Physics, and Optics
Dive into the research topics of 'CFD modeling and simulations of MHD power generation during re-entry'. Together they form a unique fingerprint.
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Longitudinal and transversal resonant tunneling of interacting bosons in a two-dimensional Josephson junction
We unravel the out-of-equilibrium quantum dynamics of a few interacting bosonic clouds in a two-dimensional asymmetric double-well potential at the resonant tunneling scenario. At the single-particle
level of resonant tunneling, particles tunnel under the barrier from, typically, the ground-state in the left well to an excited state in the right well, i.e., states of different shapes and
properties are coupled when their one-particle energies coincide. In two spatial dimensions, two types of resonant tunneling processes are possible, to which we refer to as longitudinal and
transversal resonant tunneling. Longitudinal resonant tunneling implies that the state in the right well is longitudinally-excited with respect to the state in the left well, whereas transversal
resonant tunneling implies that the former is transversely-excited with respect to the latter. We show that interaction between bosons makes resonant tunneling phenomena in two spatial dimensions
profoundly rich, and analyze these phenomena in terms of the loss of coherence of the junction and development of fragmentation, and coupling between transverse and longitudinal degrees-of-freedom
and excitations. To this end, a detailed analysis of the tunneling dynamics is performed by exploring the time evolution of a few physical quantities, namely, the survival probability, occupation
numbers of the reduced one-particle density matrix, and the many-particle position, momentum, and angular-momentum variances. To accurately calculate these physical quantities from the time-dependent
many-boson wavefunction, we apply a well-established many-body method, the multiconfigurational time-dependent Hartree for bosons (MCTDHB), which incorporates quantum correlations exhaustively. By
comparing the survival probabilities and variances at the mean-field and many-body levels of theory and investigating the development of fragmentation, we identify the detailed mechanisms of
many-body longitudinal and transversal resonant tunneling in two dimensional asymmetric double-wells. In particular, we find that the position and momentum variances along the transversal direction
are almost negligible at the longitudinal resonant tunneling, whereas they are substantial at the transversal resonant tunneling which is caused by the combination of the density and breathing mode
oscillations. We show that the width of the interparticle interaction potential does not affect the qualitative physics of resonant tunneling dynamics, both at the mean-field and many-body levels. In
general, we characterize the impact of the transversal and longitudinal degrees-of-freedom in the many-boson tunneling dynamics at the resonant tunneling scenarios.
Bibliographical note
Publisher Copyright:
© 2022, The Author(s).
ASJC Scopus subject areas
Dive into the research topics of 'Longitudinal and transversal resonant tunneling of interacting bosons in a two-dimensional Josephson junction'. Together they form a unique fingerprint.
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Percentage of Percentage Calculator
Below is a percentage of percentage calculator, which will calculate the percentage of another percentage. Enter a percentage X% which you take "of" a second percentage Y%, and the tool will tell you
the final impact.
Percentage of Percentage Calculator
What is a Percentage of Percentage?
A percentage of percentage comes up when two (or more) percentages interact in a way that you need to determine the final impact.
For example, in investing, you often have an X% claim on something, but may owe Y% in fees, or lose that Y% to some other factor (for example, in accredited investments). You won't know the true
effect of the various clauses until you work through percentages of percentages.
Example Percentage of Percentage
Let's say there is a $10,000,000 fund, and you have the rights to 35% of the gains, but pay 20% carry or performance fee. How much would you receive in a year where the fund made $2,000,000?
Stated another way, you take home 80% of 35% of the gains. Here's how it looks:
your\ take=35\%*80\% = .35*.8 = 28\%\ of\ gains\\~\\your\ distribution=.28*\$2,000,000=\$560,000
Using the Percentage of Percentages Calculator
In the entry field at the top, enter the first percentage – the percentage X that you are taking "of" the second percentage. Likewise, in the second box,
When happy, simply hit the Compute Average button below and DQYDJ will find the percentage of percentage, so you can know the final impact.
Here are our other percentage tools:
Like this type of thing? Visit our other calculators and tools.
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Farkas's Lemma and the nature of reality: Statistical implications of quantum correlations
A general algorithm is given for determining whether or not a given set of pair distributions allows for the construction of all the members of a specified set of higher-order distributions which
return the given pair distributions as marginals. This mathematical question underlies studies of quantum correlation experiments such as those of Bell or of Clauser and Horne, or their higher-spin
generalizations. The algorithm permits the analysis of rather intricate versions of such problems, in a form readily adaptable to the computer. The general procedure is illustrated by simple
derivations of the results of Mermin and Schwarz for the symmetric spin-1 and spin-3/2 Einstein-Podolsky-Rosen problems. It is also used to extend those results to the spin-2 and spin-5/2 cases,
providing further evidence that the range of strange quantum theoretic correlations does not diminish with increasing s. The algorithm is also illustrated by giving an alternative derivation of some
recent results on the necessity and sufficiency of the Clauser-Horne conditions. The mathematical formulation of the algorithm is given in general terms without specific reference to the quantum
theoretic applications.
ASJC Scopus subject areas
• General Physics and Astronomy
Dive into the research topics of 'Farkas's Lemma and the nature of reality: Statistical implications of quantum correlations'. Together they form a unique fingerprint.
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In a cohort of population, 10% of individuals that are alive at... | Filo
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In a cohort of population, 10% of individuals that are alive at the beginning of the year die each year. Try plotting such a curve graph with (i) an arithmetic Y-axis and one graph with (ii)
logarithmic Y-axis. No need to submit these graphs. What shape does the curve look in each case? What type of curve represents the data?
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Question In a cohort of population, 10% of individuals that are alive at the beginning of the year die each year. Try plotting such a curve graph with (i) an arithmetic Y-axis and one graph with (ii)
Text logarithmic Y-axis. No need to submit these graphs. What shape does the curve look in each case? What type of curve represents the data?
Updated Apr 19, 2024
Topic All topics
Subject Biology
Class High School
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Course Content
The text book for this course is `Industrial Mathematics. Modeling in Industry, Science, and Government.' by Charles R. MacCluer, Prentice Hall, 2000.
Industrial mathematics is mathematics subject to the constraints of time and money. Unlike pure mathematicians, we do not have the luxury to think for centuries about resolving a conjecture, but must
come up with a solution which makes our business more profitable in a very limited time frame. Almost without exception, the solution will involve computations. Unfortunately, we do not have the time
nor the means to develop tailored programs, so we better take advantage of available software tools. Thirdly, the reward for our work is directly linked to our ability to present our solution for a
non-mathematical and very busy readership. While these tasks seem daunting, do not worry, teamwork is essential, we collaborate.
These web pages will contain supplemental materials to the text book and the lectures, mainly concerning the computational aspects of the course.
Below are some supplemental materials to the lectures.
1. Statistical Reasoning
L-1 The four most useful distributions
L-2 Maple solution to a staffing problem
2. The Monte Carlo Method
L-3 Maple solution to the mean time between failures problem
L-4 Presentations of projects and technical writing
L-5 MTBF and servicing requests
3. Data Aquisition and Manipulation
L-6 Z-transforms and linear recursions
L-7 Filters, Stability, and plots
L-8 Filters
L-9 Control Systems Design
4. The Discrete Fourier Transform
L-10 The FFT and its application to filters
L-11 The DFT: definition, properties and filter design
L-13 Presentations of Project I
L-14 Presentations of Project I (continued)
5. Linear Programming
L-15 Linear Programming
6. Regression
L-16 Linear programming and polynomial fitting in MATLAB
L-17 Regression
7. Cost-Benefit Analysis
L-18 Fourier Series and Cost-Benefit Analysis
8. Microeconomics
L-19 Microeconomics
L-21 Midterm Exam on Chapters 1 to 8 with solutions
9. Ordinary Differential Equations
L-22 ODEs in mechanics
L-23 Linear ODEs with constant coefficients
L-24 Linear Systems
10. Frequency-Domain Methods
L-25 Frequency domains, signals, and plants
L-26 Plants, signals, and surge impedance
L-27 Filters and Bode plots
L-29 Presentations of Project II (continued)
L-30 Project II presentation (continued) and Nyquist Analysis
L-31 Nyquist plots and Control
11. Partial Differential Equations
L-32 Air Quality Modeling
L-33 Separation of Variables
L-34 PDEs in Maple
L-35 Traffic Flow Modeling
L-36 Periodic Steady State
12. Divided Differences
L-37 Numerically Solving ODEs and PDEs with Maple
13. Galerkin's method
L-38 Galerkin's method in Maple
L-39 Finite elements
14. Splines
L-41 Splines continued and review
L-42 Presentations of Project III
L-43 Presentations of Project III (continued)
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Locally efficient doubly robust DiD estimator for the ATT, with panel data — drdid_panel
drdid_panel is used to compute the locally efficient doubly robust estimators for the ATT in difference-in-differences (DiD) setups with panel data.
i.weights = NULL,
boot = FALSE,
boot.type = "weighted",
nboot = NULL,
inffunc = FALSE
An \(n\) x \(1\) vector of outcomes from the post-treatment period.
An \(n\) x \(1\) vector of outcomes from the pre-treatment period.
An \(n\) x \(1\) vector of Group indicators (=1 if observation is treated in the post-treatment, =0 otherwise).
An \(n\) x \(k\) matrix of covariates to be used in the propensity score and regression estimation. Please add a vector of constants if you want to include an intercept in the models. If
covariates = NULL, this leads to an unconditional DiD estimator.
An \(n\) x \(1\) vector of weights to be used. If NULL, then every observation has the same weights. The weights are normalized and therefore enforced to have mean 1 across all observations.
Logical argument to whether bootstrap should be used for inference. Default is FALSE.
Type of bootstrap to be performed (not relevant if boot = FALSE). Options are "weighted" and "multiplier". If boot = TRUE, default is "weighted".
Number of bootstrap repetitions (not relevant if boot = FALSE). Default is 999.
Logical argument to whether influence function should be returned. Default is FALSE.
A list containing the following components:
The DR DiD point estimate.
The DR DiD standard error.
Estimate of the upper bound of a 95% CI for the ATT.
Estimate of the lower bound of a 95% CI for the ATT.
All Bootstrap draws of the ATT, in case bootstrap was used to conduct inference. Default is NULL.
Estimate of the influence function. Default is NULL.
The matched call.
Some arguments used (explicitly or not) in the call (panel = TRUE, estMethod = "trad", boot, boot.type, nboot, type="dr")
The drdid_panel function implements the locally efficient doubly robust difference-in-differences (DiD) estimator for the average treatment effect on the treated (ATT) defined in equation (3.1) in
Sant'Anna and Zhao (2020). This estimator makes use of a logistic propensity score model for the probability of being in the treated group, and of a linear regression model for the outcome evolution
among the comparison units.
The propensity score parameters are estimated using maximum likelihood, and the outcome regression coefficients are estimated using ordinary least squares.
Sant'Anna, Pedro H. C. and Zhao, Jun. (2020), "Doubly Robust Difference-in-Differences Estimators." Journal of Econometrics, Vol. 219 (1), pp. 101-122, doi:10.1016/j.jeconom.2020.06.003
# Form the Lalonde sample with CPS comparison group (data in wide format)
eval_lalonde_cps <- subset(nsw, nsw$treated == 0 | nsw$sample == 2)
# Further reduce sample to speed example
unit_random <- sample(1:nrow(eval_lalonde_cps), 5000)
eval_lalonde_cps <- eval_lalonde_cps[unit_random,]
# Select some covariates
covX = as.matrix(cbind(1, eval_lalonde_cps$age, eval_lalonde_cps$educ,
eval_lalonde_cps$black, eval_lalonde_cps$married,
eval_lalonde_cps$nodegree, eval_lalonde_cps$hisp,
# Implement traditional DR locally efficient DiD with panel data
drdid_panel(y1 = eval_lalonde_cps$re78, y0 = eval_lalonde_cps$re75,
D = eval_lalonde_cps$experimental,
covariates = covX)
#> Call:
#> drdid_panel(y1 = eval_lalonde_cps$re78, y0 = eval_lalonde_cps$re75,
#> D = eval_lalonde_cps$experimental, covariates = covX)
#> ------------------------------------------------------------------
#> Locally efficient DR DID estimator for the ATT:
#> ATT Std. Error t value Pr(>|t|) [95% Conf. Interval]
#> -507.6113 692.9739 -0.7325 0.4639 -1865.8401 850.6175
#> ------------------------------------------------------------------
#> Estimator based on panel data.
#> Outcome regression est. method: OLS.
#> Propensity score est. method: maximum likelihood.
#> Analytical standard error.
#> ------------------------------------------------------------------
#> See Sant'Anna and Zhao (2020) for details.
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