text:
# Aliens want our water [closed]

Aliens, for some unknown reason, want all the water (H2O) in our oceans, seas, lakes, rivers, (all large bodies of water on earth), and plan to steal it from us overnight.

Luckily for us, they don't want to wreck life on this planet.

As highly advanced lifeforms, they have the technology to replace all our water with an equivalent amount of "magic liquid". We don't know why they can't make H2O - they just can't, but they can make anything else.

So the question is: Is there a replacement for H2O that the aliens can replace our water with that won't destroy life as we know it? This chemical compound should have most (if not all) of the properties of H2O.

Note: How they manage to do this is irrelevant. You can assume that by some hand-waved process, all the H2O just gets replaced with the aforementioned magic liquid instantaneously.

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

## closed as off-topic by JDługosz, Magic-Mouse, ArtOfCode, bowlturner, Dan SmolinskeJun 4 '15 at 15:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Questions about Idea Generation are off-topic because they tend to result in list answers with no objective means to compare the quality of one answer with the others. For more information, see What's wrong with idea-generation questions?." – JDługosz, Magic-Mouse, ArtOfCode, Dan Smolinske
If this question can be reworded to fit the rules in the help center, please edit the question.

• And by 'magic liquid', you mean something that actually exists, right? – DaaaahWhoosh Jun 3 '15 at 19:16
• Unless you have a simulator that can simulate the properties of a chemical formula you just made up, then yes, I would like to see a formula that currently exists in this world. I will, however, accept a formula that is speculated to work as a substitute as long as sufficient evidence is provided @DaaaahWhoosh – anon Jun 3 '15 at 19:19
• I would offer the suggestion of H2O2, since it'll eventually turn into water and oxygen, but I'm wondering if that would still kill all the fish, especially in the deep parts. – DaaaahWhoosh Jun 3 '15 at 19:20
• If there exists a liquid that is sufficiently identical to water to not disrupt the very precarious balance of the Earth's ecosystems that can be freely fabricated by the aliens, why don't they just use the fabricated liquid instead of stealing water from an inhabited world? – Frostfyre Jun 3 '15 at 19:23
• If this was ever made into a movie, it would be the next Trolls 2. – ShemSeger Jun 4 '15 at 2:29

I'll revert to my old pessimistic self and say that there's nothing that can replace good ol' H2O.

Here are some things that a replacement liquid would have to do:

1. Be denser as a liquid than a solid. This is atypical of most substances, but water is an exception. This is why only the tops of ponds freeze over and the lower layers are frozen. Otherwise, nothing would survive in winter, because ice would form and sink down, letting more water freeze. Ice fishing would be much less interesting nonexistent.
2. Be the next universal solvent. If you've taken high school biology, then you might have heard the phrase "water is the universal solvent" a couple times. Water is good because it can dissolve quite a lot of things. A related pure-chemistry-inspired property is that water plays a huge role in the reactions of acids and bases.
3. Stay liquid at room temperature, and the moderate ranges of temperatures seen on Earth. If a substance is solid at 25 degrees Celsius, then we've got a problem (namely, the oceans). It must also be possible for this substance to be gaseous at some point, facilitating the water cycle.
4. Be useable in both cellular respiration and photosynthesis. The reasons for this are pretty straightforward. These are clearly not the only forms of energy production used by organisms on Earth (e..g thermophiles use different mechanisms), but they're what you'll have to use if you want things to stay normal.
• Dihydrogen monoxide should do the trick. – Samuel Jun 3 '15 at 22:30
• Gosh, I totally forgot about Dihydrogen Monoxide! That would be perfect! I'm totally putting that into an answer. – ShemSeger Jun 4 '15 at 2:28
• @Samuel, I'm surprised no one suggested DHMO sooner! – ShemSeger Jun 4 '15 at 2:48

I believe the answer is "NO". Water is water is water. On top of that, hydrogen and oxygen are two of the most abundant elements in the universe. Stealing it from a populated planet (where you don't want to hurt the lifeforms) is madness.

• I don't see how nitrogen oxide is going to help anyone. – DaaaahWhoosh Jun 3 '15 at 19:51
• @DaaaahWhoosh :P~~~ – bowlturner Jun 3 '15 at 19:51
• Mad sci-aliens? XD – Aify Jun 3 '15 at 20:01
• I've been substituting water with Mountain Dew for years, and so far it seems to work. – IchabodE Jun 3 '15 at 21:22
• @MBurke: But Mountain Dew is at least 90% water. – jamesqf Jun 3 '15 at 23:10

Since the question specifically wants to use water from Earth (damn, these aliens are insistent), I returned to the drawing board to rethink my approach. My first reaction to this question was to refute it, but I decided to take a look at what substitutes for water were already known and, to my extreme surprise, there is one that is pretty close.

Formamide has demonstrated viability as an alternative solvent to water. It can dissolve most things that dissolve in water. This chemical was originally brought to my attention by this paper, for those who have access to The Journal of Physical Chemistry.

I am leaving this part of the answer here for posterity reasons.

Since these aliens are in dire need of water and (for whatever reason) can only use H2O despite its similarity to an easily fabricated liquid, we, as Earthlings intent on preserving our precious little pearl of a planet, should erect a giant sign in space that says:

$$\large \text{Go to Europa for Water}$$

There are several places in our solar system with water, beyond the Earth. Europa is believed to have more than twice the water volume of Earth's oceans.

• But they want OUR water! – Aify Jun 3 '15 at 20:00
• @Aify Why? Our water isn't any more special than water elsewhere in the universe. – Frostfyre Jun 3 '15 at 20:15
• Who knows? They're crazy aliens! Maybe for them, it's the last stage of a test of a chemical on life and they need mass test subjects so they're doing it here - the end result should be that life is NOT destroyed. – Aify Jun 3 '15 at 20:16
• @Aify See my new paragraphs. – Frostfyre Jun 3 '15 at 20:32
• This chemical would be so perfect if only it weren't highly corrosive ._. – Aify Jun 3 '15 at 20:34

It is a non-believable old story trope that aliens want our water. Why not just grab a chunk from the Oort cloud? I read reciently that it looks like one of Jupiter's moons appears to have a bit more water than Earth's oceans. The most common elements in the universe are, in order, Hydrogen, Helium, and Oxygen. Water is ubiquitous.

• It's not a handwave, and it's not a non-believable trope. Maybe they already took the water from the Oort cloud and elsewhere, and now have arrived at Earth needing more? Maybe they always need more because any single, scalar quantity is gonna run out sooner or later. You need a rate. Kinda like us and petroleum, isn't it. – DrZ214 Jun 10 '15 at 23:57
• As a story, people would see the aliens absconding with the icy moons and hoovering up billions of comets, and hoping and worrying that they will leve Earth alone. – JDługosz Jun 11 '15 at 0:53

# Dihydrogen Monoxide

Dihydrogen Monoxide is a clear liquid, mostly flavourless chemical that is often used in pesticide production and distribution which can effectively replace water in an ecosystem. When sprayed over crops it was discovered that DHMO had an almost identical affect on the crops growth as naturally occurring water. Shortly after pure DHMO is introduced into an ecosystem it quickly absorbs minerals in solution making it almost indistinguishable from real water that falls as precipitation!

There are a few ways to detect dihydrogen monoxide however: it is highly reactive with pure sodium, lithium and potassium (as in explosive reactive), and when poured over iron, it can cause a redox reaction that will turn the iron red and make it brittle.

But you can drink it, you can swim in it, and you can even make ice with it! To the naked eye, and without specialized equipment, it would be impossible for the average person to distinguish DHMO from H2O.


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PL EN

Journal

## Reports on Mathematical Logic

2011 | 46 | 109-132
Article title

### ON SEMILATTICE-BASED LOGICS WITH AN ALGEBRAIZABLE ASSERTIONAL COMPANION

Authors
Selected contents from this journal
Title variants
Languages of publication
EN
Abstracts
EN
This paper studies some properties of the so-called semilattice-based logics (which are defined in a standard way using only the order relation from a variety of algebras that have a semilattice reduct with maximum) under the assumption that its companion assertional logic (defined from the same variety of algebras using the top element as representing truth) is algebraizable. This describes a very common situation, and the conclusion of the paper is that these semilattice-based logics exhibit some of the good behaviour of protoalgebraic logics, without being necessarily so. The main result is that all these logics have enough Leibniz filters, a fact previously known in the literature to occur only for protoalgebraic logics. Another significant result is that the two companion logics coincide if and only if one of them enjoys the characteristic property of the other, that is, if and only if the semilattice-based logic is algebraizable, and if and only if its assertional companion is selfextensional. When these conditions are met, then the (unique) logic is finitely, regularly and strongly algebraizable and fully Fregean, this places it at some of the highest ranks in both the Leibniz hierarchy and the Frege hierarchy.
Keywords
EN
Discipline
Year
Issue
Pages
109-132
Physical description
Document type
ARTICLE
Contributors
author
• Josep Maria Font, Department of Probability, Logic and Statistics, Faculty of Mathematics, University of Barcelona, Barcelona, Spain
References
Document Type
Publication order reference
Identifiers
CEJSH db identifier
11PLAAAA10196

format: plain


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Skip to main content

# 4.5: Preparing a Production Cost Report

[ "article:topic" ]

Skills to Develop

• Prepare a production cost report for a processing department.

Question: The results of the four key steps are typically presented in a production cost report. The production cost report8 summarizes the production and cost activity within a department for a reporting period. It is simply a formal summary of the four steps performed to assign costs to units transferred out and units in ending work-in-process (WIP) inventory. What does the production cost report look like for the Assembly department at Desk Products, Inc.?

Answer:

The production cost report for the month of May for the Assembly department appears in Figure 4.9. Notice that each section of this report corresponds with one of the four steps described earlier. We provide references to the following illustrations so you can review the detail supporting calculations.

Figure 4.9 - Production Cost Report for Desk Products’ Assembly Department

a Total costs to be accounted for (step 2) must equal total costs accounted for (step 4).

b Data are given.

c This section comes from Figure 4.4.

d This section comes from Figure 4.5.

e This section comes from Figure 4.6.

f This section comes from Figure 4.7.

### How Do Managers Use Production Cost Report Information?

Question: Although the production cost report provides information needed to transfer costs from one account to another, managers also use this report for decision-making purposes. What important questions can be answered using the production cost report?

Answer:

A production cost report helps managers answer several important questions:

• How much does it cost to produce each unit of product for each department?
• Which production cost is the highest—direct materials, direct labor, or overhead?
• Where are we having difficulties in the production process? In any particular departments?
• Are we seeing any significant changes in unit costs for direct materials, direct labor, or overhead? If so, why?
• How many units flow through each processing department each month?
• Are improvements in the production process being reflected in the cost per unit from one month to the next?

### Beware of Fixed Costs

Question: Why might the per unit cost data provided in the production cost report be misleading?

Answer:

When using information from the production cost report, managers must be careful not to assume that all production costs are variable costs. The CEO of Desk Products, Inc., Ann Watkins, was told that the Assembly department cost for each desk totaled $62 for the month of May (from Figure 4.9, step 3). However, if the company produces more or fewer units than were produced in May, the unit cost will change. This is because the$62 unit cost includes both variable and fixed costs (see Chapter 5 for a detailed discussion of fixed and variable costs).

Assume direct materials and direct labor are variable costs. In the Assembly department, the variable costs per unit associated with direct materials and direct labor of $50 (=$30 direct materials + $20 direct labor) will remain the same regardless of the level of production, within the relevant range. However, the remaining unit product cost of$12 associated with overhead must be analyzed further to determine the amount that is variable (e.g., indirect materials) and the amount that is fixed (e.g., factory rent). Managers must understand that fixed costs per unit will change depending on the level of production. More specifically, Ann Watkins must understand that the \$62 unit cost in the Assembly department provided in the production cost report will change depending on the level of production. Chapter 5 provides a detailed presentation of how cost information can be separated into fixed and variable components for the purpose of providing managers with more useful information.

KEY TAKEAWAY

The four key steps of assigning costs to units transferred out and units in ending WIP inventory are formally presented in a production cost report. The production cost report summarizes the production and cost activity within a processing department for a reporting period. A separate report is prepared for each processing department. Rounding the cost per equivalent unit to the nearest thousandth will minimize rounding differences when reconciling costs to be accounted for in step 2 with costs accounted for in step 4.

Computer Application

###### Using Excel to Prepare a Production Cost Report

Managers typically use computer software to prepare production cost reports. They do so for several reasons:

• Once the format is established, the template can be used from one period to the next.
• Formulas underlie all calculations, thereby minimizing the potential for math errors and speeding up the process.
• Changes can be made easily without having to redo the entire report.
• Reports can be easily combined to provide a side-by-side analysis from one period to the next.

Review Figure 4.9 and then ask yourself: “How can I use Excel to help prepare this report?” Answers will vary widely depending on your experience with Excel. However, Excel has a few basic features that can make the job of creating a production cost report easier. For example, you can use formulas to sum numbers in a column (note that each of the four steps presented in Figure 4.9 has column totals) and to calculate the cost per equivalent unit. Also you can establish a separate line to double-check that

• the units to be accounted for match the units accounted for; and
• the total costs to be accounted for match the total costs accounted for.

For those who want to add more complex features, the basic data (e.g., the data in Table 4.2) can be entered at the top of the spreadsheet and pulled down to the production cost report where necessary.

An example of how to use Excel to prepare a production cost report follows. Notice that the basic data are at the top of the spreadsheet, and the rest of the report is driven by formulas. Each month, the data at the top are changed to reflect the current month’s activity, and the production cost report takes care of itself.

REVIEW PROBLEM 4.5

Using the information in Note 4.24 "Review Problem 4.4", prepare a production cost report for the Mixing department of Kelley Paint Company for the month ended March 31. (Hint: You have already completed the four key steps in Note 4.24 "Review Problem 4.4". Simply summarize the information in a production cost report as shown in Figure 4.9 "Production Cost Report for Desk Products’ Assembly Department".)

Answer:

(See solutions to Note 4.24 "Review Problem 4.4" for detailed calculations.)

### Definition

1. A report that summarizes the production and cost activity within a department for a reporting period.


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# Scatch space mod prob. is need a formula and answer check

1. Dec 13, 2003

### cowgiljl

i think i got it right it wants to know the speed of the command module and the time it takes to do 1 revolution
mass moon = 7.35E22 kg
radius moon = 1.74E6 m
altitude 60 mi

fromula used was Vcm^2 = GMm/Ro

the speed of the module was 1634 m/s

The time formula thans to enigma was

P = 2\pi \sqrt{\frac{a^3}{\mu}}}
T = 6511 seconds/60
T = 108 min

is this right?

2. Dec 13, 2003

### gnome

Your speed looks correct, using your value for mass of the moon.

But I don't follow what you're doing to get the period.

I'm using Kepler's 3rd law:
$$T^2 = (\frac{4\pi^2}{GM_m})r^3$$
and here r = (1.74 x 106 + 1609 x 60) so using G=6.67x10-11 and Mm=7.35x1022 I'm getting
T = 7063 s or 118 min.

(Incidentally, my book lists the mass of the moon as 7.36x1022 kg, but what's a mere 100,000,000,000,000,000,000 kg among friends?)

3. Dec 13, 2003

### cowgiljl

you re right gnome

I wrote down and typed the wrong answer but my calculations were correct. the mind was quicker than the pen

thanks again

4. Dec 13, 2003

### gnome

Sometimes its good to step back & think about what we're doing.

While I was walking my dog, I was wondering why we're using that crazy formula to get the time of 1 revolution. (Heresy -- calling Kepler's 3rd law a crazy formula!!! )

We already know the speed. Just divide the circumference of the orbit by the speed & get the same answer.]


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# Properties

 Label 12T23 Degree $12$ Order $48$ Cyclic no Abelian no Solvable yes Primitive no $p$-group no Group: $C_2 \times S_4$

# Related objects

## Group action invariants

 Degree $n$: $12$ Transitive number $t$: $23$ Group: $C_2 \times S_4$ CHM label: $S_{4}(6d)[x]2=[1/8.2^{6}]S(3)$ Parity: $1$ Primitive: no Nilpotency class: $-1$ (not nilpotent) $|\Aut(F/K)|$: $4$ Generators: (2,8)(3,9)(4,10)(5,11), (1,5,9)(2,6,10)(3,7,11)(4,8,12), (1,12)(2,3)(4,5)(6,7)(8,9)(10,11), (2,10)(3,11)(4,8)(5,9)

## Low degree resolvents

|G/N|Galois groups for stem field(s)
$2$:  $C_2$ x 3
$4$:  $C_2^2$
$6$:  $S_3$
$12$:  $D_{6}$
$24$:  $S_4$

Resolvents shown for degrees $\leq 47$

## Subfields

Degree 2: $C_2$

Degree 3: $S_3$

Degree 4: None

Degree 6: $D_{6}$, $S_4$, $S_4\times C_2$

## Low degree siblings

6T11 x 2, 8T24 x 2, 12T21, 12T22, 12T23, 12T24 x 2, 16T61, 24T46, 24T47, 24T48 x 2

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

## Conjugacy classes

 Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2, 2, 1, 1, 1, 1$ $6$ $2$ $( 2, 4)( 3, 5)( 8,10)( 9,11)$ $2, 2, 2, 2, 1, 1, 1, 1$ $3$ $2$ $( 2, 8)( 3, 9)( 4,10)( 5,11)$ $2, 2, 2, 2, 2, 2$ $6$ $2$ $( 1, 2)( 3,12)( 4, 5)( 6, 9)( 7, 8)(10,11)$ $6, 6$ $8$ $6$ $( 1, 2, 5,12, 3, 4)( 6, 9,10, 7, 8,11)$ $4, 4, 2, 2$ $6$ $4$ $( 1, 2, 7, 8)( 3, 6, 9,12)( 4,11)( 5,10)$ $3, 3, 3, 3$ $8$ $3$ $( 1, 3, 5)( 2, 4,12)( 6, 8,10)( 7, 9,11)$ $4, 4, 2, 2$ $6$ $4$ $( 1, 3, 7, 9)( 2, 6, 8,12)( 4,10)( 5,11)$ $2, 2, 2, 2, 2, 2$ $3$ $2$ $( 1, 6)( 2, 3)( 4,11)( 5,10)( 7,12)( 8, 9)$ $2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1,12)( 2, 3)( 4, 5)( 6, 7)( 8, 9)(10,11)$

## Group invariants

 Order: $48=2^{4} \cdot 3$ Cyclic: no Abelian: no Solvable: yes GAP id: [48, 48]
 Character table:  2 4 3 4 3 1 3 1 3 4 4 3 1 . . . 1 . 1 . . 1 1a 2a 2b 2c 6a 4a 3a 4b 2d 2e 2P 1a 1a 1a 1a 3a 2b 3a 2b 1a 1a 3P 1a 2a 2b 2c 2e 4a 1a 4b 2d 2e 5P 1a 2a 2b 2c 6a 4a 3a 4b 2d 2e X.1 1 1 1 1 1 1 1 1 1 1 X.2 1 -1 1 -1 1 -1 1 -1 1 1 X.3 1 -1 1 1 -1 1 1 -1 -1 -1 X.4 1 1 1 -1 -1 -1 1 1 -1 -1 X.5 2 . 2 . -1 . -1 . 2 2 X.6 2 . 2 . 1 . -1 . -2 -2 X.7 3 -1 -1 -1 . 1 . 1 -1 3 X.8 3 -1 -1 1 . -1 . 1 1 -3 X.9 3 1 -1 -1 . 1 . -1 1 -3 X.10 3 1 -1 1 . -1 . -1 -1 3


text:
# Weak-magnetism phenomena in heavy-fermion superconductors: selected μSR studies

@article{Amato2004WeakmagnetismPI,
title={Weak-magnetism phenomena in heavy-fermion superconductors: selected $\mu$SR studies},
author={A Amato and M J Graf and A de Visser and H Amitsuka and D Andreica and A. Schenck},
journal={Journal of Physics: Condensed Matter},
year={2004},
volume={16},
pages={S4403 - S4420}
}
• Published 2004
• Physics
• Journal of Physics: Condensed Matter
The behaviour of the so-called weak-moment antiferromagnetic states, observed in the heavy-fermion superconductors UPt3 and URu2Si2, is discussed in view of recent μSR results obtained as functions of control parameters such as chemical substitution and external pressure. In UPt3, the Pd substitution for Pt reveals the dynamical character of the weak-moment order. On the other hand, μSR measurements performed on samples in which Th substitutes U suggest that crystallographic disorder on the…
27 Citations
Suppression of hidden order in URu2Si2 under pressure and restoration in magnetic field
• Physics
• 2010
We describe here recent inelastic neutron scattering experiments on the heavy fermion compound URu2Si2 realized in order to clarify the nature of the hidden order (HO) phase which occurs below T0 =
Competition of ground states in URu2Si2 and UCoGe
In this thesis two heavy fermion compounds have been studied under pressure. URu2Si2 has a mysterious ground state below T0 = 17.5 K at ambient pressure. The order parameter has not been identified
Heavy-fermion behaviours in URu2Si2
• Physics
• 2005
We analyse the magnetic susceptibility χ and the specific heat C of the heavy-fermion material URu2Si2, assuming a variety of the crystal field level schemes. The heavy-fermion behaviours of χ and C
Magnetoelastic coupling in URu2Si2 : Probing multipolar correlations in the hidden order state
• Physics
Physical Review B
• 2019
Time reversal symmetry and magnetoelastic correlations are probed by means of high-resolution volume dilatometry in URu2Si2 at cryogenic temperatures and magnetic fields more than enough to suppress
Electrical Resistivity and Thermal Expansion Measurements of URu2Si2 under Pressure
• Materials Science, Physics
• 2008
We carried out simultaneous measurements of electrical resistivity and thermal expansion of the heavy-fermion compound URu 2 Si 2 under pressure using a single crystal. We observed a phase transition
Muon spin spectroscopy: magnetism, soft matter and the bridge between the two
• Physics
• 2014
The use of implanted muons to probe the spin dynamics and electronic excitations in a variety of magnetic and non-magnetic materials is reviewed and is split into three main sections, the first of


text:
Cylinder under pure compression

 Title Cylinder under pure compression Tags elasticity compression Runnng time 1 sec See also 003-cube-pure-tension CAEplex case https://caeplex.com/p/ed77e Available in HTML PDF ePub

1 Problem description

A non-dimensional cylinder of radius 1/2 and height 1 whose base rests on the x-y plane centered at the origin is subject to a nondimensional compressive pressure of 1 unit in the negative z direction acting on the face at z=1 (fig. 1). The cube has a non-dimensional unitary Young modulus E=1 and Poisson’s ratio \nu=0.3. To remove rigid-body displacemetns while obtaining pure compressive stress, instead of fixing lower-dimensional entities as in Cube under pure tension, the bottom face is set to have both tangential and radial symmetry displacement conditions.

Once again, as in Cube under pure tension, the displacements and stresses distribution within the cylinder are to be obtained. The stress \sigma_x is expected to be uniformly equal to one with the other five stress components (\sigma_y, \sigma_z, \tau_{xy}, \tau_{yz} and \tau_{zx}) to be identically equal to zero. The cylinder is expected to shrink in the z direction and to expand in both the x and y directions due to the non-zero Poisson’s ratio \nu.

The mesh is expected to be fully unstructured and not optimized as to show that the expected results are recovered with the numerical calculation independently of having a mesh composed of wedges and/or hexahedra within an azimuthal array. In this case, the actual area of the discretized surface that holds the external load does depend on the mesh coarseness, with the theoretical value of \pi/4 recovered only in the limit of infinite elements. However, the load condition in this case is an uniform pressure and not a force, so the normal stress in the axial direction is expected to be numerically equal to such pressure.

2 Geometry and mesh

The geometry is created and meshed directly from Gmsh. Only two physical surfaces are needed for the boundary conditions, namely “bottom” and “top.” The physical volume is needed to get volumetric elements in the resulting cylinder.msh file. The mesh is unstructured tetrahedral due to the reasons already explained.

SetFactory("OpenCASCADE");

Cylinder(1) = {0, 0, 0, 0, 0, 1, 0.5};

Physical Surface("bottom") = {3};
Physical Surface("top") = {2};
Physical Volume("bulk") = {1};

Mesh.ElementOrder = 2;

3 Input file

This time the input file cylinder.fin is not annotated yet again it is self-explanatory. It solves the problem and writes a VTK file.

MESH FILE_PATH cylinder.msh

E = 1
nu = 0.3

PHYSICAL_GROUP top    BC P=1

FINO_STEP

MESH_POST FILE_PATH cylinder.vtk VECTOR u v w sigmax sigmay sigmaz tauxy tauyz tauzx

Note the two special boundary conditions tangential and radial. The first one restricts displacements along the local normal of the surface. The latter forces displacements within the surface plane to be radially distributed around the surface’s barycenter.

4 Execution

$gmsh -v 0 -3 cylinder.geo$ fino cylinder.fin
$ 5 Results Figs. 2 (a), 2 (b) show \sigma_x and \sigma_y in the final deformed condition. Fig. 3 shows an interactive 3D view obtained by CAEplex where the effect of the radial symmetry condition can be seen when the displacements are warped with the slider. Figure 2: Stresses over the displaced grid and comparison with the original cube.. a — \sigma_x, b — \sigma_z Figure 3: Interactive results from CAEplex (rotate and warp it with the slider). 5.1 Check The \sigma_z(x,y,z) distribution is expected to be identically equal to minus one and all the other five stresses are expected to vanish. To verify this is the case, the following input file is built which includes the basic solution, finds the maximum and minimum values of all the six stresses and prints them out in the standard output. As handy as the radial keyword is to simplify the solution of symmetric problems, there are two catches a cognizant user needs to be aware of. On the one hand, this kind of boundary condition is not a traditional essential BC where the value of the unknown (in this case displacements) is set up to the solver’s tolerance, but a multi-freedom constrain which is implemented with a penalty method. In the general case, the tangential condition is also implemented as a multi-freedom constrain. Yet Fino is smart enough and detects that the outward normal is (0,0,-1) and then condition is translated into a traditional Dirichlet-type condition w=0. Hence the radial condition is approximately set, and the extent to which this setting applies depends on a number of things—including of course the penalty weight. This approximation propagates down to the results, as illustrated in the following check test, which reads both the solver’s tolerance (default 10^{-6}) and the penalty weight (default 10^8) from the command line. fino_reltol =$1
fino_penalty_weight = $2 INCLUDE cylinder.fin # compute min/max stresses MESH_FIND_MINMAX FUNCTION sigmax MIN sigmax_min MAX sigmax_max MESH_FIND_MINMAX FUNCTION sigmay MIN sigmay_min MAX sigmay_max MESH_FIND_MINMAX FUNCTION sigmaz MIN sigmaz_min MAX sigmaz_max MESH_FIND_MINMAX FUNCTION tauxy MIN tauxy_min MAX tauxy_max MESH_FIND_MINMAX FUNCTION tauyz MIN tauyz_min MAX tauyz_max MESH_FIND_MINMAX FUNCTION tauzx MIN tauzx_min MAX tauzx_max PRINT "stress\tmin\t\tmax" PRINT SEP "\t" "% +5e" "sigmax" sigmax_min sigmax_max PRINT SEP "\t" "% .5e" "sigmay" sigmay_min sigmay_max PRINT SEP "\t" "% .5e" "sigmaz" sigmaz_min sigmaz_max PRINT SEP "\t" "% .5e" "tauxy" tauxy_min tauxy_max PRINT SEP "\t" "% .5e" "tauyz" tauyz_min tauyz_max PRINT SEP "\t" "% .5e" "tauzx" tauzx_min tauzx_max  $ fino check.fin 1e-3 1e4
stress  min             max
sigmax  -1.861889e-03   +2.982465e-03
sigmay  -2.07812e-03     3.01866e-03
sigmaz  -1.00348e+00    -9.96333e-01
tauxy   -1.34599e-03     1.16549e-03
tauyz   -1.48900e-03     1.04906e-03
tauzx   -1.34727e-03     1.66753e-03
$fino check.fin 1e-3 1e12 stress min max sigmax -3.368149e-03 +1.916099e-03 sigmay -2.82702e-03 3.11235e-03 sigmaz -1.00313e+00 -9.96947e-01 tauxy -8.14560e-04 7.82537e-04 tauyz -1.44005e-03 1.47797e-03 tauzx -1.70246e-03 1.29679e-03$ fino check.fin 1e-9 1e4
stress  min             max
sigmax  -6.090733e-04   +4.732937e-04
sigmay  -7.18812e-04     5.77535e-04
sigmaz  -1.00135e+00    -9.97135e-01
tauxy   -6.64507e-04     6.27475e-04
tauyz   -1.08775e-03     1.17853e-03
tauzx   -1.13649e-03     1.02294e-03
$fino check.fin 1e-9 1e12 stress min max sigmax -3.826204e-03 +1.897524e-03 sigmay -2.78529e-03 2.49913e-03 sigmaz -1.00174e+00 -9.97135e-01 tauxy -7.08823e-04 1.06614e-03 tauyz -1.08773e-03 1.17850e-03 tauzx -1.65213e-03 1.02282e-03$

On the other hand, the imposition of the radial symmetry does depend on the mesh size (take a minute and think about it!). In effect, the input file check-fine.fin solves the same proble as check.fin with a finer mesh. For the default values of the tolerance and the weight, we get

$fino check.fin 1e-6 1e8 stress min max sigmax -6.090465e-04 +4.738302e-04 sigmay -7.17874e-04 5.77485e-04 sigmaz -1.00135e+00 -9.97133e-01 tauxy -6.65185e-04 6.28188e-04 tauyz -1.08804e-03 1.17926e-03 tauzx -1.13719e-03 1.02356e-03$ fino check-fine.fin 1e-6 1e8
stress  min             max
sigmax  -1.360267e-04   +1.208525e-04
sigmay  -1.17141e-04     1.03318e-04
sigmaz  -1.00021e+00    -9.99558e-01
tauxy   -1.01711e-04     1.00290e-04
tauyz   -2.23632e-04     2.59015e-04
tauzx   -2.37351e-04     2.68281e-04
\$

The conclusion is that this kind of “handy” constrain limits the precision of the stresses both through the weight of the penalty method and through the mesh size, although it is still overall good (\sim \pm 1\%) for a rather coarse mesh. Recall that in Cube under pure tension the results only depended on the precision of the linear solver. Once again, if we passed the --mumps option, the results would not depend on the first argument but they would still depend on the second one.


text:
# Determining domain of differential operators in a concrete semigroup problem

Let s>3/2, $$H^s=H^s(\mathbb{R})$$ be the Sobolev space of order $$s$$, $$B$$ be the set of bounded operators from $$H^{1/2}$$ to itself, $$u\in H^s$$ and $$A(u):=u\partial_x$$ an operator. How can I determine the domain $$D(A)$$ such that:

1.) $$U_s:=e^{-sA(u)}$$ is an operator from $$H^{1/2}$$ to $$H^{1/2}$$;

2.) $$\|U_s\|_B\leq e^{\beta s}$$, for a certain constant $$\beta$$? Here the norm is the usual norm on the space $$B$$.

As far as I discovered, this would be equivalent to prove that $$_{H^{1/2}}\geq -\beta\|\phi\|_{H^{1/2}}^2$$, for any $$D(A)$$, but I was unsuccesful in understanding why this is an equivalence.


text:
### Home > AC > Chapter 9 > Lesson 9.3.1 > Problem9-62

9-62.

Match each graph below with the correct inequality.

1.  $y>−x+2$

Which graph represents the line y = −x + 2?

Graph 3

1.  $y<2x−3$

Follow the steps in part (a).

Graph 1

1.  $y\ge\frac{1}{2}x$

Follow the steps in part (a).

1.  $y\le−\frac{2}{3}x+2$

Follow the steps in part (a).


text:
# zbMATH — the first resource for mathematics

On arrangements of plane real quintics with respect to a pair of lines. (English. Russian original) Zbl 1206.14057
St. Petersbg. Math. J. 21, No. 2, 231-244 (2010); translation from Algebra Anal. 21, No. 2, 92-112 (2009).
A real projective curve of degree $$n$$ is a real homogeneous polynomial $$C_n\in\mathbb R[X,Y,Z]$$ defined up to a nonzero constant divisor. Its set of real zeroes in the real projective plane $$\mathbb R{ \mathbb{P}}^2$$ is denoted by $$\mathbb R C_n$$.
Let $$C_1$$ and $$C'_1$$ be two distinct lines. Hence, the set $$\mathbb R{ \mathbb{P}}^2\setminus(\mathbb R C_1\cup\mathbb R C'_1))$$ consists of two connected components, each homeomorphic to an open disk, and whose closures are denoted by $$D_1$$ and $$D_2$$. In this paper the authors complete the topological classification of the quadruples $(\mathbb R{ \mathbb{P}}^2,\mathbb R C_5\cup\mathbb R C_1 \cup \mathbb R C'_1, \mathbb R C_5\cup\mathbb R C_1,\mathbb R C_1)$ under the assumptions of maximality and general position. This means the following:
(1) The quintic curve $$C_5$$ is an $$M$$-curve, that is, $$\mathbb R C_5$$ consists of an odd branch $$J$$ and $$6$$ ovals that do not surround one another.
(2) $$\mathbb R C_5,\mathbb R C_1$$ and $$\mathbb R C'_1$$ are in general position, that is, they have pairwise transversal intersection and $$\mathbb R C_5$$ does not pass through $$\mathbb R C_1 \cap \mathbb R C'_1$$.
(3) The intersections $$\mathbb R C_1\cap J$$ and $$\mathbb R C'_1\cap J$$ have $$5$$ distinct points.
This article completes the topological classification of quadruples in (i). Let us explain this more in detail. Let $$a$$ and $$b$$ be the number of arcs of $$J$$ lying on $$D_1$$ and $$D_2$$, respectively, and having one end point lying on $$\mathbb R C_1$$ and the other in $$\mathbb R C'_1$$. The classification problem was solved by the authors in case $$a=b=1$$ in their previous work [Commun. Contemp. Math. 5, No. 1, 1–24 (2003; Zbl 1077.14568)]. Moreover, the solution if $$b=5$$ and either $$a=1$$ or $$a=3$$ follows at once from the results in [G. M. Polotovskii, Sov. Math., Dokl. 18(1977), 1241–1245 (1978); translation from Dokl. Akad. Nauk SSSR 236, 548–551 (1977; Zbl 0392.14014)]. Moreover, the classification was obtained by S. Yu. Orevkov [St. Petersbg. Math. J. 19, No. 4, 625–674 (2008); translation from Algebra Anal. 19, No. 4, 174–242 (2007; Zbl 1206.14082)]. Thus the only case that remained unsolved was $$(a,b)=(1,3)$$, and this is the goal of the paper under review.
The answer to the classification problem is rather complicated to be reproduced here. I can say that there exist $$51$$ types of topological models for (i) above and that the proof is based on a similar approach to one used by the authors in its joint paper quoted above. In particular, the constructions needed to realize the existence of some configurations use Viro’s patchworking method.
The authors recognize explicitly in the paper under review the contribution of Orevkov to the final form of the classification.
##### MSC:
 14H50 Plane and space curves 14H10 Families, moduli of curves (algebraic)
Full Text:
##### References:
 [1] Anatoly B. Korchagin and Grigory M. Polotovskii, On arrangements of a plane real quintic curve with respect to a pair of lines, Commun. Contemp. Math. 5 (2003), no. 1, 1 – 24. · Zbl 1077.14568 [2] G. M. Polotovskiĭ, A catalogue of \?-splitting curves of order six, Dokl. Akad. Nauk SSSR 236 (1977), no. 3, 548 – 551 (Russian). [3] S. Yu. Orevkov, Positions of an \?-quintic with respect to a conic that maximally intersect the odd branch of the quintic, Algebra i Analiz 19 (2007), no. 4, 174 – 242 (Russian); English transl., St. Petersburg Math. J. 19 (2008), no. 4, 625 – 674. [4] S. Yu. Orevkov, Link theory and oval arrangements of real algebraic curves, Topology 38 (1999), no. 4, 779 – 810. · Zbl 0923.14032 [5] O. Ya. Viro, Gluing of algebraic hypersurfaces, elimination of singularities, and constructions of curves, Proceedings of the International Topological Conference (Leningrad, 1982), Nauka, Leningrad, 1983, pp. 149-197. (Russian) [6] O. Ya. Viro, Gluing of plane real algebraic curves and constructions of curves of degrees 6 and 7, Topology (Leningrad, 1982) Lecture Notes in Math., vol. 1060, Springer, Berlin, 1984, pp. 187 – 200. · Zbl 0576.14031 [7] -, Real algebraic manifolds with prescribed topological properties, Doctor’s Diss., Leningrad. Gos. Univ., Leningrad, 1983. (Russian) [8] A. B. Korchagin, Smoothing of 6-fold singular points and constructions of 9th degree \?-curves, Topology of real algebraic varieties and related topics, Amer. Math. Soc. Transl. Ser. 2, vol. 173, Amer. Math. Soc., Providence, RI, 1996, pp. 141 – 155. · Zbl 0858.14029 [9] A. B. Korchagin and D. E. Smith, Patchworking singularities \?_{\?} and \?_{\?} and meanders of their smoothing, Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 299 (2003), no. Geom. i Topol. 8, 193 – 217, 330 (English, with English and Russian summaries); English transl., J. Math. Sci. (N.Y.) 131 (2005), no. 1, 5366 – 5380. · Zbl 1140.14321
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
2014
03-16

The Beijing-Hangzhou Grand Canal

The Beijing-Hangzhou Grand Canal is the longest artificial canal, in which many transportation companies run their boats. Each transportation company is responsible for transporting on one leg of the river. For the regulations of the GBs, each company is only allowed to have the same type boats.

The sands of the river keep changing the depth of the water. Boats will get stranded if the water depth is less than the waterline, so the GM often digs the riverbed to maintain watercourse to transport smoothly.

Mr April wants to send a gift to Ms April for her birthday via the canal. But he do not know whether the watercourse is smooth, he turns to intelligent you for help to tell him whether he can deliver the gift or not.

In the first line ,there are 3 non-negative interger , S, H and X, S stands for the length of the river, H is the original depth of the water, and there are X transportation companies.

Then follow X lines, in the (1+i)th line ,there are 3 numbers, a , b, and c, indicate Company i is responsible for the transportation between a and b, and c is the waterline of the boats from the company.

In (X+2)th line there is an interger Q , indicating there would be Q command lines. There are 4 variations of the command.

(1)F a b h meaning for sand silting, from a to b of the river the water depth has reduce to h meters,but those places where the depth is less than h meters remain the same.

(2)D a b h indicating the GM is trying removing the sand silting and the water depth reaches h meters ,but those places where the depth is more than h meters keeps the same.

(3)O a b implying Mr April wants to know whether there is a possible ways to deliver the gift from a to b. In other words, he wants to find a way to cover the segment [a, b] or [b,a] by one or some company’s available boats.

(4)C a h saying Company a has changed all its boats to new ones which has the waterline as h.

0<S<100000, 0<=x<=100,0<=H <=1000, 0<=Q<=1000

In the first line ,there are 3 non-negative interger , S, H and X, S stands for the length of the river, H is the original depth of the water, and there are X transportation companies.

Then follow X lines, in the (1+i)th line ,there are 3 numbers, a , b, and c, indicate Company i is responsible for the transportation between a and b, and c is the waterline of the boats from the company.

In (X+2)th line there is an interger Q , indicating there would be Q command lines. There are 4 variations of the command.

(1)F a b h meaning for sand silting, from a to b of the river the water depth has reduce to h meters,but those places where the depth is less than h meters remain the same.

(2)D a b h indicating the GM is trying removing the sand silting and the water depth reaches h meters ,but those places where the depth is more than h meters keeps the same.

(3)O a b implying Mr April wants to know whether there is a possible ways to deliver the gift from a to b. In other words, he wants to find a way to cover the segment [a, b] or [b,a] by one or some company’s available boats.

(4)C a h saying Company a has changed all its boats to new ones which has the waterline as h.

0<S<100000, 0<=x<=100,0<=H <=1000, 0<=Q<=1000

10 2 2
1 6 4
5 8 2
8
O 5 7
O 3 7
D 3 5 4
O 3 7
F 6 7 1
O 3 7
C 2 1
O 3 7
4 3 2
1 2 1
3 4 2
1
O 2 3

Yes
No
Yes
No
Yes
Yes

1. 中、国、少、年四个字代表的都是一位数，9的乘法口诀中只有“一九得九“这一句的乘积是一位数，所以，很明显，乘积千位上的中=1；九九八十一，所以，乘数个位上的年=9；乘积千位上的9是最大的一位数，即使再加一也会变成两位数，所以百位上一定

2. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

3. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧


text:
13.12.2017

# Neutrino-nucleus scattering at supernova energies

Accurate estimates of nuclear responses to supernova neutrinos are needed for studies of neutrino properties and modeling of supernovae and nucleosynthesis of heavy elements. The related scattering events can be classified to neutral-current (NC) [1] and charged-current (CC) processes [2]. A striking result of [2] is that, unlike for the neutrino scattering, for the antineutrino scattering the neutrino-averaged cross section depends strongly on the hierarchy (normal or inverted) of the neutrinos. This feature of the antineutrino scattering is presented in Figure 2. Recently we have also performed a study of the charged-current scattering of neutrinos and antineutrinos from the stable even-even lead isotopes by using three different Skyrme interactions [3]. This is important for the on-going HaLO (Helium and Lead Observatory) experiment at SNOLAB in Sudbury, Canada.

[1] W. Almosly, E. Ydrefors and J. Suhonen, J. Phys. G: Nucl. Part. Phys. 42 (2015) 025106
[2] W. Almosly, E. Ydrefors and J. Suhonen, J. Phys. G: Nucl. Part. Phys 42 (2015) 095106
[3] W. Almosly, B. G. Carlsson, J. Suhonen, J. Toivanen and E. Ydrefors, Phys. Rev. C 94 (2016) 044614


text: Then input your gender and measurements below to receive a body fat index based on average values. Enter your weight. = 94 lbs. To use our ideal body fat percentage calculator, follow the below steps: It will give you the body fat, body fat percentage, army body fat, navy body fat, lean body mass, and ideal body mass. ... My Workout Trainer, and on this website is meant to be an estimate. = Weight – Fat Mass = 130 lbs. Calculate Your Health with the Body Fact Index (BFI) Body fat percentage plays a vital role in overall health. The formula for body fat percentage is as follow: Lean body mass (men) $$=\left(\text{weight}\times 1.082\right)+94.42-\text{waist}\times 5.15$$, Lean body mass (women) $$=\left(\text{weight}\times 0.732\right)+8.987+\dfrac{\text{wrist}}{3.140}-\text{waist}\times 0.157-\text{hip}\times 0.249+\text{forearm}\times 0.434$$, Body fat weight $$= \text{weight} - \text{lean body mass}$$, Body fat percentage (BFP) $$=\dfrac{\text{body fat weight}}{\text{weight}}$$. His other measurements are: As we already have all the measurements, we only need to substitute values in the above equations, which is step number 6. For women, the formula begins with a simple weight calculation: (Total bodyweight x 0.732) + 8.987 you can contact us anytime. 4.7 out of 5 stars 4,097. High body fat puts you at a higher risk of lifestyle diseases. software for managing & marketing your events. Body Fat Percentage Formula; Body Fat (women) = 495 / (1.29579 - .35004 * log 10 (Waist + Hip - Neck) + 0.22100 * log 10 (Height)) - 450; Body Fat (men) = 495 / (1.29579 - .35004 * log 10 (Waist - Neck) + 0.22100 * log 10 (Height)) - 450; Fat Mass Formula; Fat Mass = Body Fat % * Weight; Lean Mass Formula; Lean Mass = Weight - Fat Mass; Body Fat Category Below is the body fat percentage table for men and women. This measurement is only for women. Fat Caliper Help. Female Body Fat Percentage: 10% – 14%. For this example, the woman's body fat percentage is 28% percent, which is within the recommended standards. Step 1: First, calculate the lean body mass. Once you calculated your body fat percentage, you should compare it with the recommended values. Katch-McArdle This formula, sometimes called the Cunningham formula, is more accurate. Fitmatic estimates your body fat percentage by leveraging several mechanisms. Now you know your body fat percentage, but what does it mean? People with a fat percentage above a certain limit are not selected. * This measurement is necessary for females only. Sitemap www.health.harvard.edu/staying-healthy/abdominal-fat-and-what-to-do-about-it. Even after you are a part of the navy, the same health standards have to be maintained. These everyday tricks will help your body burn more calories. Find Camps & Activities for your Active Kids, 10 Surprising Reasons You're Not Losing Weight, 15-Minute Fat-Blasting Workout for Busy People. In this post, we will explain how to measure body fat percentage, body fat percentage men, body fat percentage women, body fat percentage formula, and much more. Shop: Cookie Policy No time for exercise in your busy schedule? – 36 lbs. To get the best results, measure to the nearest 1/4 inch (0.5 cm). Look for this banner for recommended activities. Need some help? The best way to get new runners off the couch and across the finish line of their first 5K. © 2020 Active Network, LLC and/or its affiliates and licensors. The Waist-to-Height ratio to body fat percentage formula comes from Relative fat mass (RFM) as a new estimator of whole-body fat percentage ─ A cross-sectional study in American adult individuals Orison O. Woolcott & Richard N. Bergman (2018) . ; select the Age. HealthStatus 0 3 0. Body Fat estimates with up to 6 models. This calculator is used to calculate the body fat percentage based on USA's Marine Corps formula. ; Enter the Neck circumference either in inches or centimeters. We’ll talk about the healthy range for men and women, and the limitations. It is actually based on correlations between several body circumference measures relative to a person's height. For men 40 to 59 years of age - 11 to 22 percent. This tool provides: body fat percentage for women, body fat percentage for men, and recommended ideal body fat precentage based on your activity levels. Footwear | Fitness Apparel | Outdoor Gear. Body Fat Caliper Formula. Though there are several ways to calculate body fat, the easiest method is to use an online body fat calculator like the one above. The average adult male body fat percentage should be between 18 and 25%. The truth is, you may have lost fat and gained muscle, but you just can’t know it. ACTIVE also makes it easy to learn and prepare for all the things you love to do with expert resources, training plans and fitness calculators. Join Active Substitute the values in lean mass formula to get the lean body mass. insights, ACTIVE Works® is the race management The Body Fat Percentage Calculator can help you find out how much fat you have, how it affects your exercise and other information. When you input body measurements into a body fat calculator like the one above, the tool uses a specific formula to estimate your percentage of fat. Sign In. You can use lbs., tone, gram, kilogram, pound, and ounce. What is my body fat percentage? Then input your gender and measurements below to receive a body fat index based on average values. Sign In, Join Active Body fat percentage (BFP) formula for adult males: BFP = 1.20 × BMI + 0.23 × Age - 16.2 Body fat percentage (BFP) formula for adult females: BFP = 1.20 × BMI + 0.23 × Age - 5.4 Body fat percentage (BFP) formula for boys: BFP = 1.51 × BMI - 0.70 × Age - 2.2 Body fat percentage (BFP) formula for girls: BFP = 1.51 × … 3 However, that is just the minimum amount. Another consideration for body fat deals with genetics. Support & Feedback Body Fat Percentage and DXA to Caliper to Conversion Methodology. Body Fat Index Calculator. It offers multiple units to measure. Enter your height in the given input box and select the unit of measurement. It can single handedly calculate your TDEE, Deficit Calories and Macronutrient Requirements without breaking a sweat. The formula to calculate your weight loss percentage is: lbs lost divided by starting weight. Here is how to take the measurements, using a soft measurement tape: Height: step on a flat surface that is perpendicular to a … BMI is a key indicator of your good health. FITINDEX Bluetooth Body Fat Scale, Smart Wireless BMI Bathroom Weight Scale Body Composition Monitor Health Analyzer with Smartphone App for Body Weight, Fat, Water, BMI, BMR, Muscle Mass - Black. Use a tape measure to determine your waist, wrist, hip and forearm circumference. Do Not Sell My Personal Information Looking great takes hard work… however, we do have access to the most useful tools nowadays, which makes everything so much simpler and stress-free (Very little thinking … Also listed are the typical body fat amounts. Today, we’ll take a look at one particular breed: Body fat percentage calculator using height and weight. The following formula is used to calculate body fat percentage from height weight and caliper measurements. Body Fat Calculator for Men and Women Six Methods include Navy Tape, Jackson/Pollock 3,4,7, Durnin and Parrillo. According to various sources, the minimum limit of body fat is around 2-5% for men 2 and 10-13% for women. Body fat percentage formula. From marketing exposure to actionable data Our body fat calculator needs four of your measurements: height, neck, waist, and hips, to determine the percentage and mass of body fat. There are lots of virtual options to keep you active and engaged — just select “Virtual Activities” as your location, and you’re ready to go. It determines the total percentage of fat in the body by taking various measurements of the body. Even here you have two methods as Estimate method and Skinfold method.. Estimate method gives you a quick estimate of your body fat percentage using your gender, weight, and the circumference of your waist in case of men. This is the range you will usually see bodybuilders in, and is not considered healthy to keep consistent. Body Fat Calculator. The essential body fat for a woman is 8%, while for a man is only 2%. Measure your forearm and note down. Updated: Our new Body Fat Calculator 2.0 can do many useful calculations including Body Fat Percentage, LBM (Lean Body Mass), FBM (Fat Body Mass) and many other body fat classification parameters fast and easily. Some methods include the Bioelectric Impedance (BIA), Hydrostatic Weighing, DEXA scanning, Skin Caliper Measurements, etc., but none of these measures is as simple as the body fat calculator. The list below comes from the American Council on Exercise and shows the average percentages in specified groups. Privacy Policy Using Navy Body Fat Percentage calculator. The body fat percentage calculator helps to calculate your body fat percentage, which is the measure that shows you how much adipose tissue you have in correlation with your bone and muscle mass. The reading is given in millimeters, which you compare to a chart with age and gender to arrive at your body fat percentage. For men the formula is : 86.01 x log 10 (waist - neck) - 70.041 x log 10 (height) + 36.76 Contrary to what some people think, body fat percentage is not calculated based on any kind of BMI calculation. This body fat percentage calculator is based on the body fat algorithms developed by the U.S. Navy, U.S. Army, U.S. Marine Corp and the YMCA. Determine your body fat percentage with our body fat calculator. You can calculate anything on Calculators.tech. All rights reserved. How much body fat do I have? Body fat percentage calculator tells you more about the body composition (muscle and fat). The formula is slightly different for men and women. FEMALES: D = (1.09949 – (0.00099 x (Triceps + Thigh + Suprailiac)) + (0.0000023 x (Triceps + Thigh + Suprailiac) 2) – (0.0001392 x Age)) MALES: Consume more calories than your body needs, you will gain weight. Given BMI, the following formulas can be used to estimate a person's body fat percentage. or Terms of Use Get a better understanding of how your body burns fat and what exercises will help you reach your goal. For men 60 to 79 years of age - 13 to 24 percent. For women, their percentage of body fat in relation to age should be between 25 and 31%. For athletics performance, 7-10% is preferred for men and 14-20% for women. Amazon's Choice for body fat percentage calculator. Then, multiply the result by 100. Careers How much fat are you carrying? Body Fat Percentage Calculator: Find some of the best techniques to calculate and track your body fat on a regular basis. Select your gender in the given input box. Check out this list of common weight loss mistakes. It is a very efficient and robust calculator that produces results in no time. Calculate the body fat weight by subtracting the lean body mass from your weight. Using Marine Body Fat Percentage calculator. 4 2 − w a i s t × 5. This is the formula we use in the calculator above if you don’t know your body fat percentage. This workout will blast fat in just 15 minutes, and all you need is a pair of dumbbells. To monitor your body in an accurate and stringent manner, using a quality navy body fat calculator is the best way out. The body fat index is not an indicator of fitness level since the calculation is made with no regard to height or weight. Lean Body Mass (men) $$=\left(\text{weight}\times 1.082\right)+94.42-\text{Waist}\times 4.15$$, $$= (40 \times 1.082) + 94.42 - 32 \times 4.15$$, Body Fat Weight $$= \text{weight} - \text{lean body mass}$$. Body Fat Percentage (BFP) $$= \dfrac{\text{body fat weight}}{\text{weight}}$$. Copyright Policy Essential fat: 10–13% (women), 2–5% (men) Athletes: 14–20% (women), 6–13% (men) or Of the 100 most populated U.S. cities, these 10 zip codes have the highest percentage of overweight residents. Determine your body fat percentage with our body fat calculator. The first fat percentage calculator provides an estimate of your fat percentage based on … Body fat percentage calculator is used to calculate body fat percentage in the human body. Measure your wrist and note down. Percentages that are too high or too low increase your risk of … With many of us still social distancing, we want to make sure you can find activities that suit your needs. Select the Gender. For specific information regarding your body fat percentage or health, please consult your physician. (See guidelines below for taking measurement.) The Online Body Fat Percentage Calculator can be used to estimate your total body fat percentage based on specific body measurements. Privacy Settings So we start from step 6. ACTIVE is the leader in online event registrations from 5k running races and marathons to softball leagues and local events. select the Age. Find out if you have a healthy body fat percentage with this calculator. ; Enter the Height either in feet/inches or centimeters. BMI Calculator - Measure your body fat% here. Please see your Privacy Rights for how your information is used. (See guidelines below for taking measurement.) If you are a woman, and your body fat percentage exceeds 31%, you are obese and thus prone to a variety of serious health conditions. Select the Gender. There are many approaches to measure body fat percentage or the amount of fat and lean mass on your body. and/or its affiliates and licensors. How to Estimate Body Fat? Women have more fat in breast tissue, waist, and thigh areas. Are you sure you want to delete this family member? All rights reserved. To calculate the body fat percentage, follow these steps carefully: A 21 years old boy Sam has a weight of 40 kg. Body Fat Calculator & Body Fat Percentage Calculator. Calculating body fat percentage isn’t always completely accurate, and there are many methods to try. Because of that, body fat percentage calculations have become incredibly popular in recent years. These 14 superfoods pack a ton of nutrition into a relatively small number of calories while filling you up at the same time. The current science states that between 7-17% body fat is optimal for general health for men and 14-25% for women. Use a tape measure to determine your waist, wrist, hip and forearm circumference. The result might not be accurate as the one you would get from an actual body composition test, but it can give you an estimate. 0 8 2) + 9 4. #1 – Body Fat Caliper: Skin Fold Method The “skin fold” method measures your body fat percentage by pinching your fat with your fingers then measuring the thickness with a body fat caliper. Cookie Settings. You exercise and eat right, but those unwanted pounds still aren’t coming off. Running Shoes|Fitness Apparel|Sports, Daily Deals: Lean body mass (women) Is my body fat percentage normal? This measurement is only for women. 1 5. Now substitute the body fat weight and your weight in body fat percentage formula to get the body fat percentage. Consume less calories than your body needs, you will lose weight.. Our Fat Loss calculator is by far the most Powerful Weight Loss tool one can get their hands on. There is also such thing as too little body fat, or otherwise known as malnourished. 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# NCERT Solutions for Class 9 Maths chapter 12 Exercise 12.1 - Heron’s Formula

Hey, are you a class 9 Student and Looking for Ways to Download NCERT Solutions for Class 9 Maths chapter 12 Exercise 12.1? If Yes then you are at the right place.

Here we have listed Class 9 maths chapter 12 exercise 12.1 solutions in PDF that is prepared by Kota’s top IITian’s Faculties by keeping Simplicity in mind.

If you want to score high in your class 9 Maths Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

### Download The PDF of NCERT Solutions for Class 9 Maths chapter 12 Exercise 12.1 "Heron’s Formula"

#### All Questions of Chapter 12 Exercise 12.1

Once you complete the chapter 12 then you can revise Ex. 12.1 by solving following questions

Q1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is $180 \mathrm{~cm}$, what will be the area of the signal board ?

Q2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 \mathrm{~m}, 22 \mathrm{~m}$ and $120 \mathrm{~m}$ (see Fig.). The advertisements yield as earning of Rs. 5000 per $\mathrm{m}^{2}$ per year. A company hired one of its walls for 3 months. How much rent did it pay ?

Q3. There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN" (see Fig.). If the sides of the wall are $15 \mathrm{~m}, 11$ $\mathrm{m}$ and $6 \mathrm{~m}$, find the area painted in colour.

Q4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Q5. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.

Q6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of triangle.


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An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wages earners 586 648 Mean of monthly wages Rs.5253 Rs.5253 Variance of the distribution of wages 100 121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Asked by Abhisek | 1 year ago |  112

##### Solution :-

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= $$\sqrt{100}$$

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=$$\sqrt{121}$$

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

Answered by Pragya Singh | 1 year ago

### Related Questions

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Solid angle subtended by a circle

I don't know if I understand this correctly. Let's say that we have a point source of particles and it is pointed at a circular detector. The radius of the detector is 5cm and the distance to the detector is 10cm. I therefore conclude that:

$$\Omega = S/r^2 = \pi r_{circle}^2/r^2 = \pi/4.$$

What is wrong with my reasoning?

• Let me present an example based on your reasoning: Suppose that the radius of the detector is some huge value like 500 km but the distance to the detector is still 10 cm. According to the approach you used above, $\Omega$ would be some huge number, whereas we know that the solid angle $\Omega$ can never be greater than $4\pi$. (And, in fact, for this example $\Omega$ should be very close to $2\pi$). What went wrong? – user93237 May 9 '17 at 2:18
• @SamuelWeir I guess that I need to define the surface area differently? – loltospoon May 9 '17 at 2:21
• Well, I also know that $\Omega = \int \int sin \phi d\theta d\phi$, but I have no clue how to use this – loltospoon May 9 '17 at 2:22
• What should the limits be in that integral? – Brian Moths May 9 '17 at 2:25
• @NowIGetToLearnWhatAHeadIs That's what I couldn't figure out. At first I thought it was $\theta = [0,\pi /2]$ and $\phi = [0,2\pi]$, but this just gave me $2\pi$, which I don't think is right – loltospoon May 9 '17 at 2:33

If you take an object and project it radially outward onto the surface of a sphere the area of that projection divided by the square of the sphere's radius is the solid angle. That's the general definition. In this case, because the disk is oriented so that its surface normal is parallel to the radial direction the easiest sphere to choose for measuring solid angle is one that contains the circle. The radius of that sphere will then be: $$R_{\mathrm{sphere}} = \sqrt{r_{\mathrm{circle}}^2 + d^2},$$ where $d$ is the distance to the circle.
The area from the sphere that the circle projects onto can be found by choosing polar coordinates, with the center of the polar angle ($\theta=0$) corresponding with the circle's axis. In those coordinates the area can be found with the double integral: \begin{align} A &= \int_0^{2\pi} \operatorname{d}\phi \int_0^{\tan^{-1}(r_{\mathrm{circle}} / d)} \operatorname{d}\theta\ R_{\mathrm{sphere}}^2 \sin \theta \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left[-\cos \theta \vphantom{\int}\right]_{\theta=0}^{\tan^{-1}(r_{\mathrm{circle}} / d)} \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left(1 - \cos\left[\tan^{-1}\left(\frac{r_{\mathrm{circle}}}{ d}\right)\right]\right) \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left(1 - \frac{d}{\sqrt{r_{\mathrm{circle}}^2 + d^2}}\right) \end{align} This gives a solid angle of: $$\Omega = 2\pi \cdot \left(1 - \frac{d}{\sqrt{r_{\mathrm{circle}}^2 + d^2}}\right).$$
Confirm for yourself that this obeys the correct limits ($d \rightarrow \infty \Rightarrow \Omega \rightarrow 0$ and $r_{\mathrm{circle}} \rightarrow \infty \Rightarrow \Omega \rightarrow 2\pi$). Also be careful of numerical issues when actually doing calculations with this formula because it suffers from a similar numerical instability to the quadratic formula when $d\gg r_{\mathrm{circle}}$.


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# Notes: Online Optimization

The online optimization problem is defined as follows:

• For each time $t = 1,\dots, T$
1. Learner makes a decision $d_t \in \mathcal{D}$
2. Observes the cost function $c_t$ and incurs cost $c_t(d_t)$
• The goal is to minimize the regret: the difference between the algorithm’s total cost and the cost of always using a single best decision $d^*$
• We now consider online linear optimization
• The decision space $\mathcal{D} \subseteq R^n$
• The cost function is a linear function: $c_t(d_t) = d_t \cdot s_t$, where $s_t\in S\subseteq R^n$
• Assume we have access to an oracle $M$, such that $M(s) = \arg min_{d\in \mathcal{D}} d\cdot s$
• $d_t = M(s_{1:t-1})$, where $s_{1:t-1} =\sum_{i=1}^{t-1} s_i$
• The algorithm fails even for a simple 2-dimension case
• Follow the Perturbed leader: $FPL(\epsilon)$
1. Pick $p_t$ uniformly from $[0, \frac{1}{\epsilon}]^n$
2. Choose $M(s_{1:t-1} + p_t)$
• Another variant: $FPL^*(\epsilon)$
1. Pick $p_t = x$ with density $d\mu(x) \propto e^{-\epsilon ||x||_1}$
2. Choose $M(s_{1:t-1} + p_t)$
• Assume
• $|d-d'| \le D$,  $\forall d,d' \in \mathcal{D}$.
• $|d \cdot s| \le R$,  $\forall d\in \mathcal{D}, s\in S$.
• $|s_t| \le A$,  $\forall t\in \{ 1,\dots, T \}$
• Theorem 1:
• Set $\epsilon = \sqrt{\frac{D}{ART}}$, then$FPL(\epsilon)$ has cost $\le OPT + \epsilon \cdot ART + \frac{D}{\epsilon}$
• Set $\epsilon = \min \{ \frac{1}{2A}, \sqrt{\frac{D(\ln n + 1)}{A\cdot OPT}} \}$ , then $FPL^*(\epsilon)$ has cost $\le (1+\epsilon)OPT + \frac{4AD(\ln n +1)}{\epsilon}$
• Lemma 1:
• Consider the “Be the leader” algorithm (impossible in practice, only for analysis): use $d_t = M(s_{1:t} )$ at time $t$
• $\sum_{t=1}^T M(s_{1:t}) \cdot s_t \le M(s_{1:T}) \cdot (s_{1:T})$
• We prove this by induction:
• $\sum_{t=1}^{T+1} M(s_{1:t}) \cdot s_t \le M(s_{1:T}) \cdot s_{1:T} + M(s_{1:T+1}) \cdot s_{T+1} \le M(s_{1:T+1}) \cdot s_{1:T+1}$
• Lemma 2: “Be the perturbed leader” has a small regret
• $\sum_{t=1}^{T} M(s_{1:t} + p_t) \cdot s_t \le M(s_{1:T}) \cdot s_{1:T} + D\sum_{t=1}^T |p_t - p_{t-1}|_\infty$.
• pf: Suppose the cost is $s'_t = s_t + p_t - p_{t-1}$
• $\sum_{t=1}^T M(s'_{1:t}) \cdot s'_t \le M(s'_{1:T}) \cdot s'_{1:T}$
• $\sum_{t=1}^T M(s_{1:t} + p_t) \cdot (s_t + p_t - p_{t-1}) \le M(s_{1:T} + p_T) \cdot (s_{1:T} + p_T) \le M(s_{1:T}) \cdot (s_{1:T} + \sum_{t=1}^T (p_t - p_{t-1}))$
• $\sum_{t=1}^T M(s_{1:t} + p_t) \cdot s_t \le M(s_{1:T}) \cdot s_{1:T} + \sum_{t=1}^T \left( M(s_{1:T}) - M(s_{1:t} + p_t) \right) (p_t - p_{t-1}) \le M(s_{1:T}) \cdot s_{1:T} + D\sum_{t=1}^T |p_t - p_{t-1}|_\infty$
• Lemma 3:
• The distributions over $s_{1:t-1} + p_t$ and $s_{1:t} + p_t$ are similar: they are both distributions over cubes. If they overlap on a fraction $f$ of their volume, then $E(M(s_{1:t-1} + p_t)\cdot s_t) \le E(M(s_{1:t}+p_t)) + (1-f) R$
• Lemma 4:
• For any $v\in R^n$, the cubes $[0, \frac{1}{\epsilon}]^n$ and $v + [0, \frac{1}{\epsilon}]^n$ overlap in at least a $(1-\epsilon |v|_1)$ fraction.
• Proof of Theorem 1:
• $E(cost(FPL(\epsilon))) =\sum_{t=1}^T E(M(s_{1:t-1}+p_t)\cdot s_t)$
• (using Lemma 3) $\le \sum_{t=1}^T E(M(s_{1:t}+p_t)\cdot s_t) + \epsilon ART$
• (using Lemma 2) $\le OPT + \epsilon ART + \frac{D}{\epsilon}$
• Online Convex Optimization
• We now consider the case where the cost function $c^t$ are convex
• At each time $t$, choose the decision $x^t\in K$, where $K$ is a convex body of diameter $D$.
• Let $\nabla c^t(x) = \left( \frac{dc^t(x)}{dx_1}, \dots, \frac{dc^t(x)}{dx_n} \right)$ be the gradient of the cost function.
• $x^{t+1} = Proj_K \left( x^t - \eta_t \nabla c^t(x^t) \right)$
• Facts:
1. For a convex function $f$, $f(y) - f(x) \ge \nabla c(x) \cdot (y-x)$
2. $\forall a\in K$, $||Proj_K (y) - a|| \le || y -a ||$
• Theorem:
• Let $x^* = \arg min_{x\in K}\sum_{t=1}^T c^t(x)$.  Assume $||\nabla c^t|| \le G$.
• If set $\eta_t = \frac{D}{G\sqrt{T}}$, then $\sum_{t=1}^T \left[ c^t(x^t) - c^t(x^*) \right] \le DG\sqrt{T}$.
• If set $\eta_t =\frac{1}{\sqrt{t}}$, then $\sum_{t=1}^T \left[ c^t(x^t) - c^t(x^*) \right] \le (\frac{D^2}{2} + G^2)\sqrt{T}$.
• Proof:
• Let $\phi_t = ||x^t - x^*||^2$
• $\phi_{t+1} -\phi_t = ||x^{t+1} - x^*||^2 - ||x^t - x^*||^2 = ||Proj_K \left( x^t - \eta_t \nabla c^t(x^t) \right) - x^*||^2 - ||x^t - x^*||^2 \le || x^t - \eta_t \nabla c^t(x^t) - x^*||^2 - ||x^t - x^*||^2 = \eta_t^2||\nabla c^t(x^t)||^2 - 2\eta_t \nabla c^t(x^t) (x^t - x^*)$
• $\nabla c^t(x^t) (x^t - x^*) \le \frac{\eta_t}{2} G^2 - \frac{1}{2\eta_t} (\phi_{t+1} - \phi_t)$
• By Fact 1, we have $c^t(x^t) - c^t(x^*) \le \nabla c^t(x^t) (x^t - x^*)$
• Set $\eta_t =\eta = \frac{D}{G\sqrt{T}}$, then $\sum_{t=1}^T \left[ c^t(x^t) - c^t(x^*) \right] \le \frac{\eta T}{2} G^2 + \frac{D^2}{2\eta} \le DG\sqrt{T}$
• Set $\eta_t = \frac{1}{\sqrt{t}}$, then $\sum_{t=1}^T \le \frac{G^2}{2}\sum_{t=1}^T \eta_t + \frac{\phi_1}{2 \eta_1} + \sum_{t=2}^T \phi_t ( \frac{1}{\eta_t} - \frac{1}{\eta_{t-1}}) \le \frac{G^2} {2}\sum_{t=1}^T \eta_t + \frac{D^2}{2\eta_T}$
• $\sum_{t=1}^T \frac{1}{\sqrt{t}} \le 1 + \int_{t=1}^T \frac{dt}{\sqrt{t}} \le 2\sqrt{T}$
• In HW4, we prove that FPL also works for online convex optimization, by using the gradients as the linear cost functions.


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# Multivariate Normal Distribution¶

The multivariate Normal distribution has the probability density function (PDF):

$f(x\mid \mu ,\Sigma) = (2\pi)^{-\frac{k}{2}} |\Sigma|^{-\frac{1}{2}} \exp \left(-\frac{1}{2}(x-\mu)^\top \Sigma^{-1}(x-\mu) \right)$

where $$\mu \in \mathbb{R}^k$$ is the mean vector and $$\Sigma \in \mathbb{R}^{k\times k}$$ is a positive definite covariance matrix.

## Usage¶

e["Distribution"][*index*]["Type"] = "Multivariate/Normal"

## Configuration¶

These are settings required by this module.

Mean Vector
• Usage: e[“Mean Vector”] = List of real number

• Description: Means of the variables.

Sigma
• Usage: e[“Sigma”] = List of real number

• Description: Cholesky Decomposition of the covariance matrix.

Name
• Usage: e[“Name”] = string

• Description: Defines the name of the distribution.

Random Seed
• Usage: e[“Random Seed”] = unsigned integer

• Description: Defines the random seed of the distribution.

Range
• Usage: e[“Range”] = gsl_rng

• Description: Stores the current state of the distribution in hexadecimal notation.

## Default Configuration¶

These following configuration will be assigned by default. Any settings defined by the user will override the given settings specified in these defaults.

{
"Mean Vector": [],
"Name": "",
"Random Seed": 0,
"Range": "",
"Sigma": []
}


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# Tag Info

10

Short of having a 'reasonable' predictive model for expected returns and the covariance matrix, there are a couple lines of attack. Shrinkage estimators (via Bayesian inference or Stein-class of estimators) Robust portfolio optimization Michaud's Resampled Efficient Frontier Imposing norm constraints on portfolio weights Naively, shrinkage methods ...

8

Perhaps you may want to consider article by D. Levine - Modeling Tail Behavior with Extreme Value Theory who gives practicale example on how EVT can be used to calculate probabilities on returns in tails with use of the Pickands-Balkema-de Haan Theorem and generalized Pareto distribution. It also contains some criterias and points on other methods that can ...

8

One approach is Conditional Value at Risk (CVaR) a.k.a. Expected Shortfall (ES). It does, as you suggest, take into account the whole set of returns. However, instead of traditional VaR which asks "what is the worst 1% or 5% loss I can expect" in a given time frame, conditional VaR asks "assuming I sustain losses of at least 95% or 99% (and perhaps am ...

8

Both answers from Shane and Vishal Belsare make sense and detail different models. In my experience, I have never been satisfied by a unique model since the majority of papers out there can be split in two categories: Those that predict the mean component of the problem. Those that predict the variance component of the problem. The ideal (to read ...

6

You raise a very important point, which unfortunately doesn't have a simple answer. Black-Litterman addresses the allocation problem by allowing you to provide a prior within a bayesian framework. It doesn't really tell you how to produce the prior itself. But more importantly, it doesn't address the fundamental problem: it's difficult to accurately ...

6

Here's a partial answer: This partly depends on the return characteristics. One way to look at this is to analyze the skewness and kurtosis of the returns. Most strategies have a negative skewness, which roughly means that they have mostly consistent small positive returns, with the occasional large negative return. Alternatively, some strategies have ...

5

If you don’t have any specific model which describes the behavior of the asset being traded, you can estimate the empirical distribution of returns by backtesting your momentum strategy. Then you can adjust this estimate during your strategy’s lifetime from your trading results. Additionally you can enhance this by accounting for different market regimes ...

5

You are trying to apply the Kelly Criterion, supposedly to maximize how aggressively to bet, and you are having trouble when the Kelly Value turns negative. The naive answer to your question is that when your kelly value turns negative, then $f=\frac{bp-q}{b}$ turning negative means the instantaneous expected return is negative, which means you should not ...

4

If the equation satisfies all the assumptions of OLS, particularly homoscedasticity and no autocorrelation in the errors, then the expected return for the equation you laid out is $E[r_{future}|r_{history},x_{news}]=\alpha+\beta_1r_{history}+\beta_2x_{news}+\beta_3r_{history}*x_{news}$ If the unconditional expected return is zero (as is likely to be ...

3

What a great question -- it touches on many issues at the core of quantitative finance. This answer might be a lot more than you bargained for, but it's too interesting to pass up. References Mostly, this subject falls somewhere at the intersection of these three highly-interrelated topics: risk-neutral valuation, rational pricing and the fundamental ...

3

Another possible approach is taking views a la Black-Litterman. There is a 2006 paper "Incorporating Trading Strategies in the Black-Litterman Framework" that discusses the methodology in more detail. There are several practical issues that one should consider when implementing a momentum strategy with optimization. I would pay careful attention to the ...

3

Tail risk represents the probability that the magnitude of returns on an asset/portfolio will exceed some threshold (usually three standard deviations) on the normal curve. If you visualize a normal curve on standard axes, the tail on the left side corresponds to an extreme low return and the tail on the right side corresponds to an extreme high return. In ...

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Returns-based analysis cannot calculate the expected return of a trading system. It yields nonsensical results and is not suited to this particular calculation. Consider a game where every time you play, you win 25% twice and lose 40% once. There are basically three permutations of this game. Represented in R vectors: first <- c(.25, .25, -.4) second ...

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You can't add returns. You must multiply them. In your example above where daily returns are 25%, 25%, and -40% To compute expected return from a return series, simply use this formula: return = product( 1+return); in the case of you example this yields: return = (1.25 * 1.25 * .6) = .9375 To get the expected daily return use the geometric mean: ...

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It seems that your real question is: is the PFP (Price Formation Process) diffusive from intraday to weekly sampling rate? It is a very good question since on intraday, some academics found some multifractal features into intraday returns, meaning that the PFP is not a Geometric Brownian Motion at small scales (even considering stochastic volatility). You ...

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This may or may not be helpful, since I don't have anything to point you to that specifically addresses the high skewness of the distribution you mention. However, this sounds like it is probably an idiosyncratic risk, and that certainly has bearing on whether or not it would be priced. In the standard capital asset pricing model, the marginal investor ...

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Isn't this a simple mathematical rule? $$\Delta r_{t}=r_{t} - r_{t-1} = ln(p_{t}) - ln(p_{0}) - ln(p_{t-1}) + ln(p_{0})=ln(\frac{p_{t}}{p_{t-1}})$$ i.e. logarithmic or continuously compounded return. As a result: E(\Delta r_{t})=\frac{1}{T}\sum_{t=1}^{T}\Delta r_{t} = ...

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text:
# Aiming a Laser Project

So we hot glued two servos together and the to a mirror.

We’re aiming a laser at it. A blindingly powerful laser

Essentially Servo B controls the angle in the x-y plane of the mirror and servo A controls the z angle to that plane.

The mirror law is $v_2=v_1 - 2(v_1 \cdot n)n$

Where all are unit vectors. $v_1$ is the incoming, $n$ is the mirror normal and $v_2$ is the outgoing unit vector.

We fixed the laser to the base so

$v_1 = \hat{y}$

The outgoing ray must hit the ceiling (at a height R) at position x,y.

$v_2 = (x,y,R) \frac {1} {\sqrt{x^2+y^2+R^2}}$

Here’s a nice observation:

$v_2 - \hat{y} \propto n$

So we have an algorithm for finding n right there. Find v2, subtract off 1 and then normalize the resulting vector to get $latex \hat{n}$.

Then finally we can write n in terms of the angles $\alpha,\beta$, which heavily depend on our conventions of where angle 0 is and whether the servo spin clockwise or counterclockwise.

I believe ours came out to be something like:

$n = (\cos(\alpha)\sin(\beta), -\cos(\alpha)\cos(\beta) ,\sin(\alpha))$

In all honesty, we coded her up, then fiddled with minus signs until it was working. Not necessarily a bad way of going about things. Find the things to think about and find the things to just try.

Then here is the code: