text:
# Homework Help: Some angular momentum problems

1. Feb 24, 2009

### nns91

1. The problem statement, all variables and given/known data

1. There is m1 on an incline connected to m2 by a string through the pulley at the top of the incline. The incline is frictionless and the string pases through the center of mass of each block. The pulley has a moment of inertia I and a radius r. (a) Find the net torque acting on the system. (b) Write an expression for the total angular momentum of the system about the center of the pulley when the masses are moving with a speed v. (c) Find the acceleration of the masses from your results for parts (a) and (b) by setting the net torque equal to the rate of change of the angular momentum of the system

2. Relevant equations

Torque=dL/dt
L=I*w
T=F*r

3. The attempt at a solution

1. So I used Newton's 2nd law to get the acceleration as:
(m2*g*sin$$\theta$$- m1*g)/ (m1+m2+(1/2)M)

However, M which is mass of the pulley is not given. I can make M=2I/r^2

I still don't get the right answer for net torque. It should be rg(m2*sin$$\theta$$-m1)

Anyone know if I am on the right track ?? Any help will be appreciated

2. Feb 24, 2009

### Delphi51

I'm picturing m1 on a ramp at angle theta with the horizontal so that the force on it along the ramp is m1*g*sin(theta). Also, m2 hanging straight down from the pulley, with force m2*g.
But this would give a torque of (m2*g - m1*g*sin(theta))*r
or rg(m2 - m1*sin(theta)).

I must be mistaken about where the masses are?

3. Feb 24, 2009

### nns91

m2 should be on the ramp and m1 should be hanging.

How about other parts. I think I get part a now.

4. Feb 24, 2009

### Delphi51

L = Iw for the turning pulley.
plus mvr for the moving masses.

5. Feb 24, 2009

### nns91

For part a, just to make sure, why are tensions not taken into account ?

6. Feb 24, 2009

### Delphi51

The tension in the string just transfers the force from one object to another.
If you think of the whole thing (strings and masses) being one object you don't need to bother with tension. Usually.

If we ignore the moment of inertia of the pulley in this problem, we could think of it as one object and write
m1*g - m2*g*sin(A) = (m1 + m2)*a

or we could write T - m2*g*sin(A) = m2*a for the mass on the ramp and
m1*g - T = ma for the hanging mass.
If you add these two equations together, the T's annihilate and you get the first equation 4 lines up.

Of course none of this helps in your problem where you have moment of inertia to deal with.
EDIT: actually, calculating "a" this way is a big help. If we temporarily let I = 0 in the acceleration we get using the torque - angular momentum approach, it simplify to
a = (m1*g = m2*g*sin(A))/(m1 + m2) from the equation above.

7. Feb 24, 2009

### nns91

Thanks. I could solve it now

Another problem:

A spaceship rotates about its longitudinal axis at 6 rev/min. The occupants wish to stop this rotation. They have small jets mounted tangentially, at a distance R=3m from the axis, as indicated, and can eject 10g/s of gas from each jet with a nozzle velocity of 800 m/s. For how long must they turn on these jets to stop the rotation ? To rotational inertia of the ship around its axis (assumed to be constant) is 4000 kg*m^2

Any hint on how I should approach this one ?

8. Feb 25, 2009

### Delphi51

the force exerted by one of those "jets" is F = d/dt(mv) = v*dm/dt
where dm is the mass ejected in time dt and v is the exhaust velocity.
You should be able to work out the torque from that and the radius.
Do you have a formula relating torque and inertia (analogous to F = ma)?

9. Feb 25, 2009

### nns91

how do I calculate alpha though ?

Should it be m*dv/dt instead of v*dm/dt ??

10. Feb 25, 2009

### Delphi51

Actually it is d/dt (mv) = dm/dt*v + m*dv/dt
Since the velocity of he exhaust is constant, the second term is zero so
F = v*dm/dt

To calculate alpha (angular acceleration), you need some formula analogous to F = ma relating the Torque to the angular acceleration. Do you have any formulas that say "Torque = ..." ?

11. Feb 25, 2009

### nns91

I know Torque= I*alpha.

I have I but Toruqe and alpha are unknown so how can I calculate ?

12. Feb 25, 2009

### Delphi51

You can figure out the torque caused by those jets since you know the force each one exerts.


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# Ubuntu – Eclipse has two icons on unity launcher when I only want one

eclipseunity

I recently upgraded to Eclipse Luna and since then I've had trouble launching it from the Unity Launcher. I've modified my old .desktop file located in ~/.local/share/applications, then added it to my launcher and now it looks like this:

How can I fix this?

For reference, here's my eclipse.desktop file:

#!/usr/bin/env xdg-open
[Desktop Entry]
Version=1.0
Type=Application
Terminal=false
Path=/home/jinkim/eclipse
Icon=/home/jinkim/eclipse/icon.xpm
Exec=/home/jinkim/eclipse/eclipse
Name=Eclipse Luna
Comment=This is an IDE
Icon=eclipse

• It seems there a line in the .desktop file that I was missing. In particular, the crucial line seems to be:

StartupWMClass=Eclipse

More info on that here: What represent the StartupWMClass field of a .desktop file?

So, I have found what works for me below:

#!/usr/bin/env xdg-open
[Desktop Entry]
Name=Eclipse
GenericName=Eclipse Java IDE
Comment=Eclipse Java IDE
Categories=Utility;Developer Tools;IDEs;Java;
StartupWMClass=Eclipse
StartupNotify=true
Terminal=false
Type=Application
X-GNOME-FullName=Eclipse Java IDE

The steps I took to get this working are:

1. Use a text editor to create eclipse.desktop, and paste the above into it

2. Move eclipse.desktop into ~/.local/share/applications

3. Allow it to be executable: chmod +x eclipse.desktop

4. Double-click eclipse.desktop which should appear in Nautilus as Eclipse

5. In Unity launcher, right-click and select "Lock to Launcher"

format: plain


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# Embedding of algebraic surfaces

If I am not mistaken there is a theorem that says any curve $C$ can be embedded in $\mathbf{P}^3$. What can be said about surfaces? Do we have a theorem like:

All surfaces can be embedded in $\mathbf{P}^{N}$ for some fixed $N$.

And if not what is the simplest counterexample? A counterexample would be an infinite sequence of surfaces: $\{S_1, S_2, \cdots \}$ and infinite sequence of strictly increasing positive integers: $\{n_1, n_2, \cdots \}$ such that $S_i$ can not be embedded in $\mathbf{P}^{n_i}$.

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A related question: mathoverflow.net/questions/14177/… –  Georges Elencwajg Jan 20 '11 at 13:25

All smooth projective surfaces can be embedded in $\mathbb{P}^5$ using a (linear) projection. This is a classical theorem in algebraic geometry. In general, it is known that any smooth projective variety of dimension $n$ can be embedded into $\mathbb{P}^{2n+1}$.

The proof runs as follows: The secant variety $Sec(X)$ (the closure of points on secants on $X$) and the tangent variety $Tan(X)$ (the closure of the union of tangent spaces) both have dimension $\le 2n+1$, so if $X$ lies in a big projective space $\mathbb{P}^N$, successive projections from points outside $Sec(X)\cup Tan(X)$, gives you a morphism $\mathbb{P}^{2n+1}$ which is injective, finite, unramified hence a closed embedding.

This result makes surfaces in $\mathbb{P}^4$ an interesting study. A remarkable theorem of Ellingsrud-Peskine says that surfaces in $\mathbb{P}^4$ are very special: those not of general type belong to finitely many families, hence one has some hope of classifying them all. For example, Hartshorne conjectured that the degree of rational surfaces in $\mathbb{P}^4$ is bounded (it is now belived that this bound is equal to 12).

There are however non-projective surfaces, which cannot be embedded in any projective space, see e.g., this question.

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Nonprojective surfaces are singular. If you allow singularities, there is no bound on the embedding dimension, even for projective surfaces (or one-point schemes, for that matter). –  Laurent Moret-Bailly Jan 20 '11 at 15:07
"Nonprojective surfaces are singular". Why? A non-algebraic complex torus is certainly smooth as a complex Kaehler manifold. Probably you meant "non-projective, proper algebraic" or something like this. –  Francesco Polizzi Jan 20 '11 at 16:31
@Francesco: Indeed, I was assuming from the start that we were talking about proper algebraic surfaces. Smooth proper algebraic surfaces are known to be projective; together with compactification and resolution of singularities, this implies that smooth algebraic surfaces are quasiprojective. –  Laurent Moret-Bailly Jan 20 '11 at 16:40

The explanation above can be expanded with the following classical fact. Consider the Veronese surface $V$ in $\mathbb{P}^{5}$ which is the image of $\mathbb{P}^{2}$ via the embedding induced by $\mathcal{O}_{\mathbb{P}^{2}}(2)$. Note that $\mathbb{P}^{5}$ parametrizes conics and $V$ parametrizes conics of rank $1$ i.e. squares of linear forms. Since the general conic can not be written as sum of two squares the variety of secant lines $Sec(V)$ has dimension $4$ instead of the expected dimension $5$. So $Sec(V)$ is a cubic hypersurface in $\mathbb{P}^{5}$ singular along $V$. This means that if $p\in \mathbb{P}^{5}$ is a general point the projection

$$\pi_{p}:V\rightarrow \mathbb{P}^{4}$$

defines an isomorphism between $V$ and a smooth surface of degree $4$ in $\mathbb{P}^{4}$. Indeed Severi proved that the Veronese surface $V$ is the only surface in $\mathbb{P}^{5}$ which can be isomorphically projected in $\mathbb{P}^{4}$.

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# Is it worth it to pay for a certificate from edX?

1. Feb 27, 2015

I am doing an online course on edX, but the policy is that i will not get a certificate if i dont pay for it. So are certficates from edX really worth it? Are they recognised by academic institutions?

2. Feb 27, 2015

### Greg Bernhardt

The backing of MIT and Harvard is promising. The certificates won't transfer as school credits. What do you mean by recognized by academic institutions? With only being 3 years old I wonder how many academic institutions know much about edX.

http://www.nopaymba.com/pay-verified-statement-accomplishment/

3. Feb 27, 2015

Thank you, the link was very useful. I dont think i will pay, i am motivated enough to work through the course material.

4. Mar 4, 2015

### CalcNerd

You did not specify what type of courseware you are planning to take.

If you were trying to show a mastery of statistics for example, you could look at certification through ASQ (American Society of Quality). All of their certifications are well regarded (I believe this organization is 20 years or older). There are other types of professional certifications as well, depending upon your area of interest.

And there are online programs as well.

Let us know what your area of interest is.

5. Mar 4, 2015

I am taking 8.05x masterinng quantum mechanics. Its already too late to pay for the certificate anyway, but tell me your opinion.

6. Mar 4, 2015

### CalcNerd

You are probably smart by not paying for that.

What would help you best upon finishing some of these type of classes is a published White paper to reference when you need to prove your mastery of the subject.

At the very least it will help you in future graduate work, if you can apply it to your studies or research. Not all coursework will be relevant to you in the near or often even the far future aside from personal satisfaction of knowledge gained.

7. Mar 4, 2015

### Staff: Mentor

I don't remember ever seeing the term "white paper" in this context. What would be in that paper, and where would he publish it?

8. Mar 4, 2015

### CalcNerd

http://physicsworld.com/cws/channel/whitepapers

This is not something I would suggest for someone taking anything but Phd quality coursework, but 8.05 X indicates doctorial level type study. (I am editing, mea culpa) This is an intermediate class and the rest of my comments DO NOT apply. But it is good information for someone who may pursue upper level independent study.

This isn't for some undergrad who believes they are formulating new unproven frontiers of science. There really is no other alternative if someone wants to demonstrate a knowledge of his advanced independent studies than to publish.

Last edited: Mar 4, 2015
9. Mar 7, 2015

### jack476

I'm taking that too :)

I don't normally pay for MOOCs. Sometimes I do, but it's because I enjoy the program and want to support it, not because I think it will actually get me anything. I know edX has a scheme for some of the courses where you can decide how much to donate (I'll usually leave $10 or$20) and on Coursera they're not terribly expensive. The groups that produce MOOC content do so for free, and if they can't expect some kind of compensation then they're going to stop. And as the upshot, showing universities that this is a potentially profitable venture will encourage the production of more high-quality content.

OTOH, I also have seen edX courses where the fee for the verified certificate is more than the per hour tuition rate at my own college. Let me think, no.

10. Mar 7, 2015


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Boise Lumber has decided to enter the lucrative prefabricated housing business. Initially, it plans 1 answer below »

Boise Lumber has decided to enter the lucrative prefabricated housing business. Initially, it plans to offer three models: standard, deluxe, and luxury. Each house is prefabricated and partially assembled in the factory, and the final assembly is completed on site. The dollar amount of building material required, the amount of labor required in the factory for prefabrication and partial assembly, the amount of on-site labor required, and the profit per unit are as follows.
Standard Model Deluxe Model Luxury Model
Material $6,000$8,000 $10,000 Factory Labor (hr) 240 220 200 On-Site Labor (hr) 180 210 300 Profit$3,400 $4,000$5,000
For the first year's production, a sum of \$8,200,000 is budgeted for the building material; the number of labor-hours available for work in the factory (for prefabrication and partial assembly) is not to exceed 209,000 hr; and the amount of labor for on-site work is to be less than or equal to 231,000 labor-hours. Determine how many houses of each type Boise should produce to maximize its profit from this new venture. (Market research has confirmed that there should be no problems with sales.)
 standard model houses deluxe model houses luxury model houses

Rajeswari
Std X Model Del Y Lux mod Z Available Material 6000 8000 10000 8200000 Fac labr hours 240 220 200 209000 Onsite lab hours 180 210 300 231000 Profit 3400 4000 5000 Objective is to maximise Profit Z 3400x+4000y+5000z Subject to : 6000x+8000y +10000z <= 8200000 Or 3x+4y+5z <=4100 240x+220y+200z <=209000 Or 12x+11y+10z <= 10450 180x+210y+300z <=231000 Or 6x+7y+10z<= 7700...

Plagiarism Checker

Submit your documents and get free Plagiarism report

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text:
• D

F Nous contacter

0

Single angle

O

Single angle Distributions of Frobenius of elliptic curves #4

Auteurs : Jones, Nathan (Auteur de la Conférence)
CIRM (Editeur )

Résumé : In the 1970s, S. Lang and H. Trotter developed a probabilistic model which led them to their conjectures on distributional aspects of Frobenius in $GL_2$-extensions. These conjectures, which are still open, have been a significant source of stimulation for modern research in arithmetic geometry. The present lectures will provide a detailed exposition of the Lang-Trotter conjectures, as well as a partial survey of some known results.

Various questions in number theory may be viewed in probabilistic terms. For instance, consider the prime number theorem, which states that, as $x\rightarrow \infty$ , one has
$\#\left \{ primes\, p\leq x \right \}\sim \frac{x}{\log x}$
This may be seen as saying that the heuristic "probability" that a number $p$ is prime is about $1/\log p$. This viewpoint immediately predicts the correct order of magnitude for the twin prime conjecture. Indeed, if $p$ and $p+2$ are seen as two randomly chosen numbers of size around $t$, then the probability that they are both prime should be about $1/(\log t)^2$, which predicts that
$\#\left \{ primes\, p\leq x : p+2\, is\, also\, prime \right \}\asymp \int_{2}^{x}\frac{1}{(\log t)^2}dt \sim \frac{x}{\log x}$
In this naive heuristic, the events "$p$  is prime" and "$p+2$ is prime" have been treated as independent, which they are not (for instance their reductions modulo 2 are certainly not independent). Using more careful probabilistic reasoning, one can correct this and arrive at the precise conjecture
$\#\left \{ primes\, p\leq x : p+2\, is\, also\, prime \right \} \sim C_{twin}\frac{x}{(\log x)^2}$,
where $C_{twin}$  is the constant of Hardy-Littlewood.
In these lectures, we will use probabilistic considerations to study statistics of data attached to elliptic curves. Specifically, fix an elliptic curve $E$  over $\mathbb{Q}$ of conductor $N_E$. For a prime $p$ of good reduction, theFrobenius trace $a_p(E)$ and Weil $p$-root $\pi _p(E)\in \mathbb{C}$ satisfy the relations
$\#E(\mathbb{F}_p)=p+1-a_p(E)$,
$X^2-a_p(E)X+p=(X-\pi _p(E))(X-\overline{ \pi _p(E)})$.
Because of their connection via the Birch and Swinnerton-Dyer conjecture to ranks of elliptic curves (amongother reasons), there is general interest in understanding the statistical variation of the numbers $a_p(E)$ and $\pi_p(E)$, as $p$ varies over primes of good reduction for E. In their 1976 monograph, Lang and Trotter considered the following two fundamental counting functions:
$\pi_{E,r}(x) :=\#\left \{ primes\: p\leq x:p \nmid N_E, a_p(E)=r \right \}$
$\pi_{E,K}(x) :=\#\left \{ primes\: p\leq x:p \nmid N_E, \mathbb{Q}(\pi_p(E))=K \right \}$,
where $r \in \mathbb{Z}$ is a fixed integer, $K$ is a fixed imaginary quadratic field. We will discuss their probabilistic model, which incorporates both the Chebotarev theorem for the division fields of $E$ and the Sato-Tatedistribution, leading to the precise (conjectural) asymptotic formulas
(1) $\pi_{E,r}(x)\sim C_{E,r}\frac{\sqrt{x}}{\log x}$
$\pi_{E,K}(x)\sim C_{E,K}\frac{\sqrt{x}}{\log x}$,
with explicit constants$C_{E,r}\geq 0$ and $C_{E,K} > 0$. We will also discuss heuristics leading to the conjectureof Koblitz on the primality of $\#E( \mathbb{F}_p)$, and of Jones, which combines these with the model of Lang-Trotter for $\pi_{E,r}(x)$ in order to count amicable pairs and aliquot cycles for elliptic curves as introduced by Silvermanand Stange.
The above-mentioned conjectures are all open, although (in addition to the bounds mentioned in the previous section) there are various average results which give evidence of their validity. For instance, let $R\geq 1$ and $S\geq 1$be an arbitrary positive length andwidth, respectively, and define
$\mathcal{F}(R,S):= \{ E_{r,s}:(r,s)\in \mathbb{Z}^2,-16(4r^3+27s^2)\neq 0, \left | r \right |\leq R\:$ and $\left | s \right | \leq S \}$,
where $E_{r,s}$ denotes the curve with equation $y^2=x^3+rx=s$. The work of Fouvry and Murty $(r=0)$, and of David and Pappalardi $(r\neq 0)$, shows that, provided min $\left \{ R(x), S(x) \right \}\geq x^{1+\varepsilon }$, one has
(2) $\frac{1}{\left |\mathcal{F}(R(x),S(x)) \right |} \sum_{E\in \mathcal{F}(R(x),S(x))} \pi_{E,r}(x) \sim C_r \frac{\sqrt{x}}{\log x}$
where $C_r$ is a constant. We will survey this and other theorems on average, and then discuss the nature of the associated constants $C_{E,r},C_{E,K}$ etc. We will discuss the statistical variation of these constants as $E$ varies over all elliptic curves over $\mathbb{Q}$, and use this to confirm the consistency of (2) with (1), on the level of the constants

Keywords : Galois representation - elliptic curve - trace of Frobenius - Chebotarev density theorem - Sato-Tate conjecture - Lang-Trotter conjecture

11G05 - Elliptic curves over global fields
11R44 - Distribution of prime ideals

 Informations sur la Vidéo Réalisateur : Hennenfent, Guillaume Langue : Anglais Date de publication : 09/10/14 Date de captation : 20/02/14 Collection : Research talks Format : quicktime ; audio/x-aac Durée : 01:04:12 Domaine : Number Theory Audience : Chercheurs ; Doctorants , Post - Doctorants Download : http://videos.cirm-math.fr/2014-02-20_Jones_part4.mp4 Informations sur la rencontre Nom du congrès : Jean-Morlet Chair - Doctoral school : Frobenius distribution on curves / Chaire Jean-Morlet - Ecole doctorale : distribution de Frobenius sur des courbesOrganisteurs Congrès : Kohel, David ; Ritzenthaler, Christophe ; Shparlinski, IgorDates : 17/02/14 - 28/02/14 Année de la rencontre : 2014 URL Congrès : http://shparlinskikohel.weebly.com/abstr...

Voir aussi

Bibliographie

• Abstract
• conference2.jpg

Z


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# New type of series!

Number Theory Level 5

Given that $\sum_{p=prime}\dfrac{(f+g)(p)}{p^s}=\sum_{p=prime}\dfrac{(f*g)(p)}{p^s}$ where $$f* g$$ is a dirichlet convolution then this is always true (assuming both sides converges) if f,g are

×


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# How to ask hyperref works only with href

How to ask hyperref works ONLY with \url and \href commands?

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You want to use hyperref only for \url and \href commands, but not all other features like PDF bookmarks, section links etc., is it? There is the url package for once. Not sure if it also provides \href. –  Martin Scharrer Aug 3 '11 at 20:48
All of your questions have one or several responses that seem to answer them, please consider marking one of them as ‘Accepted’ by clicking on the tickmark below their vote count. This shows which answer helped you most, and it assigns reputation points to the author of the answer (and to you!). –  doncherry Aug 17 '11 at 21:25

\usepackage[implicit=false]{hyperref}

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Thank you! Why I didn't find it in the documentation –  newuser Aug 4 '11 at 8:59

Not sure exactly what you are asking, but does this help:

   \documentclass{article}
\usepackage{hyperref}

\begin{document}


text:
Expii

Oxides as Acids and Bases — Overview & Properties - Expii

Acidic oxides form between an oxygen and nonmetals. Basic oxides form between oxygen and metals. Some oxides are amphoteric and act as both an acid and a base.


text:
I could not find a simple demonstrative example of *insert title here*. I am leaving this out here for future reference.

Note that a recursive implementation for deCasteljau’s is not efficient, since it results in unnecessary multiple computation of some intermediary points.

def deCasteljau(points, u, k = None, i = None, dim = None):
"""Return the evaluated point by a recursive deCasteljau call
Keyword arguments aren't intended to be used, and only aid
during recursion.

Args:
points -- list of list of floats, for the control point coordinates
example: [[0.,0.], [7,4], [-5,3], [2.,0.]]
u -- local coordinate on the curve: $u \in [0,1]$

Keyword args:
k -- first parameter of the bernstein polynomial
i -- second parameter of the bernstein polynomial
dim -- the dimension, deduced by the length of the first point
"""
if k == None: # topmost call, k is supposed to be undefined
# control variables are defined here, and passed down to recursions
k = len(points)-1
i = 0
dim = len(points[0])

# return the point if downmost level is reached
if k == 0:
return points[i]

# standard arithmetic operators cannot do vector operations in python,
# so we break up the formula
a = deCasteljau(points, u, k = k-1, i = i, dim = dim)
b = deCasteljau(points, u, k = k-1, i = i+1, dim = dim)
result = []

# finally, calculate the result
for j in range(dim):
result.append((1-u) * a[j] + u * b[j])

return result

A demonstration of the above function

import numpy as np
import pylab as pl
import math

# insert deCasteljau function definition here

points = [[0.,0.], [7,4], [-5,3], [2.,0.]]

def plotPoints(b):
x = [a[0] for a in b]
y = [a[1] for a in b]
pl.plot(x,y)

curve = []

for i in np.linspace(0,1,100):
curve.append(deCasteljau(points, i))

plotPoints(curve)
pl.show()

#### For Rational Bezier Curves

With a small modification, same function can be used for rational Bezier curves

def rationalDeCasteljau(points, u, k = None, i = None, dim = None):
"""Return the evaluated point by a recursive deCasteljau call
Keyword arguments aren't intended to be used, and only aid
during recursion.

Args:
points -- list of list of floats, for the control point coordinates
example: [[1.,0.,1.], [1.,1.,1.], [0.,2.,2.]]
u -- local coordinate on the curve: $u \in [0,1]$

Keyword args:
k -- first parameter of the bernstein polynomial
i -- second parameter of the bernstein polynomial
dim -- the dimension, deduced by the length of the first point
"""
if k == None: # topmost call, k is supposed to be undefined
# control variables are defined here, and passed down to recursions
k = len(points)-1
i = 0
dim = len(points[0])-1

# return the point if downmost level is reached
if k == 0:
return points[i]

# standard arithmetic operators cannot do vector operations in python,
# so we break up the formula
a = rationalDeCasteljau(points, u, k = k-1, i = i, dim = dim)
b = rationalDeCasteljau(points, u, k = k-1, i = i+1, dim = dim)
result = []

# finally, calculate the result
for j in range(dim+1):
result.append((1-u) * a[j] + u * b[j])

# at the end of first and topmost call, when the recursion is done,
# normalize the result by dividing by the weight of that point
if k == len(points)-1:
for i in range(dim):
result[i] /= result[dim]
# dimension is also the index with the weight

return result

We can demonstrate by e.g. comparing the algorithm’s results with a circular arc

import numpy as np
import pylab as pl
import math

# insert rationalDeCasteljau function definition here

points = [[1.,0.,1.], [1.,1.,1.], [0.,2.,2.]]

def plotPoints(b):
x = [a[0] for a in b]
y = [a[1] for a in b]
pl.plot(x,y)

curve = []

# limit to 5 points to show the difference with analytic solution
for i in np.linspace(0,1,5):
curve.append(rationalDeCasteljau(points, i))

plotPoints(curve)

# plot the actual circular arc
arc_x = np.linspace(0,1,100)
arc_y = []

for i in arc_x:
arc_y.append(math.sqrt(1-i*i))

pl.plot(arc_x, arc_y)

pl.show()

I am not actually working on Bezier curves, but NURBS. My reference for studying is The NURBS Book by Piegl and Tiller, which is excellent so far.


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# Homework Help: Static and kinetic friction of a crate

1. Mar 10, 2009

### dcangulo

a dockworker loading crates on a ship finds that a 20kg. crate, intially at rest on a horizontal surface, requires a 75-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60-N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

how do i find the coefficients? there is a table in the book with coefficients but it doesnt have one with wood on floor.

2. Mar 10, 2009

### tnutty

K + U = K_o + U_o

what is k,u,k_o,u_o ?

3. Mar 10, 2009

### Staff: Mentor

You're supposed to solve for the coefficients using the information given in the problem, not look them up in some table.

Hint: How does static and kinetic friction relate to the normal force?

4. Mar 10, 2009

### dcangulo

$$\mu$$=f/N

where N is the normal force and f would be the 20kg?

Last edited: Mar 10, 2009
5. Mar 10, 2009

### Staff: Mentor

Good. But realize there's a difference between static and kinetic friction.

N is the normal force. What does the normal force equal? Hint: It's related to the 20 kg mass.

f is the friction force. Hint: The static and kinetic friction forces are given in the problem statement.

6. Mar 10, 2009

### Oscar Wilde

You need to solve for the two values of friction. Static friction being the frictional force that one must "overcome" to put the object into motion, and kinetic friction being the frictional force that must be "overcome" to sustain the objects motion.

7. Mar 11, 2009

### dcangulo

ok so what i did was find the normal force of the static friction which i believe was N=mg which is N=20(9.8) and then i did the little $$\mu$$=f/N with $$\mu$$=75/196

and that found me, what i believe to be the static friction.

Now how do i find kinetic friction coefficient??

8. Mar 12, 2009

### Staff: Mentor

Good! The normal force is just the weight of the crate. The maximum static friction force was given as 75 N. You combined those facts to find the static friction coefficient.

Exactly the same way. What's the kinetic friction force? Does the normal force change?

9. Mar 12, 2009

### dcangulo

i think kinetic friction force is 60... or do i subtract 75-60 and that'll give me the kinetic friction??

this is what i did

75-f$$_{}s$$=0
f$$_{}s$$=75

this is the only thing i can think of

10. Mar 12, 2009

### Staff: Mentor

Right.
No. Kinetic and static friction are two different things, so don't combine them.

But you calculate the coefficient of friction in the exact same manner for both:
μ = F/N, where F is the friction force and N is the normal force.


text:
# NCERT Solutions for Class 8 maths Chapter 1 : rational Numbers Exercise 1.1

## NCERT Solutions for Class 8 maths Chapter 1 Exercise 1.1

In this page we have NCERT Solutions for Class 8 maths Chapter 1 : rational Numbers for Exercise 1.1. Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1:
Using appropriate properties find:

(i)Using commutativity property of rational number
Now using Distributivity property

(ii) Using commutativity property of rational numbers

Now using Distributivity property

Question 2
Write the additive inverse of each of the following:
(a) $\frac {2}{8}$
(b) $\frac {-5}{9}$
(c) $\frac {-6}{-5}$
(d) $\frac {2}{-9}$
(e) $\frac {19}{-6}$

(i) $\frac {2}{8}$
Additive inverse of $\frac {2}{8}$= $\frac {-2}{8}$
(ii) $\frac {-5}{9}$

Additive inverse of $\frac {-5}{9}$ = $\frac {5}{9}$
(iii) $\frac {-6}{-5} =\frac {6}{5}$

Additive inverse of $\frac {-6}{-5}$= $\frac {-6}{5}$
(iv)$\frac {2}{-9}=\frac {-2}{9}$

Additive inverse of $\frac {2}{-9}$ =$\frac {2}{9}$
(v) $\frac {19}{-6}= \frac {-19}{6}$

Additive inverse of $\frac {19}{-6}$ =$\frac {19}{6}$

Question 3
Verify that - ( - x) = x for.
(i) x =11/15 (ii) x=-13/17
-x=-11/15 As (11/15) + (-11/15) =0
The above also represent that additive inverse of (-11/15) is (11/15) or we can say that
-(-11/15)=11/15
Or x=-(-x)
(ii) x=-13/17
-x=-13/17 As (-13/17) + (13/17) =0
The above also represent that additive inverse of (13/17) is (-13/17)) or we can say that
-(13/17) = -13/17
Or x=-(-x)

Question 4
Find the multiplicative inverse of the following.
(a)$-13$
(b)$\frac {-13}{19}$
(c) $\frac {1}{5}$
(d) $\frac {-5}{8} \times \frac {-3}{7}$
(e) $-1 \times \frac {-2}{5}$
(f) $-1$

(a) $-13$
Multiplicative inverse of $-13$=$\frac {-1}{13}$

(b)$\frac {-13}{19}$
Multiplicative inverse of $\frac {-13}{19}$ =$\frac {-19}{13}$

(c)$\frac {1}{5}$
Multiplicative inverse of $\frac {1}{5}$=$5$

(d) $\frac {-5}{8} \times \frac {-3}{7}=\frac {15}{56}$
Multiplicative inverse of $\frac {-5}{8} \times \frac {-3}{7}$ =$\frac {56}{15}$

(e) $-1 \times \frac {-2}{5}= \frac {2}{5}$
Multiplicative inverse of $-1 \times \frac {-2}{5}$ =$\frac {5}{2}$

(f) $-1$<
Multiplicative inverse of $-1$=-1

Question 5 :
Name the property under multiplication used in each of the following:

1. 1 is the multiplicative identity
2. Commutatively
3. Multiplicative inverse

Question 6:
Multiply 6/13 by the reciprocal of -7/16
Reciprocal of -7/16= -16/7
So $\frac {6}{13} \times (- \frac {16}{7}) =- \frac {96}{91}$

Question 7 Tell what property allows you to compute

Associativity

Question 8
Is 8/9 the multiplicative inverse of $-1 \frac {1}{8}$
Why or why not?
$-1 \frac {1}{8}$
$=\frac {-9}{8}$
Now
$\frac {8}{9} \times (-\frac {9}{8}) = -1$
So it is not the multiplicative inverse

Question 9 Is 0.3 the multiplicative inverse of $3 \frac {1}{3}$
Why or why not?
$3 \frac {1}{3}$
=10/3
Now
$\frac {3}{10} \times \frac {10}{3} =1$
So it is the multiplicative inverse

Question 10
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.

1. 0
2. 1 and -1
3. 0
Question 11 :
Fill in the blanks.
(i) Zero has __________ reciprocal.
(ii) The numbers __________ and __________ are their own reciprocals
(iii) The reciprocal of - 5 is __________.
(iv) Reciprocal of (1/x) where is x ≠ 0 __________.
(v) The product of two rational numbers is always a __________.
(vi) The reciprocal of a positive rational number is __________.

1. No
2. 1 and -1
3. -1/5
4. X
5. Rational Number
6. Positive rational number


text:
# How does friction act on a body, if only 2 regions on it are rough? [closed]

While tackling an Olympiad question, it came to my mind that friction need not act in the same direction at all points on a body. I thought of using integration to evaluate the net frictional force, but was stopped by this statement.

Frictional force between 2 bodies, doesn't depend on the surface area in contact.

If that is the case, then how does the force act on a body like the following, where only the narrow strip marked is rough and hence friction can act only there? What are the individual frictional forces at the points A and B?

The image shows a hollow cylinder, mass $M$ and radius $R$ from top view. The cylinder is placed with its flat end in contact with the ground. Only the gray shaded line is rough, id est, friciton acts only there. The cylinder is rotating about the topmost point. Ignore the $v$ in the diagram.

## closed as unclear what you're asking by ja72, Kyle Kanos, ACuriousMind♦, Neuneck, DanuAug 26 '15 at 9:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Your question is unclear to me. Friction acts on each little patch of contact in a way that always opposes motion. – ja72 Aug 21 '15 at 18:06
• Yes, I understand that. What I don't get, is this. If suppose the little patch is of area $dA$ and has height $h$ and density $\rho$, then will the friction force acting on that small patch be $F_{fr} = \mu dm g = \mu g \rho h dA$ ? If yes, then that would imply that the force IS dependent on the surface area in contact. If no, then how can the force be evaluated? – Aritra Das Aug 21 '15 at 18:20
• @ja72 or would the force at either point A and B be dependent on the mass of the entire cylinder? – Aritra Das Aug 21 '15 at 18:44

As for the case you represent, the cylinder will perceive friction forces opposed to its rotation movement, which will be tangential to the circle, and opposed to the linear velocities in these points. So it will either stay static, or it will rotate w.r.t its center, depending on the actual values of $v$ and $\mu$.


text:
# How to align coordinate systems of Inset and enclosing Graphics?

Suppose we have some plot with AspectRatio not being Automatic:

iIS = 350;
iPR = {{0, 25}, {0, 100}};
iIP = {{20, 5}, {15, 5}};
inset = ListPlot[Prime[Range[25]], ImageSize -> iIS, PlotRange -> iPR,
AxesStyle -> Bold, ImagePadding -> iIP, Frame -> True]

Now we wish to place it as Inset inside of a larger Graphics object (with Frame->True) in such a way that the coordinate systems of both plots exactly coincide with each other and the original plot will not be cropped.

What is the right way to align coordinate systems of Inset and enclosing Graphics exactly?

The ideal solution would allow setting arbitrary PlotRange, ImageSize and ImagePadding for the enclosing Graphics.

(This quesiton is a more demanding variation of How to make Inset graphics maintain relative sizes when combined.)

• Is this the same topic or are there any subtleties? mathematica.stackexchange.com/q/73522/5478 – Kuba May 17 '15 at 6:42
• @Kuba Yes, that thread is related but my question is substantially different: 1) the OP of that question was satisfied by approximate solution while I need exact match; 2) I consider more general case where AspectRation is not Automatic and both the inset and enclosing graphics have non-zero ImagePadding. – Alexey Popkov May 17 '15 at 7:19
• I put a link to that question in your own question that it may serve as a starting point or simpler case depending on circumstance of the reader. – Mr.Wizard May 17 '15 at 10:01
• I'm exploring a way to solve this for completely arbitrary graphics and I'm running into a problem that, for instance, resizing your inset does not resize the ImagePadding of the inset (valid and expected behavior, but somewhat complicates things). Does your question, nonetheless imply, that we know the padding of the inset anyway? Is it necessarily given as a fixed amount of printer points, or could it be in Scaled or ImageScaled coordinates? Could it be Automatic as well? – LLlAMnYP May 17 '15 at 12:46
• @LLlAMnYP Unfortunately ImagePadding accepts only explicit values in printer's points or All. Life would be much easier if this option would accept scaled specification. I assume that ImagePadding for the inset is specified in printer's points and is known. – Alexey Popkov May 17 '15 at 13:13

As per the comment section of the OP I assume that the image padding is a constant number of printer points, though it is not necessarily known. I use the rasterize trick to obtain the size of the plotting range in printer points:

printerPointsPlotRange =
(#[[2]] - #[[1]] &)@
(Rasterize[Show[#, Epilog ->
{Annotation[Rectangle[Scaled[{0, 0}], Scaled[{1, 1}]],
"Two", "Region"]}], "Regions"][[-1, 2]]) &

Similarly, my own implementation of ImageDimensions. It gives the size of the rectangle defined by Rectangle[ImageScaled[{0,0}],ImageScaled[{1,1}]] in printer points, which does not always match ImageDimensions if ImageSize and AspectRatio lead to conflicting results:

realImageDimensions =
(#[[2]] - #[[1]] &)@
(Rasterize[Show[#, Epilog ->
{Annotation[Rectangle[ImageScaled[{0, 0}], ImageScaled[{1, 1}]],
"Two", "Region"]}], "Regions"][[-1, 2]]) &

I implement a modified version of PlotRange to account for PlotRangePadding.

realPlotRange =
Module[
baserange = (#[[2]] - #[[1]] &) /@ PlotRange[#], range},

range = (baserange + padding) /.
{a_ Scaled[b_] :> Scaled[a b], Scaled[a_] + Scaled[b_] :> Scaled[a + b]} /.
{a_ + Scaled[b_] :> a/(1 - b)};
range] &

It appears to fail for mixed specifications, such as {{1,Scaled[.1]},{Scaled[.02],Scaled[.02]}}. However if the left and right paddings are both given either in Scaled form or in the coordinates of the plot and the same holds for the top and bottom padding specs, the function works fine.

plotRangeRatio = realPlotRange[#1]/realPlotRange[#2] &

Setting everything up:

SetOptions[Plot, {GridLines -> Automatic, Frame -> True}]
gr1 = Plot[Sin[x], {x, 0, 4}, PlotRange -> {{-1, 5}, {-2, 2}},
PlotRangeClipping -> False, ImageSize -> {600, 400},
gr2 = Plot[Cos[x], {x, -.5, 4}, ImageSize -> {400, 300},
ImagePadding -> 30, PlotRange -> {{-.5, 4}, {-1, 1}}]

And producing the image:

Show[gr1, Epilog ->
Inset[Show[gr2,
ImageSize ->
plotRangeRatio[gr2, gr1] printerPointsPlotRange[gr1] +
(realImageDimensions[gr2] - printerPointsPlotRange[gr2]),
AspectRatio ->
(Last[#]/First[#] &)@
(plotRangeRatio[gr2, gr1] printerPointsPlotRange[gr1])],
{0, 0}, {0, 0}, Automatic]]

The main assumption here is that the image padding is fixed and does not change upon resizing the image. It should then be always equal to

 (realImageDimensions[gr2] - printerPointsPlotRange[gr2])

where I do not even care, how wide the padding is on which side, all I care is how much bigger the image is, compared to the size of the plot range. So for plots with ImagePadding of 30, or {{5,55},{20,40}} or {{35,25},{50,10}} the above code will in all cases return {60,60}.

Upon aligning the coordinates the aspect ratio of the graphic being enclosed may change depending on the coordinate scales of the enclosing graphic. It is thus calculated beforehand and set to the appropriate value.

• I appreciate this solution very much, it is great! Your realPlotRange currently fails in some cases, you may easily implement more universal analog on the base of my completePlotRange function instead. But I appreciate your straightforward approach in the realPlotRange implementation. – Alexey Popkov May 17 '15 at 18:21
• @AlexeyPopkov Please do elaborate, in which cases realPlotRange fails, I'd like to explore alternatives. – LLlAMnYP May 17 '15 at 20:05
• For example it fails for Graphics[{}, Frame -> True, PlotRange -> {{0, 1}, {0, 1}}, PlotRangePadding -> Scaled[.1]], it also fails for any graphics for which the function PlotRange fails, for example: Graphics[{GeometricTransformation[Rectangle[], RotationTransform[1]]}, Frame -> True]. – Alexey Popkov May 18 '15 at 1:26
• You can shorten the implementation of realPlotRange through an undocumented function: realPlotRange[gr_] := #2 - #1 & @@@ Chartingget2DPlotRange[gr] – J. M. will be back soon Jun 28 '16 at 12:01
• @J.M. Chartingget2DPlotRange is more advanced currently (v. 10.4.1) than it was in version 8 but it is still Kernel-based and so will fail in many cases including the second example from my previous comment here. It also seemingly gives slightly wrong value for the first example from that comment. – Alexey Popkov Jun 28 '16 at 20:09

My way of thinking so far:

1. According to the Documentation, when the third argument of Inset is Automatic, the inset will have its original size inside of enclosing graphics. Its a good start.

2. The inset has non-zero ImagePadding (needed for the frame ticks), so some additional space must be added inside of the plot range of the enclosing graphics via PlotRangePadding. Probably PlotRangePadding -> Scaled[.1] is a good value to start with.

3. The enclosing graphics must have larger ImageSize. With PlotRangePadding -> Scaled[.1] the plot range covers a fraction 1-2*0.1 = 0.8 of the whole horizontal plotting range. So the image size must be iIS/0.8 (where iIS is ImageSize of the inset): ImageSize -> iIS/.8.

4. When we increase overall image size we must scale ImagePadding proportionally in order to keep original scales for the axes. So we must set ImagePadding -> iIP/.8.

What we get with these settings:

gr = Graphics[{Inset[
Show[inset, GridLines -> Automatic], {0, 0}, {0, 0}, Automatic]},
Frame -> True, PlotRange -> iPR, AspectRatio -> 1/GoldenRatio,
ImageSize -> iIS/.8, PlotRangePadding -> Scaled[.1],
FrameStyle -> Red, ImagePadding -> iIP/.8, GridLines -> Automatic,
GridLinesStyle -> Directive[Gray, Dashed]]

This method works but requires a lot of manual tuning of options. I am still looking for a better alternative.

## UPDATE 1: ImageScaled as fourth argument of Inset allows ImageSize to be arbitrary

Instead of adjusting ImageSize of enclosing graphics it is better to specify the size of the inset in scaled coordinates relative to the size of the enclosing graphics. Then the latter becomes arbitrary. As it is mentioned in p.3 above, the inset will have size 0.8 of the size of the enclosing graphics when PlotRangePadding -> Scaled[.1], so the third argument of Inset must be ImageScaled[.8]:

gr2 = Graphics[{Inset[
Show[inset, GridLines -> Automatic], {0, 0}, {0, 0},
ImageScaled[.8]]}, Frame -> True, PlotRange -> iPR,
AspectRatio -> 1/GoldenRatio, PlotRangePadding -> Scaled[.1],
FrameStyle -> Red, ImagePadding -> iIP/.8, GridLines -> Automatic,
GridLinesStyle -> Directive[Gray, Dashed]]

With this modification of the original approach we still must fix AspectRatio, PlotRange and ImagePadding of the enclosing graphics but now ImageSize is arbitrary! I'm still loooking for a way to make at least PlotRange of the enclosing graphics arbitrary too.

## UPDATE 2: ImagePadding "features" and a way to arbitrary PlotRange

Further experimentation showed that ImagePadding of the Inset isn't scaled by the fourth argument of Inset but is applied unscaled. This can be proven as follows:

iIS = 350;
iPR = {{0, 100}, {0, 100}};
iIP = {{20, 5}, {15, 5}};
inset = ListPlot[Range[100], ImageSize -> iIS,
PlotRange -> {{0, 90}, {0, 90}}, AxesStyle -> Bold,
Frame -> True, AspectRatio -> 1/GoldenRatio];
gr3 = Graphics[{Inset[Show[inset, GridLines -> Automatic], {0, 0}, {0, 0},
ImageScaled[.8]]}, Frame -> True, PlotRange -> iPR,
AspectRatio -> 1/GoldenRatio, PlotRangePadding -> Scaled[.1],
FrameStyle -> Red, ImagePadding -> iIP/.8, GridLines -> Automatic,
GridLinesStyle -> Directive[Gray, Dashed], ImageSize -> iIS]

In the above I have reduced the PlotRange of the inset by 10%. Correspondingly I must increase ImagePadding of the inset in order to keep the scale of the intrinsic coordinate system the same. The perfect match is achieved when I add 25.5 to the horizontal padding. This is calculated as follows:

0.1*(0.8*350 - 25)

25.5

where 0.8 is from ImageScaled[.8] (the fourth argument of Inset), 350 - horizontal ImageSize of the inset, 25 = 20 + 5 - horizontal ImagePadding, and 0.1 - 10% shortening of the horizontal PlotRange. In the above formula horizontal ImagePadding isn't multiplied by 0.8 (as one could expect) because scaling isn't applied to ImagePadding.

This feature is very unfortunate because it makes the plot not scalable again: for keeping the perfect match we must fix ImageSize of the enclosing graphics. But knowing this feature, we now can make PlotRange of the inset arbitrary!

One should also take into account that when the ImageSize -> w specification is used, the vertical ImagePadding seems to be ignored!

• Alexey, if you don't mind my asking, what is the point you are trying to achieve? I can't think of a reason one would want to have an inset that takes most of the space in the surrounding plot, so I am having a hard time thinking of solutions. Could you tell us a bit more about what you would like to do with it? – MarcoB May 17 '15 at 3:14
• This approach can have interesting applications in infographics but original motivation was much more utilitarian: currently in Mathematica we have no direct way to control the distance between tick labels and frame and it came to mind that one possible way is to create an Inset only with frame and place it inside of another graphics which has only tick labels. But actually I already have used it in situations where I need to have two coordinate systems at the same time on the plot exactly aligned with each other. – Alexey Popkov May 17 '15 at 3:29
• Cool! Thank you, I was sure that I was missing something here, and I was quite curious to see where this was going. Thank you for commenting on this! – MarcoB May 17 '15 at 3:31

Here is a solution that allows dynamic resizing of the image with a mouse.

As shown by @LLlAMnYP, in order to get Inset graphics to line up with the coordinate system of the enclosing Graphics object, it is necessary to know both the absolute PlotRange and ImagePadding of the inset graphics, as well as the absolute PlotRange and ImagePadding of the enclosing Graphics. This is because the PlotRange and the ImagePadding of a graphic are independently controllable. Consider the following diagram:

To line up with the coordinate system of the enclosing Graphics object, the ratio of the plot range point sizes of the Inset object and the enclosing Graphics object must be the same as the ratio of their absolute plot ranges. The total inset size will then be the plot range point size plus the image padding. Consider the horizontal dimension. If we set:

• ratio = ratio of absolute plot ranges of the inset and enclosing graphics
• inset = total image padding (left + right) of the inset object
• size = plot range point size of the enclosing graphics object

Then the horizontal size of the inset (in points) should be:

ratio * size + inset

or in Scaled coordinates:

Scaled[ratio + inset / size]

Notice that for nonzero inset, both of these expressions (and the equivalent ImageScaled version) depend on the size of the Image. This means that the size of the inset must change in a nontrivial way when the size of the enclosing graphics object is modified (e.g. by using the mouse). Therefore, to obtain a graphic where the inset graphics remain aligned to the enclosing graphic when the size is modified, one needs to use Dynamic.

All that remains is to determine the necessary graphics information for the Inset and enclosing Graphics objects. This can be done by using the graphicsInformation function from my answer to question 2091. Using this function, we can construct a dynamically updated graphics object where Inset objects are aligned correctly inside of an enclosing Graphics object even when the Graphics object is resized:

Options[overlayGraphics] = Options[Graphics];

overlayGraphics[prim_:{}, g:{__Graphics}, opts:OptionsPattern[]] := Module[
{insetInfo, insetRanges, graphRange, graphInfo},

insetInfo = graphicsInformation /@ g;

insetRanges = "PlotRange" /. insetInfo;
graphRange = Replace[OptionValue[PlotRange],
Except[{{_?NumericQ,_?NumericQ},{_?NumericQ,_?NumericQ}}] :> combinedRange[insetRanges]
];
graphInfo = graphicsInformation @ Graphics[prim, PlotRange -> graphRange, opts];
graphRange = "PlotRange" /. graphInfo;

With[
{
ratios = ScalingTransform[1/xyRange[graphRange]][xyRange/@insetRanges]
},

DynamicModule[{w,h},
With[
{
]
},

{w,h} = "ImageSize" /. graphInfo;
Graphics[
{
prim,
insets
},

PlotRange -> graphRange, PlotRangePadding -> 0,
ImageSize -> Dynamic[{w,h}], opts
]
]
]
]
]

combinedRange[p_] := MinMax /@ Transpose[p]

xyRange[{{a_,b_}, {c_,d_}}] := {b-a, d-c}

toInset[
gr_Graphics,
pr_,
{xratio_, yratio_},
{width_, height_}
] := Inset[
{0,0},
{0,0},
Dynamic @ Scaled[{xratio + (r+l)/width, yratio + (b+t)/height}]
]

Some implementation details.

1. I insert the explicit absolute plot range (with PlotRangePadding->0) and image padding into the inset and enclosing graphics so that any Automatic adjustments to PlotRange and ImagePadding are avoided.

2. I use AspectRatio->Full for the Inset objects so that the 4th argument of the Inset controls the full size of the inset.

3. I use AspectRatio->Full and explicit w and h ImageSize values for the enclosing Graphics object so that the resizing boxes directly control the ImageSize.

Here is overlayGraphics applied to your example inset:

overlayGraphics[
{},
{
Show[inset, GridLines->Automatic]
},
Frame->True, FrameStyle->Red, PlotRange->{{-5, 35}, {-10, 110}},
ImageSize->500, AspectRatio->1/GoldenRatio, GridLines->Automatic
]

Here is another example:

overlayGraphics[
{Red, Line[{{-1,-1.5}, {0,0}}]},
{
Plot[Sin[x], {x, 0, Pi}],
Graphics[{Circle[{-1, -1.5}, .5]}, ImagePadding->{{1, 2}, {3.3, 4.4}}]
},
AspectRatio -> .5, ImageSize -> 600, GridLines->Automatic, ImagePadding->20
]

• Very slick! Happy to be the first to upvote this. :-) – Mr.Wizard Mar 2 '17 at 2:53


text:
### Double Integral Transformation (Ellipse Region) Example 2

You may be given a transformation to use to solve a double integral, and sometimes the region of integration is a shape, such as a parallelogram or ellipse.

Use the transformation $$x = 2u$$ and $$y = 3v$$ to solve the integral $$\int_{}^{} {\int_R^{} {{x^2}dA} }$$, where R is the ellipse $$9{x^2} + 4{y^2} = 36$$.

Immediately substitute x and y into the equation for the ellipse.

$9{x^2} + 4{y^2} = 36$

$9{(2u)^2} + 4{(3v)^2} = 36$

$9(4{u^2}) + 4(9{v^2}) = 36$

$36{u^2} + 36{v^2} = 36$

${u^2} + {v^2} = 1$

By using the transformation we got a unit circle, just a regular circle with its center in the origin and radius of the root of 1, which is just one. Next we should find the Jacobian because this transformation has multiple variables.

$J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{array}} \right|$

$J = \left| {\begin{array}{*{20}{c}}2&0\\0&3\end{array}} \right|$

$J = (2)(3) - (0)(0)$

$J = 6$

What does the Jacobian here tell us? It tells us $$dxdy = 6dudv$$.

That's good. So now we can make our double integral. Remember that we're integrating over a unit circle for the u and v variables. How can we write that out? Well, you can say v goes from -1 to 1, and u goes from $$- \sqrt {1 - {v^2}}$$ to $$\sqrt {1 - {v^2}}$$.

$I = \int_{ - 1}^1 {\int_{ - \sqrt {1 - {v^2}} }^{\sqrt {1 - {v^2}} } {4{u^2} \cdot 6dudv} }$

And when integrating over a circle, use polar coordinates! See polar coordinate examples for more details, but we'll show all the steps here.

$u = r\cos (\theta )$

$v = r\sin (\theta )$

$dudv = rdrd\theta$

In polar coordinates, the radius goes from 0 to 1. Theta goes from 0 to $$2\pi$$ since it's an unrestricted full circle. Now our integral will look like:

$I = \int_0^{2\pi } {\int_0^1 {24{{(r\cos (\theta ))}^2}rdrd\theta } }$

$I = \int_0^{2\pi } {\int_0^1 {24{r^3}{{\cos }^2}(\theta )drd\theta } }$

$I = \int_0^{2\pi } {\frac{{24}}{4}} {(1)^4}{\cos ^2}(\theta ) - \frac{{24}}{4}{(0)^4}{\cos ^2}(\theta )d\theta$

$I = \int_0^{2\pi } {6{{\cos }^2}(\theta )d\theta }$

Now you can use trigonometric identities, such as double angle formulas.

$I = 3\int_0^{2\pi } {(1 + \cos (2\theta ))d\theta }$

$I = 3\left[ {(2\pi ) + \frac{{\sin (2(2\pi ))}}{2} - 0 - \frac{{\sin (2(0))}}{2}} \right]$

$I = 6\pi$

Please see other double integral transformation examples below:

Double Integral Transformation (Parallelogram Region) Example 1

Double Integral Transformation (Triangle Region) Example 3


text:
## Monday, October 12, 2015

### 52 Things: Number 51: What is the security model for ID-based encryption, and describe one IBE scheme.

This is the latest in a series of blog posts to address the list of '52 Things Every PhD Student Should Know To Do Cryptography': a set of questions compiled to give PhD candidates a sense of what they should know by the end of their first year. This week we introduce Identity-Based Encryption.

In public key cryptography, if Alice wants to send a message to Bob, she needs his public key. Typically this will be some very long bitstring (encoding, for example, the product of two large primes).

Suppose instead that Alice could use Bob's name, or email address, as his public key. Practically speaking, this is very convenient: there are no very long strings for Alice to obtain and remember and she doesn't need to verify that some seemingly random string is in fact Bob's public key (and not, for example, Charlie's public key). But to facilitate this, one needs Identity Based Encryption, or IBE.

In IBE, there is an entity called the Private Key Generator, or PKG. The PKG is able to compute Bob's private key using his ID (e.g. his email address) and a master key. Once Bob has authenticated himself to the PKG, he can ask for his private key and then, once he has it, he can decrypt any messages that have been encrypted under his ID.

There is an issue here, though. With the master key, the PKG can generate private keys for any agent it likes. Therefore the PKG can decrypt any messages intended for any agent. This is called key escrow, and it means that you must either trust the PKG not to read your encrypted messages or else not care that it does. In a company, though, senior management often has the right to read your (work) emails, and so IBE can be an appropriate solution for internal correspondence.

Formally, an IBE scheme consists of four algorithms: setup, extract, encrypt and decrypt.

Setup takes a security parameter and outputs the (secret) master key and the (public) system parameters, such as the message and ciphertext spaces.

Extract takes an ID and a master key and returns a private key corresponding to the ID.

Encrypt takes a message and an ID, and returns a ciphertext.

Decrypt takes a ciphertext and a private key, and returns a message.

Boneh and Franklin gave an IBE scheme in this 2003 paper. They prove, under an assumption similar to assuming that the CDH problem is hard, that their scheme is IND-ID-CCA secure in the Random Oracle Model. This means that (assuming all hash functions are random oracles), any attacker, running in polynomial-time with respect to the security parameter, wins the following security game with probability that is only negligibly (with respect to the security parameter) more than 1/2:

First, the attacker
• can request the private keys corresponding to any ID
• can request decryptions of any ciphertexts under any ID.
Then, the attacker chooses two messages $m_0, m_1$ and an ID $ID^*$ that does not occur in the list of IDs for which he has requested the corresponding private key. The attacker then receives the encryption $c^*$ of $m_b$ under $ID^*$ (where the bit $b$ is chosen uniformly at random). Then the attacker
• can request the private keys corresponding to any ID apart from $ID^*$
• can request decryptions of any ciphertexts under any ID apart from $\left ( c^*, ID^* \right)$
and outputs a bit $b'$. We say the attacker wins if $b' = b$.

The scheme given by Boneh and Franklin relies on a non-degenerate bilinear map $e: G_1 \times G_1 \to G_2$, where $G_1$ is a group of prime order $q$, which we write additively, and $G_2$ is a group, also of order $q$, which we write multiplicatively. They instantiate this map with the Weil pairing on elliptic curves, but we omit details here. All that matters is bilinearity: $e \left( aP, bQ \right ) = e \left( P, Q \right)^{ab}$ for any $a, b \in \mathbb{Z}_q$.

There's not enough space here to describe the scheme in full, but essentially the master key is some non-zero $s \in \mathbb{Z}_q$ and the private key corresponding to $ID$ is $sH \left(ID \right)$, where $H$ is a hash function sending bitstrings to elements of $G_1$. There are public generators $P$ and $P_{pub} = sP$ of $G_1$.

To encrypt $m$ under $ID$, one selects a random string $\sigma$ and XORs $m$ with a hash of $\sigma$, creating $c_m$. Then $M$ and $\sigma$ are hashed together, giving a non-zero element $r \in \mathbb{Z}_q$. Finally one computes the pairing $e \left( H \left( ID \right), P_{pub} \right)$, raises it to the power $r$, hashes this and XORs with $\sigma$, creating $c_{ID}$. The triple $\left( rP, c_{ID}, c_m \right)$ is the ciphertext.

With the private key $d = sH\left(ID \right)$, one decrypts the triple $\left (U, V, W \right)$ as follows: compute $e \left( d, U \right)$, which, by bilinearity, will equal $e \left( H \left( ID \right), P_{pub} \right)^r$ if the ciphertext was genuine. So one XORs $V$ with the hash of the pairing to obtain $\sigma$. Then XORing $W$ with the hash of $\sigma$ will give $m$. To check that this is the intended message, one verifies that $\sigma$ and $m$ hashed together gives $r$ such that $U = rP$.


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d the area of pentagon ABCDE in which BL ⊥ AC,
Question:

Find the area of pentagon ABCDE in which BL ⊥ ACDM ⊥ AC and EN ⊥ AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

Solution:

Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{AEN})+($ Area of trapezium EDMN $)+($ Area of $\Delta \mathrm{DMC})$ $+($ Area of $\Delta \mathrm{ACB})$

$=\left(\frac{1}{2} \times \mathrm{AN} \times \mathrm{EN}\right)+\left(\frac{1}{2} \times(\mathrm{EN}+\mathrm{DM}) \times \mathrm{NM}\right)+\left(\frac{1}{2} \times \mathrm{MC} \times \mathrm{DM}\right)+\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)$

$=\left(\frac{1}{2} \times \mathrm{AN} \times \mathrm{EN}\right)+\left(\frac{1}{2} \times(\mathrm{EN}+\mathrm{DM}) \times(\mathrm{AM}-\mathrm{AN})\right)+\left(\frac{1}{2} \times(\mathrm{AC}-\mathrm{AM}) \times \mathrm{DM}\right)+\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)$

$=\left[\left(\frac{1}{2} \times 6 \times 9\right)+\left(\frac{1}{2} \times(9+12) \times(14-6)\right)+\left(\frac{1}{2} \times(18-14) \times 12\right)+\left(\frac{1}{2} \times 18 \times 4\right)\right] \mathrm{cm}^{2}$

$=(27+84+24+36) \mathrm{cm}^{2}$

$=171 \mathrm{~cm}^{2}$

Hence, the area of the given pentagon is $171 \mathrm{~cm}^{2}$.


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# Electrical and Electronic Engineering - Theses

## Search Results

Now showing 1 - 2 of 2
• Item
Energy efficient wireless system design
Kudavithana, Dinuka ( 2015)
The demand for telecommunication networks is increasing rapidly. Wireless access is a major contributor to this trend. On the other hand, wireless is considered as a least energy efficient transmission medium mainly due to its unguided nature. The general focus of increasing wireless system energy efficiency is on reduction of the transmit power. However, this strategy may not save energy in short distance communication systems as the processing energy in hardware becomes more significant compared to the transmit radio energy. This thesis focuses on looking at the energy consumption of wireless systems by modeling the energy consumption as a function of several parameters such as receiver SNR, RF bandwidth, information rate, modulation scheme and code rate. We propose energy models for synchronization systems and other digital signal processing modules by considering the computational complexity of the algorithm and the required circuitry. Initially we focus on the synchronization aspects of wireless receivers. We study various algorithms on symbol timing recovery, carrier frequency recovery and carrier phase recovery and compare the performance in order to identify the suitable algorithms to operate at different SNR regions. We then develop energy models for those synchronization sub-systems by analyzing the computational complexity of circuitries based on a number of arithmetic, logic and memory operations. We define a new metric - energy consumption to achieve a given performance as a function of SNR - in order to compare the energy efficiency of different estimation algorithms. Next, we investigate the energy-efficiency trade-offs of a point-to-point wireless system by developing energy models of both the transmitter and receiver that include practical aspects such as error control coding, synchronization and channel equalization. In our system, a multipath Rayleigh-fading channel model and a low-density parity check (LDPC) coding scheme are chosen. We then develop a closed-form approximation for the total energy consumption as a function of receiver SNR and use it to find a minimum-energy transmission configuration. The results reveal that low SNR operation (i.e. low transmit power) is not always the most energy efficient strategy, especially in short distance communication. We present an optimal-SNR concept which can save a significant amount of energy mainly in short-range transmission systems. We then focus on cooperative relay systems. We investigate the energy efficiency trade-offs of single--relay networks by developing energy models for two relay strategies: amplify-and-forward (AF) and detect-and-forward (DF). We then optimize the location and power allocation of the relay to minimize the total energy consumption. The optimum location is found in two-dimensional space for constrained and unconstrained scenarios. We then optimize the total energy consumption over the spectral efficiency and derive expressions for the optimal spectral efficiency values. We use numerical simulations to verify our results. Finally, we focus on energy efficiency of multi-relay systems by considering a dual-relay cooperative system using DF protocol with full diversity. We propose a location-and-power-optimization approach for the relays to minimize the transmit radio energy. We then minimize the total system energy from spectral efficiency perspective for two scenarios: throughput-constrained and bandwidth-constrained configurations. Our proposed approach reduces the transmit energy consumption compared to an equal-power allocated and equidistant-located relay system. Finally, we present an optimal transmission scheme as a function of distance by considering single-hop and multi-hop schemes. The overall results imply that more relays are required as the transmission distance increases in order to maintain a higher energy efficiency.
• Item
Fundamental energy requirements of information processing and transmission
Angley, Daniel Michael ( 2015)
This thesis investigates fundamental limits on the energy required to process and transmit information. By combining physical laws, such as the second law of thermodynamics, with information theory, we present novel limits on the efficiency of systems that track objects, perform stochastic control, switch communication systems and communicate information. This approach yields results that apply regardless of how the system is constructed. While the energy required to perform an ideal measurement of a static state has no known lower bound, this thesis demonstrates that this is not true for noisy measurements or if the state is evolving stochastically. We derive new lower bounds on the energy required to perform such tracking tasks, including Kalman filtering. The goal of stochastic control is usually to reduce the entropy of the controlled system. This is also the task of a Maxwell demon, a thought experiment in which a device or being reduces the thermodynamic entropy of a closed system, violating the second law of thermodynamics. We demonstrate that the same arguments that `exorcise' Maxwell's demon can be used to find lower bounds on the energy consumption of stochastic controllers. We show that the configuration of a switching system in communications, that directs input signals to the desired outputs, can be used to store information. Reconfiguring the switch therefore erases information, and must have an energy cost of at least $k_B T \ln(2)$ per bit due to Landauer's principle. We then calculate lower bounds on the energy required to perform finite-time switching in a one-input, two-output MEMS (microelectromechanical system) mirror switch subject to Brownian motion, demonstrating that the shape of the potential that the switch is subject to affects both the steady-state noise and the energy required to change the configuration. Finally, by modifying Feynman's ratchet and pawl heat engine in order to perform communication instead of doing work, we investigate the efficiency of communication systems that operate solely using the temperature difference between two thermal reservoirs. The lower bound for the energy consumption of any communication system operating between two thermal reservoirs, with no channel noise and using equiprobable partitions of heat energy taken from these reservoirs, is found to be $\frac{T_H T_C}{T_H-T_C} k_B \ln(2)$, where $T_H$ and $T_C$ are the temperatures of the hot and cold reservoir, and $k_B$ is Boltzmann's constant.


text:
Having spent the past couple weeks wrestling with what to write about I’ve settled on this helpful tidbit. My constant need to refer to this basic workflow is some indication that (a) it’s useful, and (b) I really ought to have better recall. Although there are helpful command-line tools out there that steamroll over much of what is to follow, I myself am in a bit of Homebrew limbo, not to mention the following is arguably more enlightening.

Aim of the game: Create a repo from a command-line interface and push the local repo to GitHub. This gives us that much more control in setting up the initial directory structure before starting out on a project. Having run a git init in our root directory and made our first commit we can get started.

First we need to create the GitHub repo from the command-line. To do this we use the unix curl command, which allows us to interact with the server. Following the GitHub API we write,

where we specify our GitHub username and the repo name in place of ‘USER’ and ‘REPO’ respectively. The option -u specifies the username (and password) to be used, and -d indicates a POST request. At a minimum we should specify our username and the repo name. We can easily include more information in the POST request as required. We can also specify further authentication following the -u parameter if desired. Following exactly as above, we will be prompted to type our password after entering the curl command.

where ‘USER’ and ‘REPO’ are our GitHub username and the new repo name, as above. This creates an alias for the URL specified, in all subsquent push commands we need only specify our remote with origin, which will be understood as the full URL.
It should be said that this step require you to already have set up an SSH key with your GitHub account, if not you can follow the steps here. You can just as easily push to an HTTPS URL, which is this case would correspond to, https://github.com/USER/REPO.git.


text:
# Viscosity

Written by Jerry Ratzlaff on . Posted in Fluid Dynamics

## Viscosity

Viscosity ( $$\eta$$ (Greek symbol eta) ) is the measure of the internal friction/resistance to the flow of a liquid. Lower viscosity fluids flow easily in pipes where high viscosity fluids have a have a higher pressure drop.  Viscosity of fluids is typically temperature dependent and is not affected as dramatically by pressure as gas viscosity. In fact, typically as a liquid temperature increases, the velocity decreases. When a gas temperature increases, the viscosity increases!

There are two different types of viscosity, dynamic viscosity and kinematic viscosity. Higher the viscosity the more resistance to flow. Lower the viscosity the less resistance to flow. Temperature also plays a part, lower the temperature, higher the resistance. Higher the temperature, lower the resistance. A Viscometer meter can be used to measure viscosity.

## Coefficient of Viscosity

Coefficient of viscosity ( $$\eta$$ (Greek symbol eta) ) (also called viscosity coefficient) is the tangential friction force required to preserve a unit velocity gradient between two parallel layers of liquid of unit area.  The greater the coefficient of viscosity the greater the force required to move the layers at a velocity.

### Coefficient of Viscosity Formula

$$\eta = \frac { Fl } { Av }$$          $$coefficient \; of \; viscosity \;=\; \frac { tangential force \;\;x\;\; distance \; between \; the \; layers } { area \;\;x\;\; velocity }$$

Where:

$$\eta$$ (Greek symbol eta) = coefficient of viscosity

$$F$$ = tangential force

$$l$$ = distance between the layers

$$A$$ = area

$$v$$ = velocity

Solve for:

$$F = \eta \frac { Av } { l }$$

## Dynamic Viscosity

Dynamic viscosity ( $$\mu$$ ) (also called absolute viscosity of simple viscosity) is the force required to move adjacent layers that are parallel to each other at different speeds.  The velocity and shear stress are combined to determine the dynamic viscosity.

### Dynamic Viscosity Formula

$$\mu = \frac {\tau} {\dot {\gamma}}$$          $$dynamic \; viscosity \;=\; \frac { shear \; stress } { shear \; rate }$$

Where:

$$\mu$$ (Greek symbol mu) = dynamic viscosity

$$\tau$$ (Greek symbol tau) = shear stress

$$\dot {\gamma}$$ (Greek symbol gamma) = shear rate

## Kinematic Viscosity

Kinematic viscosity ( $$\nu$$ ) is the ratio of dynamic viscosity to density or the resistive flow of a fluid under the influance of gravity.

### Kinematic Viscosity Formula

$$\nu = \frac {\mu} {\rho}$$          $$kinematic \; viscosity \;=\; \frac { dynamic \; viscosity } { density }$$

Where:

$$\nu$$ (Greek symbol nu) = kinematic viscosity

$$\mu$$ (Greek symbol mu) = dynamic viscosity

$$\rho$$ (Greek symbol rho) = density

## Newton's Law of Viscosity

Newton's law of viscosity states that shear stress between adjacent fluid layers is porportional to the velocity gradients between the two layers.  The ratio of shear stress to shear rate is a constant for a given temperature and pressure.

### Newton's Law of Viscosity Formula

$$\tau = \mu \frac {d \nu}{dy}$$          $$shear stress = dynamic \; viscosity \;\;x\;\; rate \; of \; shear \; deformation$$

Where:

$$\tau$$ (Greek symbol tau) = shear stress

$$\mu$$ (Greek symbol mu) = dynamic viscosity

$$\frac {d \nu}{dy}$$ = rate of shear deformation

## Reduced Viscosity

### Reduced Viscosity Formula

$$\mu_r = \frac {\mu_i} {c}$$          $$reduced \; viscosity \;=\; \frac { relative \; viscosity \; increment } { mass \; concentration }$$

Where:

$$\mu_r$$ (Greek symbol mu) = reduced viscosity

$$\mu_i$$ (Greek symbol mu) = relative viscosity increment

$$c$$ = mass concentration

$$\mu$$ (Greek symbol mu) = dynamic viscosity

$$\mu_s$$ (Greek symbol mu) = viscosity of a solution used

Where:

$$\mu_i = \frac { \mu \;-\; \mu_s } { \mu_s }$$

## Relative Viscosity

Relative viscosity ( $$\eta_r$$ ) is the ratio of the viscosity of a solution to the viscosity of a solution used.

### Relative Viscosity Formula

$$\eta_r = \frac {\eta} {\eta_s}$$          $$relative \; viscosity \;=\; \frac { viscosity \; of \; a \; solution } { viscosity \; of \; a \; solution\; used }$$

Where:

$$\eta_r$$ (Greek symbol eta) = relative viscosity

$$\eta$$ (Greek symbol eta) = viscosity of a solution

$$\eta_s$$ (Greek symbol eta) = viscosity of a solution used

## Viscosity Index

Viscosity index ( $$VI$$ ) is a measure of a fluid's sensitivity to change in viscosity with change in temperature.

### Viscosity Index Standards

• ASTM Standards
• ASTM D2116 - Standard Method for Conversion of Kinematic Viscosity to Saybolt Universal Viscosity or to Saybolt Furol Viscosity
• ASTM D2270 - Standard Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100°C
• ASTM STP168 - ASTM Viscosity Index Calculated from Kinematic Viscosity
• ASTM STP532 - Viscosity Testing of Asphalt and Experience with Viscosity Graded Specifications

## Viscosity of Slurry

The viscosity of slurry ( $$\eta_s$$ ), slurry being a mixture of liquids and solids, is comparitive to the viscosity of the liquid phase.

### Viscosity of Slurry Formula

$$\eta_s = \frac {\mu} {\eta_r}$$          $$slurry \; viscosity \;=\; \frac { dynamic \; viscosity } { relative \; viscosity }$$

Where:

$$\eta_s$$ (Greek symbol mu) = slurry viscosity

$$\mu$$ (Greek symbol mu) = dynamic viscosity

$$\eta_r$$ (Greek symbol eta) = relative viscosity

## Viscosity of a Mixture

This calculates the effective viscosity of gasses.

### Viscosity of a Mixture formula

$$\mu = \frac {\sum ({\mu_i y_i \sqrt{M_i}})} {\sum ({y_i \sqrt{M_i}})}$$

Where:

$$\mu$$ (Greek symbol mu) = dynamic viscosity of the gas mixture

$$\mu_i$$ (Greek symbol mu) = dynamic viscosity of a gas component

$$y_i$$ = mole fraction or percent of gas component

$$M_i$$ = molecular weight of gas component


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# maths genie percentages solutions

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# Understanding a comment by Thurston

In page 359 (right after Theorem 2.3) of the following paper

Thurston, William P., Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Am. Math. Soc., New Ser. 6, 357-379 (1982). ZBL0496.57005.

W. Thurston states (and I quote)

A hyperbolic structure on the interior of a compact manifold $$M^3$$ has finite volume if and only if $$\partial M^3$$ consists of tori, with the single exception of Example 2.1, which has no hyperbolic structure of finite volume.

His example 2.1 is simply the product of the 2-torus by the closed interval, $$M = \mathbb{T}^2\times I$$.

What did Thurston had in mind when he stated that? Why did he left the solid torus $$\mathbb{D}\times\mathbb{S}^1$$ out of this, say, classification? I ask that to understand if it was just a blunder or if he actually had a reason to exclude this case (which can be obviously obtained by the quotient $$\mathbb{H}^3/\{\tau\}$$, where $$\tau$$ is a parabolic translation in $$\mathbb{H}^3$$)?