Extruding outwards from a cylinder Im trying to extrude points outwards like in this video at 24:50   This is what is happening.
Extrude Faces along Local Normals I am trying to extrude faces of a cylinder along local normals. 3dsmax can do it easy, couldn't find how to do it in blender.  The extrude region option extrudes it as shown in the first example. I want it done like the 2nd example.
What is serial voting and how does it affect me? I just noticed that I lost a bunch of points from my reputation score, and I used the "reputation" tab on my user profile page to try and track down the cause.  During my investigation, I noticed there was an unusual event of type "reversal". In the normal place of a question title, it says "voting corrected".     What does this mean, and what caused it?  Did I do something wrong? Why did I lose all of that reputation? Is the system punishing me for leaving too many good answers?  Is there any way to earn it back? Ubuntu 14.04 Wireless not working: no network interface. Atheros [168c:003e] device Is there a C++ decompiler? What is the correct pronunciation of the word “processes”? Moderncv has a line of empty space between address and phone/email, like this:   MWE:  \documentclass[12pt]{moderncv} \moderncvstyle{banking} \name{Sherlock}{Holmes} \address{221B Baker Street}{London}{U.K.} % Your current address \phone[mobile] {+44~(0)71234 00000}    % Your mobile phone number \email{superdetective@mail.com}  \begin{document} \makecvtitle \end{document}   How do I remove/decrease that space by shifting the phone and email line up? Harry is a half-blood and Ginny is a pure-blood, so what does that make his children? I've been told it means they are pure-bloods but I can't be sure. What do you think? What is the preferred way to remove spaces from a string in C++? I could loop through all the characters and build a new string, but is there a better way? How can I correct a reviewer's opinion? My edit to an answer was accepted by two of the three reviewers, with the third rejecting with a claim that part of my edit was incorrect.  In fact, my proposed edit was entirely and demonstrably correct, and the reason given for rejecting is invalid.  How can I let that reviewer know about their error, and have them change their review to "Accept" - changing stats for both of us?  Please note:  This is not the same question as asking for an appeal of an edit that was rejected.  It's asking what I can do about one of the three reviewers, a dissenter who voted against my edit based on their incorrect information. Where can I get the original Unix? I have a function with the same name, but with different signature in a base and derived classes. When I am trying to use the base class's function in another class that inherits from the derived, I receive an error. See the following code:  class A {     public:     void foo(string s){}; };  class B : public A {     public:     int foo(int i){}; };  class C : public B {     public:     void bar()     {         string s;         foo(s);     } };   I receive the following error from the gcc compiler:  In member function `void C::bar()': no matching function for call to `C::foo(std::string&amp;)' candidates are: int B::foo(int)   If I remove int foo(int i){}; from class B, or if I rename it from foo1, everything works fine.   What's the problem with this? In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it? I am a new user of Ubuntu 12.04.   I installed Ubuntu and Windows XP, and I want to use both operating systems. When I restart or boot my PC the GRUB menu does not appear.  How can I solve this problem? SQL command based on PHP strstr or LIKE Get and Set Screen Resolution I installed Ubuntu 12.10 and entered a password in SeaHorse. I would like to remove the password of "Default" but I get this error :  GDBus.Error:org.freedesktop.Secret.Error.NoSuchObject: The collection does not exist   Is there a fix or a way to bypass modify the password ? I am trying to prove the following statement but have trouble  comprehending/going forward with some parts! Here is the statement:     If $A$ is any set, then  $|A|$ $&lt;$ $|P(A)|$   Here is what I have so far:  We need to show that there is an injection from $A$ to $P(A)$ but not a surjection.   A natural choice for an injection is the function $ f(x)$ $=$ $\{x \}$, which in plain English, takes any element $x$ (that is in $A$) and sends it to the one-element set $\{x \}$. Thus $f(x)$ is injective!  To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from $P(A)$) to elements (from $A$) right?  If the above is correct, my problem arises here. I am not sure how to prove that $f$ is not surjective. Unfortunately, I am easily confused by notation so please explain in English.  Thank you in advance!! :) What does the author mean by 'back to back' and 'hidden city' ticketing hacks? What are the browser and resolution stats for SO users? Google Analytics (and some log parsers) will give you breakdowns on usage of different browsers, different versions of those browsers and screen resolutions.  I'm just curious to know what SO users user. Any chance we could get some stats on that? You may view this as commercial information, in which case I understand (as i'ts relevant to marketing/advertising).  I'm also curious to know the # of users who read and/or post from mobile devices and if so which ones.  UPDATE: Please refrain from personal anecdotes ("I use..."). I'm interested in actual usage, not personal stories. Thankyou. Let A,B be in $M_{mxn}(\mathbb{R})$  Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0}  I started with a proof by contradiction, since we know that rank(A + B) ≤ rank(A) + rank(B). I assume that if rank(A + B) &lt; rank(A) + rank(B), then col(A) ∩ col(B) = {0}. I realize that we don't have to do the > case. However I am a little confused on how to show col from rank. Polynomial bounded real part of an entire function Let $f(z)$ be an entire function whose real part is bounded by a polynomial in $|z|$. Does it follow that $f(z)$ is a polynomial?  Or, without loss of generality and more suggestively $$(f(z)=\sum_{k=0}^\infty a_k z^k \quad \land \quad \operatorname{Re}(f(z))\leq |z|^n)\quad \forall z\in \mathbb{C} \qquad \Rightarrow \qquad f(z)=\sum_{k=0}^n a_k z^k \quad\forall z\in \mathbb{C}$$