Which one is correct and why?     If you knew how bad things were, you would not go on vacation right now.   or     If you knew how bad things are, you would not go on vacation right now.   This is assuming that "things" are actually bad right now (that's not up for discussion).
I have a basic grammar question about the past tense. If I met a guy yesterday, which tense should I use about his name today?  For example,     "Yesterday I met a guy, his name is/was John."
Does wide angle equivalent in crop sensor skew image? I see many posts on both this forum and elsewhere the discuss the use of a wide angle on a crop sensor.  For example, they discuss how at 14mm lens is 14mm on a 35mm film or full frame, but with a 1.6x crop sensor it is an effective 28mm.  My question is, is a 14mm on a crop sensor the same image as a 28mm on a full frame.  Or does "effective" include some other conotation.  I read that a 14mm can create distortion on a full frame and some vignette, so people seem to say using it on a crop takes out some of those edge problems. My confusion is how the lens is actually taking in the extra image.  Reading kenrockwells blog I learned that wide angle keeps lines straight, while fish eye actually skews the lines.    So with this as my understanding, I am still a bit confused how a lens that is curved in order to get a greater angle of view, will be the same as a cropped version of a less curved lens.  Or does "effective" only refer to the objects to the sides but not to the distortion of the image.  Or, if it is different, is it so small that for photography it does not matter?  A technically description would be welcome as long as a basic understanding of the situation.  Thanks!  Edit: To be more clear after the comments,  I am specifically talking about the end resulting 2d projection from 3d space.  I have read both those previous answers.  the first one is closer to what I am talking about.  However, it is still confusing for me.  I have experience with 3d modeling and projection matricies, which might be confusing me with this.  But for example, i dont understand the images having a 50 mm lens being drawn Starting from the sensor, so it shows a different fov for crop and full frame . If the lens is the same, it takes the same amount of information from the world, which makes a ray from a certain degree off the last lens projected into a smaller space, there is no projection from the sensor, so drawing the line from the sensor doesnt make sense to me.  The distance of the sensor to the rear of the lens must have some impact on the light projected but is not represented in the images.  Further, again from that first answer it says cropping is the same as zoom, but from what i understand from how perspective works, it is different, since a wide angle will project lines differently than a small fov to the same size, so cropping the center of a wide angle and zooming into something is very different.  This is easily recreated in an opengl application with varying fov projection matrix, i imagine in painting class people learn this as well.  A fisheye would be a good example, since if cropping was the same as zooming, a fish eye lens would keep the center exactly like a normal lens and then a gradient weighted towards the outside would rapidly create a warped perspective, but from what I see, it is even.  To me those images just look like its comparing cropped with full frame in regards to orthogonal projections. If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not? In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?  I tried by assuming a set is equal to its interior points and contains its limit points.  A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?  Anyway, it must then be unbounded.  If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$. Can command substitution be nested in variable substitution? How do we define capillary number in 2D (two dimensions) In 3D the capillary number $Ca$ is defined as: $$Ca=\frac{\nu \rho U}{\gamma}$$ where $\nu$ is the kinematic viscosity ($m^2/s$), $U$ is the velocity ($m/s$)and $\gamma$ is the interfacial tension ($N/m$) and $\rho$ is the density in $kg/m^3$ . However in 2D I will have $\rho$ in units of $kg/m^2$ (2D simulation), so how to find capillary number? How to avoid ArrayIndexOutOfBoundsException or IndexOutOfBoundsException? How to handle relative urls correctly with a nginx reverse proxy How to fix the weird surface in fluid object Cannot determine size of graphic I'm trying to include graphics in my Latex-file, which I compiled with latex+dvipdf on OS X. Latex however returns this error: "Cannot determine size of graphic"  My graphic is exported from PowerPoint, so I have tried both .pdf and .png. I get the same errors for both.  Sample code that doesn't work:  \begin{figure}[htb] \begin{center} \leavevmode \includegraphics[width=0.8\textwidth]{graph.png} \end{center} \end{figure}   If I have to manually set some measurements of the image, please tell me how I can find out those measurements. I need the image to take up about 80% of text width and be centered. Thanks. I am trying to host some static content in an Amazon S3 bucket and to have that content available through my own domain.  The bucket in question is called "e.tl-dr.org" and I have "index.html" set as the index document. I can visit  and I see the contents of index.html (just "test"). This works fine.  I also have added the CNAME entry for "e.tl-dr.org" pointing to "s3.amazonaws.com" from within my registrar's site. (I am not using the AWS Route 53 DNS services, which is what most tutorials about this seemed to be geared at.) So now "" correctly shows the file, but just the bare domain only gives me S3's standard 403 message.  Is there anything else I need to set up to get this working? When you use the allowframebreaks option in beamer then you get additional frames if the content is too high for one frame. The vertical spacing gets disturbed when you use a list (itemize, enumerate, description), for example, see the following animation. I used the t option (for top alignment) for the frame.     \documentclass[aspectratio=169]{beamer}   \begin{document}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title} \setlength{\mylength}{\baselineskip} \begin{itemize} \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \end{itemize}  \end{frame}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title}  Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\  \end{frame}  \end{document}    I thought of a solution, when I read  question where they explain the different length in lists. The main idea is to check if the current frame is not the first frame of a frame continuation and add the needed vertical space. I thought that \topsep is the needed length but apparently, it is not quite right. It ist almost perfect but it still "dances" a little bit. Which additional length do I need? Here's the animation and the used code.     \documentclass[aspectratio=169]{beamer}  \setbeamertemplate{frametitle continuation}{%     (\insertcontinuationcount)%     \ifnum\insertcontinuationcount&gt;1%     \vspace*{\topsep}%     \else%     %     \fi% }  \begin{document}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title}  \begin{itemize} \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \end{itemize}  \end{frame}  \end{document}   Related        Definition of \itemize in beamer  In beamerbaselocalstructure.sty I found  \renewcommand{\itemize}[1][]{%   \beamer@ifempty{#1}{}{\def\beamer@defaultospec{#1}}%   \ifnum \@itemdepth &gt;2\relax\@toodeep\else     \advance\@itemdepth\@ne     \beamer@computepref\@itemdepth% sets \beameritemnestingprefix     \usebeamerfont{itemize/enumerate \beameritemnestingprefix body}%     \usebeamercolor[fg]{itemize/enumerate \beameritemnestingprefix body}%     \usebeamertemplate{itemize/enumerate \beameritemnestingprefix body begin}%     \list       {\usebeamertemplate{itemize \beameritemnestingprefix item}}       {\def\makelabel##1{%           {%             \hss\llap{{%                 \usebeamerfont*{itemize \beameritemnestingprefix item}%                 \usebeamercolor[fg]{itemize \beameritemnestingprefix item}##1}}%           }%         }%       }   \fi%   \beamer@cramped%   \raggedright%   \beamer@firstlineitemizeunskip% }   But I do not understand where the different kinds of spaces are defined that cause the "dancing".  "Evil" Stuff in beamer  Motivated by the comment of @cfr I searched the beamer documentation for "evil" stuff -- this is what I have found (beamer guide version 3.41):   Do not use subsubsections, they are evil. Never ever use the evil option shrink.  Using this [again shrink] option is very evil. The use of this [allowframebreaks] option is evil. $G \times H \cong G \times K$ , then $ K \cong H$ I already know that if groups $G,H,K$ are finitely generated abelian groups, following is true.  If $G\times K$ is isomorphic to $H\times K$, then $G$ is isomorphic to $H$. I prove this by uniquness of factorization of finitely generated abelian groups.  My questions are   If group $G$ is finite,$G\times K$ is isomorphic to $H\times K$, then $G$ is isomorphic to $H$? Can you give me a easiest proof and it can be proved by projection function on external direct product? If $G,H,K$ are groups.  If $G\times K$ is isomorphic to $H\times K$, then $G$ is isomorphic to $H$. This statement is false. What is a counterexample?   *$\times$ is external direct product Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:  Given that $P(B) = 0$,  $$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$  Is this probability vacuously $0$ of $1$? Can we show that it's one or the other? Did spacetime start with the Big bang? There is by now a captcha-problem Is it true that $0.999999999\ldots=1$? LaTeX macro expander Failed to Login : Invalid session (Try restarting your game) Minecraft Error When I try to connect to servers or join up on lan worlds in Minecraft it says "Failed to Login : Invalid session (Try restarting your game)". Multiplayer used to work fine but now keeps coming up with this error. Can anybody help? How to encrypt external devices using a cross-platform solution? Why do Ruby setters need "self." qualification within the class? Somewhere on Stack Exchange I saw the equation     $$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$   I had never seen this before, so I started trying to prove it. Without success...  Can anyone explain me (so actually prove) why this equation is true?  And can we say the same when replacing the '$2$' by any integer number '$a$'?