How much zoom is needed for a distance of 600 ft? How much zoom is needed for a distance of 600ft?
What is the formula for percent of frame filled at a specific distance and focal length? I am thinking about buying a new lens specifically to take photos of distant targets (birds, bears, etc). I currently own the Canon 70-300 IS USM, which is a nice lens, but I find that if I want to take a photo of anything further away than about 10 meters, the target does not fill a large amount of the frame (meaning I have to significantly crop in post production).  I am currently looking at the Canon prime 400mm, however I would like to determine at what distance an object will fill a reasonable proportion of the frame.  Is there a formula or rule of thumb I can apply that will help me in this situation? I realise the size of the target I am shooting will play a role here, so if we need to make an assumption about the size of the target please let me know.
How to find and replace using sed on records starting with Specific number Consistency of an order statistic in exponential distribution Each Stack Exchange site should provide users the option of using a dark mode design/theme.  Why is this feature missing?  The major IDEs are already a letdown regarding dark color schemes. I have notifications in the OSX Notification Center, as for most of my apps I am using the setting, "Show in Notification Centre: 10 Recent Items" This is a good setting but rather than have a very full Notification Center, I would still like to clear it from time to time with 1 click.  I am using Yosemite 10.10.3 What is the correct way of installing BOINC software? I try to understand why for a connected Lie group $G$ the first fundamental group $\pi_1(G)$ is abelian, and mainly why the second fundamental group is trivial $\pi_2(G)=0$?  Thanks for anyone who give me references for a 'simple proof' of these results How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?  It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me. Residue theorem with $\frac{1}{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) d\theta$. The following integral should be doable using the residue theorum:     $$\frac1{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) \,d\theta$$ It happens sometimes to receive or see a very good answer that arrives very late and that would deserve more upvotes. What can be done to give a boost to them? I would happily spend some of my reputation points to be able to give them such boost. This is not merely a question of reputation, but the idea is to let the answer &quot;bubble up&quot; faster so that it can be seen easily by other people to look into question. I hope I explained my question so that is understandable. Most of you got the point and there are some interesting proposals like the one that proposes to put randomly an answer at top. But I do have not enough experience to say if there is one that could solve the issue or not. I would be curious to know what Jeff thinks about it, if it feels there is something that can be improved or not. Would a wormhole be able to move? What makes matter travel along geodesics? The relativistic explanation of gravity is geometric, the motion of a body in a field of space-time distortion can be described as being at rest and travelling along a geodesic of that field, but why does it travel along that geodesic? What causes it to move? Can't mount an IDE HDD - "can't read superblock" error message I'm trying to mount my HDD,  /dev/sdb, which is connected via an enclosure but every time I try do so I get an error message   mount: /dev/sdb: can't read superblock   How do I solve this problem and mount my HDD? Browse local copies of websites when possible (Firefox with slow internet connection) I am on a slow internet connection. I would like to use Firefox to load local copies of websites when it exists and is relatively recent (say one week).   An example usage would be browsing a lot of Wikipedia webpages, coming back to the same ones on a regular basis.  I would like to avoid having to manually specify to save single pages on each website I visit (which is what I guess I would need to do with a plugin like ). And I would also like to avoid having to switch between "work offline" or "work online".   Any suggestions? Converting transparent gif / png to jpeg using java I'm pulling items out of the DOM with jQuery and want to set a property on an object using the id of the DOM element.   Example  const obj = {}  jQuery(itemsFromDom).each(function() {   const element = jQuery(this)   const name = element.attr('id')   const value = element.attr('value')    // Here is the problem   obj.name = value })   If itemsFromDom includes an element with an id of "myId", I want obj to have a property named "myId". The above gives me name.  How do I name a property of an object using a variable using JavaScript? While revising for exams, I came across a question where at the end of it, we had to determine whether the below series was convergent or divergent:  $$\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $$  Unfortunately, other than knowing that as the series $ \sum_{n=1}^\infty \frac{(-1)^n}{n} $ converges, this series will most likely converge as the $(2 + (-1)^n)$ terms are bounded, I can't see a way of using this to determine if it converges or not.  Here's a list of some of the other ideas I've tried, which don't seem to get me anywhere:   Generalising the series - By this, I mean replacing the $-1$ for $z$ and then try and use the ratio test to see whether $-1$ is inside or outside or on the boundary of the circle of convergence - however, when I did so, I believe that the ratio test is inconclusive for all values of $|z|$, which while disappointing, is an interesting feature of the series.  - While this is usually a nice tool to use for when dealing with awkward sums, I can't see any good choice of sequences $ (a_n) $ and $(b_n)$ to choose to use. Summing consecutive terms - By this, I mean evaluating the series by looking at the sum of    $$\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k(2+(-1)^{2k})} + \frac{(-1)^{2k-1}}{(2k-1)(2+(-1)^{2k-1})}$$  $$ \Rightarrow \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{3(2k-1)}$$  $\quad$ but then as we are left with parts of the harmonic series this doesn't seem like a nice way to  $\quad$evaluate it.  Other than that, I'm completely out of ideas, so any new ones (or ways to make my old ones work) would be much appreciated! We need a profanity filter I searched SE.Meta and was unable to find such question.  Someone just swore at us in a  as an  (warning: profanity, may be NSFW), sure, Arqade (profanity in phrases in games) or some SE (ELU, ELL or Anime.SE maybe?) can sometimes contain profanity as a small parts of question.  I think that a part should be added to the review system that will automatically hide questions/answers with profanity, just like the close vote system, the question/answer will be on hold and will be voted to be shown or be deleted.  Tip, if this gets implemented, u can be typed as \/ or v, and most other can be written as leet (l = 1 etc). Maybe it can be called "suspicious activity", and can be triggered when there is too many number or Unicode characters. (I can write hi as |-| |, |-|I, |-|l, l-li, hî, 1-1i, H1 and many more) Find the limit $ \lim_{x \to \infty} \frac{e^{x}}{(1+\frac{1}{x})^{x^{2}}} $ Find the limit $$ \lim_{x \to \infty} \frac{e^{x}}{(1+\frac{1}{x})^{x^{2}}}. $$  I know that the limit is of indeterminate form type $\frac{\infty}{\infty}$, but it seems using L'Hopital's rule directly does not help here.  Do I somehow use the fact that $\lim_{x \to \infty} (1+\frac{1}{x})^{x}=e$? I only ask questions. Am I unwanted? Possible Duplicate:          I've been  for maybe a year to get answers to questions I couldn't find on my own.  I always make an attempt first to find the answer on my own.  As of this posting, I've asked 14 questions.  I only once gave an answer, and it was to a non-technical question, and only because I was really bored and had too much coffee to drink.  I wasn't intending to answer any other questions in the future, because on on my best days I'm only a mediocre programmer, and there are some real geniuses here on StackOverflow who can answer other peoples' questions faster and more accurately than I can.  It actually is my nature.  I am an Aquarian.  We are asker-listeners.  Today I discovered Meta Stack Overflow and started browsing around.  I became familiar with the concept of the "Help Vampire".  I seem to qualify.  I continued reading more posts on the topic and got the distinct impression that people who only ask, and never answer, are less than welcome here, rather despised.  So, should I just go away?  Or should I start contributing answers just to have a answer-to-question ratio of 3-to-1 (I read somewhere that's the average) even though I am less than one eighth as knowledgeable as the 10K+ guys and gals?    Or should I continue as I have begun?  I mean, doesn't a 0-questions-asked user like Jon Skeet need a Blankman?  Yin and yang, you know? The 'nofollow' attribute is set in user profile website link... why? I understand that from an SEO perspective it is beneficial for Stack Overflow to have the  on the website link for user profiles. Especially if people will just create an account to get a backlink. What I think however, is that contributing members of Stack Overflow should be awarded with this backlink. I think that if someone contributes and has some decent reputation, then they should not have nofollow set on their website link in their user profile. Considering the amount of users that are backlinking to Stack Overflow with flair, I really think that Stack Overflow should hook us up with some search engine love.