Why is my smoke simulation not showing up with the default renderer? As per the picture, the fire/smoke is not showing up with the blender internal renderer. Is this a common issue?  Are there any obligatory steps to make it show like it does in the preview?  Thank you.
Smoke Generator does not work When I try to run the smoke simulator in Blender 2.62 I get no smoke at all. When I attempt to render, the only output is a black screen. What am I doing wrong here that is making the smoke simulation not render?  Is it a mistake I made, or a problem with Blender version 2.62?
How do I remove the proprietary ATI drivers? I've installed 11.10 and the proprietary ATI drivers using "additional drivers" The performance of my system is absolutely awful and it shouldn't be. I tried to remove the proprietary drivers using the Additional Drivers tool and it appears to remove them. However after I reboot I cant get back into my desktop properly (the panel and launcher go missing). This doesn't seem to be an isolated problem in 11.XX.  covers how to restore the desktop (panel and launcher), but the guide doesn't fix my problem though.   Whenever I do sudo unity --reset it runs through its normal processes until it hangs at setting update "run_key" and never gets past that. I must reinstall the proprietary drivers using jockey-text or jockey-gtk in order to get back to my proper desktop.   Interestingly enough the system performance seems improved while it is in its "broken" state (missing panel and launcher).  I think restoring the default drivers may solve my problems but I cant figure out how to do it. Why is Leia okay with being tracked to Yavin? In A New Hope, after Han and Luke fight off the TIE Fighters while escaping from the Death Star, these exchanges happen:             INT. DEATH STAR - CONTROL ROOM             Darth Vader strides into the control room, where Tarkin is             watching the huge view screen. A sea of stars is before him.                                   TARKIN                      Are they away?                                   VADER                      They have just made the jump into                       hyperspace.                                   TARKIN                      You're sure the homing beacon is                       secure aboard their ship? I'm taking                       an awful risk, Vader. This had better                       work.             INT. MILLENNIUM FALCON - COCKPIT             Han, removes his gloves and smiling, is at the controls of             the ship. Chewie moves into the aft section to check the             damage. Leia is seated near Han.                                   HAN                      Not a bad bit of rescuing, huh? You                       know, sometimes I even amaze myself.                                   LEIA                      That doesn't sound too hard. Besides,                       they let us go. It's the only                       explanation for the ease of our                       escape.                                   HAN                      Easy... you call that easy?                                   LEIA                      They're tracking us!                                   HAN                      Not this ship, sister.             Frustrated, Leia shakes her head.                                   LEIA                      At least the information in Artoo is                       still intact.                                   HAN                      What's so important? What's he                       carrying?                                   LEIA                      The technical readouts of that battle                       station. I only hope that when the                       data is analyzed, a weakness can be                       found. It's not over yet!   Leia doesn't even know if there is a weakness, let alone whether they'll find one. If she (correctly) realizes that they're being tracked, why doesn't she suggest that the Millennium Falcon go somewhere besides Yavin 4 instead of just frustratedly shaking her head? I know the plans , but it seems unlikely that continuing to fly to the secret Rebel base in the ship she's certain is being tracked is the only option. They are already in hyperspace when this conversation happens, but they apparently have time to talk for a little while after, so presumably they'd have time to stop and reroute. What did Bill do to make Hanzo mad? I have a Windows 7 64-bit installation on my current PC, and I want to move it to a new PC .. I do NOT have a problem with licensing, as the new system already has its own new Windows 7 license, which I intend to use.  I want to use my existing installation, because it has 3 years worth of installed office related software development software, which installing again could take weeks !  I'm assuming that I can use Windows 7's Backup and Restore feature to backup a system image to a network location, then restore that network stored system image on the new PC ?  Again, like I said, I don't have a WIndows 7 licensing issue, as both machines are corporate provided and come with their own licences. win 10 home how to rename my home directory and account login Do synonyms have exactly the same meanings as each other in all contexts? I have a question about synonyms: I'm wondering for quite a while if synonyms always stand exactly for the same thing. Is there sometimes a little difference in meaning? Let's use clumsy and clunky for example. Can I always replace clunky with clumsy? Since I'm not a native speaker, as you may have noticed, I cannot tell them apart. According to my dictionary, they mean exactly the same. Is it always true?  If the question has already been ask, sorry I couldn't find it.  Edit: Thank you very much! Every answer has helped a lot and thank you so much for it. I really appreciate it. I've voted all your answer because every answer helped me in a different way. I've accepted Barrie England's answer because it was the best one in my case. But all of yours were really good and helpful :) In CodeGear Delphi 2007, how can I turn specific warnings and hints off? I am attempting to turn off H2077 - Value assigned to 'varname' never used. Does lower resolution result in better signal-to-noise ratio, given the same sensor size? With the release of the Canon EOS R5/R6, there have been many saying that they prefer the 20MP R6 camera over the 45MP R5 camera, due to its lower price but also because of &quot;better low-light performance.&quot; In camera comparison sites, I've also seen the sentiment that bigger pixels (lower resolution) = better low light performance. However, is this a myth? Here is my reasoning for this. Let's say that there are two full-frame sensors, one 20MP sensor and a 80MP sensor. According to many people, the 80MP sensor would be terrible for low-light photography since the pixels are smaller. However, each 2x2 block in the 80MP sensor could be mapped, and in post averaged to a 1x1 pixel, effectively making the image into a 20MP image. This could also be done in-camera as an option. Would this image have just the same amount of usable information as the image from the 20MP sensor? In other terms, since both sensors are 35mm sensors (i.e. equal area), the amount of light falling onto the sensor is the same. So, at the end of the day, there will be no difference between a 20MP image and a 20MP image that has been downsampled from 80MP, right? So their low-light performance would be the same. You even get an advantage from having the option to get 80MP photos if you wanted. Why do people say that a lower-resolution camera has better low-light performance, when it should theoretically be the same as one with higher resolution (all other factors equal)? Note: this question is about sensors with the same size but with different pixel density. Hence, it is different from  asking about different sensor sizes. I am aware that a full-frame camera will have better low-light signal due to its bigger area than an APS-C sensor, but what about the similarly sized sensors? Is a join optimized to a where clause at runtime? If a $\{a_n\}$ diverges, so $a_n \rightarrow + \infty$, how to find sequence $\{b_n\}$ such that $\sum |b_n|<\infty$ but $\sum |a_n||b_n|$ diverges? If we are given any sequence of real numbers $\{a_n\}$ diverges, so $a_n \rightarrow + \infty$, how can we find a sequence $\{b_n\}$ such that $\sum |b_n|$ converges but $\sum |a_n||b_n|$ diverges?  I want to use this fact in another problem but don't immediately see how to prove it. Where and how can I view skype logs (Chat and Voice history)? Where and how can I view my skype logs (Chat and Voice history)? Does it get saved somewhere in PC by default? Can I choose a folder location myself to start saving voice and chat history? I have a Non Root User abc and want to give the user sudo rights only to restart, start or stop - mysqld service. i.e sudo service mysqld restart. How can I do it so?     I would like to do the same to usergroup too!! Writing string at the same position using Console.Write in C# 2.0 Logit or arcsine transformation for proportion in meta-analysis When you reach 10k rep, you get to see other users deleted answers. I just think it's dishonest as a system to call a feature "delete," but show that to totally random strangers in the backroom. Due to the nature of the rep, given enough time, all of the regular users would eventually reach 10k rep, no matter how insensitive or jerk they are.  There's usually a good reason why people hit "delete" to their own work. They regret writing something totally bullshit or maybe they spoke too much about their own work that they shouldn't have.  First of all, is there really a practical reason why 10k rep user should view self-deleted answers? Does the benefit really outweigh the creepiness? (Imagine if total strangers could see your deleted messages in Gmail?)  What suggestions could we make to change it? Borrowing from the email metaphor, I think it makes sense for users to "empty trash" so the items gets seriously deleted. Imho, however the deletion is implemented is not important. It could be a soft delete, SQL Delete, or physical destruction of metal plates as long as the user or the 10k rep user no longer sees the deleted item.  What are the workarounds, if we can't change the behavior?  The least we could do is stop calling it "delete" if it's actually not a "delete." Call it "hide from public" or "actually not delete" or something.  Related user voice:        Edit: Recovery of a deleted-but-good answer is a practical benefit for the community, but I see that as creepy dangerous feature. IMHO, each user should have the rights to take back his or her perfectly good answer, and that the protecting privacy outweighs the benefit or creepiness.   I understand that it's all possible to find things out from Google cache or wherever, but why promote this as default behavior? As 10k rep user, I see others deleted answers by default. @codinghorror wrote:     But like any version control system, deletions are illusory.   Anything and everything that's implemented by human is more or less illusory including phone system and the medical records. Should the government start tapping into phone lines and scanning for words? Remember, this is a question of should had we given the opportunity to design a system. Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove  $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$  This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.   $$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$  First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?  Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$  which leads to  $$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$  Since $(n-1)$ is a constant, we can throw it out front of the integral:  $$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$  I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$  $$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$  According to Wikipedia as noted , this simplifies to:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$  Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?  Assuming knowledge of equation 5, I see how to rewrite it as  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x  - (n-1) I_{n} \tag{6}$$  and solve for $I_n$. I had tried exploiting the fact that  $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$  and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$  which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$  but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$ Prove that if $X$ and it's closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected. Can anyone give me a proof of the statement that if $X$ and its closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected?  Thank you. Could gravity accelerate light? Gravity causes anything with energy to accelerate toward the source. Black holes, for example, have such strong gravity that they pull in light and don't let any escape. But can acceleration still apply to light? The speed of light is constant, of course, but why are photons affected by gravity yet aren't accelerated by it?  Edit: My main question is why photons aren't affected in the same way as most other particles. I'm perfectly aware that it cannot surpass lightspeed, but I want to know what makes it unaffected by acceleration while other particles are affected. Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot Connection between rank and matrix product I have a problem understanding the following:     Let $A$ be an $m \times n$ matrix and let t $\in \mathbb{N}$. Prove that   $\operatorname{rank}(A)\leq t$ if and only if there exists an $m \times t$ matrix $B$ and a $t \times n$ matrix $C$ so that $A = BC$.   I know what a rank is but I can't make a connection between the rank and the existence of two matrices such that $A = BC$.