I'm trying to recreate these F holes in the body without cutting into the bodies mesh  The F holes and body are two separate meshes.   I am completely lost as to how to cut out a hole in my main body mesh without messing it up.  Is there any way to use the F hole mesh to make a hole in the body mesh?
Coming from SketchUp one of the tools I miss most is the circle draw tool which lets you draw a circle on any face and then you could extrude it in/out. The only tool that I know of that comes close to this is the knife tool but this would be very impractical to cut out with the knife  Example:    Is there any tool in blender that can match this function? also how could you do this with squares?
Is it true that $(R\times S)[G]\cong R[G]\times S[G]$? I am working with dimensionality reduction algorithms. Linear Discriminant Analysis (LDA) is a supervised algorithm that takes into account the class label (which is not the case of PCA for example). I am using Python to do a comparative study between some algorithms.  Why with two classes (k = 2), regardless of the data dimensionality, LDA gives one dimension, i.e. the new subspace is composed of only one dimension? For example if I try to project my dataset (of 100 attributes and 2 classes) to 10-dimensional space by using LDA(n_components=10) I only obtain 1 dimension?  For k&gt;2 I can obtain the number of discriminant vectors that I specified as n_components. How can I get n_components dimensions for k=2? Show that $d_2$ defined by $d_2(x,y)=\frac{|x-y|}{1+|x-y|}$ is a metric I am pretty sure that this is such a stupid, stupid question, but how do you prove that $d_2(x,y)={|x-y|\over {1+|x-y|}}$ satisfying the third condition to be a metric, which is the triangle inequality.   I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me. I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine). I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here? Google chrome is reading characters backwards Is particle superposition reflected in the particle's gravitational footprint? Object selection box in addon In my addon panel, how can I have an object selection box?  And what data type would I use to store the the selected object?  It should look like this: Merge two arrays as key value pairs in PHP Does a 2100W angle grinder need an earth pin? I have recently purchased a Makita 2100W angle grinder. It has just the live (L) and neutral (N) connectors unlike the one I previously used, which had a third protected earth (PE) connection.   Would it be better if I changed the connections to a 3 pin connector? Why does a RegExp with global flag give wrong results? If $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$ If $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$     My  idea:  $$b_{n}=\min(a_{n},\tfrac{1}{n})\Longrightarrow b_{n}\le a_{n},b_{n}\le\tfrac{1}{n}$$ then I don't know how to go on. If I have a 18-55mm lens, is there a point in buying a 35mm prime lens? I have the Nikon kit lens that came with a D5100 — the kit lens is  . I was thinking of getting a fast lens and I am looking to buy the .  Should I consider 35mm or instead buy a 55-200mm for better range? I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$.  One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th edi, p442) is as following: $$ \langle x|[x,p]|y \rangle = \langle x|xp-px|y \rangle = (x-y) \langle x|p|y \rangle. $$ On the other hand,  $ \langle x|[x,p]|y \rangle = i \langle x|y \rangle = i \delta (x-y)$. Thus $$ (x-y) \langle x|p|y \rangle = i \delta(x-y). \tag{1} $$  We use $(x-y) \delta(x-y) = 0$. Take the derivative with respect to $x$; we have $\delta(x-y) + (x-y) \delta'(x-y) = 0$. Thus $$ (x-y) \delta'(x-y) = - \delta(x-y). \tag{2} $$  Comparing Eqs. (1) and (2), we identify $$ \langle x|p|y \rangle = -i \delta'(x-y). \tag{3} $$  In addition, we can add $\alpha \delta(x-y)$ on the right-hand side of Eq. (3), i.e. $$ \langle x|p|y \rangle = -i \delta'(x-y)  + \alpha \delta(x-y), $$ and $[x,p] = i$ is still satisfied. We can also add $$ \frac{\beta}{\sqrt{|x-y|}}\delta(x-y)$$ on the RHS of Eq. (3).  Here $\alpha$ and $\beta$ are two real numbers.  My question is, what is the most general expression of $\langle x|p|y \rangle$? Can we always absorb the additional term into a phase factor like Dirac's quantum mechanics book did?  Thank you very much in advance. How to make equally sized windows in corners of the screen? Could you please help me computing    $\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor$? Could the Olympians use modern celestial weapons? Instead of using short range celestial bronze (CB) weapons like swords and arrows why not make advanced weapons like machine guns or grenades with CB shrapnel in it? When thinking about travelling, I often have to think about a place to visit, and then look up visa rules for that particular country. This can be a fairly painful process because official government sites can often be hard to find, and advice on (say) guidebooks or Wikitravel tends to be generic.  Is there any resource where I can enter my nationality/residency, and then look up a list of visa rules for various countries? What is the formula for armor/magic resist reduction? As far as I can tell, it's something like this:  damage reduction = armor / (armor + 100)   Is this accurate? Let’s take an example of a movie site. So a well structured set of URLs for this site can be:     Movie page - example.com/movies/the-matrix Review for movie - example.com/movies/the-matrix/reviews Trailers for movie - example.com/movies/the-matrix/trailers Specific trailer for movie - example.com/movies/the-matrix/trailers/trailer1   where it’s possible to navigate the resource even by directly deleting/appending resource name seamlessly.  For SEO purpose I read somewhere that for two given pages, Google will prefer the page which have search query up in the URL hierarchy among other things. So for the given 2 URLs and the query "the matrix trailer"    example.com/movies/the-matrix/trailers example.org/trailers/the-matrix/    `     Google might prefer the second URL as it has the keyword "trailer" at first level comparing to first URL which has "trailer" at 3rd level (a resource deeper in URL hierarchy may be less important).    If it is correct, one may be tempted to create URL hierarchy somewhat like      Movie page - example.com/movies/the-matrix Review for movie - example.com/reviews/the-matrix/ Trailers for movie - example.com/trailers/the-matrix/ Specific trailer for movie - example.com/trailers/the-matrix/trailer1     which will break the seamless hierarchy.    My questions are:   Is it true that resource deeper in URL hierarchy is tend to be less important when comparing to URLs of other site?   Is sacrificing meaningful, seamless hierarchy worth for this SEO optimization? Can I get from Gatwick airport to London by Oyster card? I read many contradictions as to whether one can use a Oyster card from Gatwick Airport to central London.  What is the most time-efficient way to go from Gatwick to "Old Billingsgate" (Monument tube station is nearby)? Public transportation is preferred over taxi.