Understanding blender collection, instancing from Sketchup side of view I'm trying to migrate to Blender from Sketchup.  There, instancing is managed simply by 'group' and 'component'.  Groups are just separate coordinate system while components make object an instance. So when you want to separate singular object you make group, else to make linked objects you make component. Groups can have components and components also do. There is no depth restriction so a group can have groups which have component which have another components that have a group, and so on.  This kind of structuring is very crucial in architectural modeling as they tend to have duplicated objects and become very hierarchical.  I've read this and that about blender instancing, grouping, tied playing with boxes for days but still things are not clear. I think one of the cause is that parenting and grouping is separate in Blender.   So I'm asking here hopping someone could give a golden answer (once again) that would weave everything vividly and reveal real power of Blender specific hierarchy structure represented by 'collection'.  Is there anyone coming from Sketchup side and managed to understand instancing in Blender? How should I reroute my thinking?
Are Groups or Sketchup-like components feasible in Blender? Apart from testing it myself, I've seen a few posts  and , including from Sketchup users like me, sort of circle around this topic but never ask this question directly. How can I overcome these problems in DupliGroups or Collection Instances?  moving its origin seems to require moving to World origin (and using up a layer for clarity) &lt;=&gt; linked meshes have their own local origin which I can move data around in Edit Mode (easy enough)  It is impractical to link up DupliGroup data to previously created meshes because the origin mismatch as above, and the mesh data that remains in it and needs to be deleted somehow  Linked meshes don't have the above two problems (can be 'blocked in' and then detailed up all at once) but I can't add objects to them, only faces. Say I'd like to add handles as linked meshes, because handles are used elsewhere in the model too.    How feasible would it be to make a linked object type with deeper hierarchy? Say an object that would look at its children and simply make every other instance have the same children. It would also mean dragging and dropping in outliner would be the simplest way to manage it by the looks of it.
To what extent is WhatsApp's statement on secure messaging realistic? Why Was It 'Essential' That Voldemort Kill Harry Potter? In the spirit of a  I would like to know if there's a basis for understanding why there  that have non-zero electric charge but zero spin?  Can such a quantum theory be written down and self-consistent? Do the current symmetries of our present QFTs not allow such a particle? How to model this odd-shaped distribution (almost a reverse-J) Give an example of three positive integers m, n, and r The spell was cast without words, simply a flick of the Death Eater's wand, and a 'streak of purple flame' that lanced across her chest. It was later mentioned that the spell was weakened because of the lack of a verbal component, but still 'did enough damage as it was'. She was in the infirmary healing from the damage for some time. What spell was this? Is is possible to move all windows (or all not minimized windows) form one workspace to another?   I know I can move one window to other workspace with Shift+Ctrl+Alt+arrow, but it will move only that one focused window. a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance! Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why? If $f'(x)\le g'(x)$, prove $f(x)\le g(x)$ I have to do the following exercise:     Let $f$ and $g$ two differentiable functions such that $f(0)=g(0)$ and $f'(x)\leq g'(x)$ for all $x$ in $\mathbb{R}$. Prove that $f(x)\leq g(x)$ for any $x\geq0$.   Now, I know this is true because the first derivative of a function is the angular coefficient of the function in a point $x$. So, $f'(x)\leq g'(x)$ means, in other words, that the function $g(x)$ grows faster than $f(x)$. I think this is the base for a more formal proof, could someone help me to figure out a more formal proof? Changing an element's ID with jQuery Will acceleration rate of expansion of space become faster than speed of light? How to login to gamecenter at PewDiePie's Tuber Simulator? How do I login to the Gamecenter on PewDiePie's Tuber Simulator, if you have not done it when they asked? Effect of linear terms on a QFT I was told when first learning QFT that linear terms in the Lagrangian are harmless and we can essentially just ignore them. However, I've recently seen in the linear sigma model, \begin{equation}  {\cal L} = \frac{1}{2} \partial _\mu \phi _i \partial ^\mu \phi _i - \frac{m ^2 }{2} \phi _i \phi _i  - \frac{ \lambda }{ 4} ( \phi _i \phi _i ) ^2  \end{equation}  with $m ^2 =-\mu^2 &gt; 0$, adding a linear term in one of the fields $\phi_N$, does change the final results as you no longer have Goldstone bosons (since the $O(N)$ symmetry is broken to begin with).   Are there any other effects of linear terms that we should keep in mind or is this the only exception of the "forget about linear terms" rule? Multiplicative nature of the separability degree In what follows, let $E / F$ be an algebraic extension, $h(x),f(x)\in F[x]$ polynomials, $h(x)$ irreducible.   Definitions.   We say $h(x)$ is separable if it has not repeated factors. We say $f(x)$ is separable if all its irreducible factors are separable. We say $\alpha\in E$ is separable over $F$, if its minimal polynomial over $F$ is. The extension $E/F$ is separable if all the elements of $E$ are separable over $F$. We define the field $$\newcommand\sclo{\operatorname{Scl}}\sclo_F E= \{\alpha\in E : \text{ $\alpha$ is separable over $F$ }\}$$ as the separable closure of $F$ in $E$. The separability degree of the extension $E/F$ is defined as $$[E:F]_s = [\sclo_F E:F].$$   Now, suppose the extension $E/F$ is finite.  My question is: how can one proof that with these definitions, if $B$ is an intermediate field of the extension $E / F$, then $$[E:F]_s = [E:B]_s[B:F]_s?$$  I know that the purpose of the separability degree is to count in how many ways we can extend certain injective homomorphism to an automorphism. If one starts defining separability degree this way, as done in the book by Lang, the result follows without much difficulty.  I wonder how to achieve it in this way. I came upon an interesting way to relatively quickly compute modular exponentiation with large numbers. However, I do not fully understand it and was hoping for a better explanation.  The method basically states if you have $x^y \bmod N$, it can be computed by repeatedly multiplying by $x$ modulo $N$:  $$x \bmod N \rightarrow x^2 \bmod N \rightarrow x^4 \bmod N \rightarrow x^8 \bmod N \rightarrow \ldots \rightarrow x^{2[\log y]} \bmod N$$  This is supposed to calculate in $\mathcal{O}(\log_2 N)$ time.  An example is given:  $$x^{25} = x^{11001_2} = x^{10000_2} * x^{1000_2} * x^{1_2} = x^{16} * x^8 * x^1$$  I don't really understand how this is faster. Aren't you just doing the same thing? You're just breaking down the exponents into representations of base $2$. Also where does $x^{2[\log y]}$ come from? If you're breaking it down into exponents of two where does a logarithm come from? (Aren't logs representations of numbers in exponents of $10$?)  I'm sorry if this appears to be a really stupid question - I have minimal knowledge in math.  Also, I wasn't sure if this belongs here or on stack exchange, so give me the word and I'll move it. Creaking from cranks/spindle. How to fix? Get URL of ASP.Net Page in code-behind How do you convert a number to a string showing dollars and cents?  eg: 123.45    =&gt; '$123.45' 123.456   =&gt; '$123.46' 123       =&gt; '$123.00' .13       =&gt; '$0.13' .1        =&gt; '$0.10' 0         =&gt; '$0.00' In Windows, if you right-click a .bat file, there's an "Edit" option, which opens the .bat file with notepad. How do I make Windows use Notepad++ instead?  Note that this is different from the question . In that question, user asks for a way to make Notepad++ the default opener for specific file types. This is not what I want here, since I expect that double-clicking on .bat files executes them.