I am using Blender as the basic 3d tool to design the paper craft (like origami, but with die-cut).  My question is that is it possible for me to flatten the mesh with the polygon undistorted? So that I can print it out on paper and reconstruct the 3d model.
Is there a way to take a mesh, cut it with seams or sharp edges and flatten it, so that the faces stay the same shape and size?  Making paper models is one aspect, where you could use that technique.  No, I'm afraid SZ0Return is not going to do it properly.  And misusing the UV-unwrap is only a last option.    Note, while it's an option to revert to Blender 2.49 which had a script to do this, I'd like a solution that I can use inside the current workspace.
I was solving a few problems from a textbook and I came across this one:     Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through   the point $P=(12,3)$   I could find the tangency points $$ but my solution was completely messy (I had four equations that I had to go back-and-forth with.  Is there an elegant/simple solution to that?  Tangency points:  $$P_1 = (0,3)$$ $$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$ Determinant of a $2\times 2$ block matrix I would like to know the proof for:  The determinant of the block matrix $$\begin{pmatrix} A &amp; B\\ C&amp; D\end{pmatrix}$$ equals $$(D-1) \det(A) + \det(A-BC) = (D+1) \det(A) - \det(A+BC)$$ when $A$ is a square matrix, $D$ is a scalar, $C$ is a row vector and $B$ is a column vector.  I appreciate your help. "Speed" of Gravity and Speed of Light Read/write to Windows registry using Java How is it possible to read/write to the Windows registry using Java? "On their back" or "on their backs"? After the therapy, eight children (43%) became able to crawl/move on their back.   Or should I use "on their backs"?   Singular because each child only has one back, or plural because we're dealing with eight backs? Evaluate:  $$\lim_{x\to{0}}\bigg(\frac{2}{x^3}.(\arcsin{x}-\arctan{x})\bigg)^{2/x^2}$$  I can just expand $\arcsin{x}$ and $\arctan{x}$ using their taylor expansions, but is there any other method? Isomorphism between $\Bbb R$ and $\Bbb R(X)$? My questions are:     $1.$ Is there a field morphism $\Bbb R(X) \hookrightarrow \Bbb R$ ?      $2.$ If the answer to $1.$ is "yes", are $\Bbb R$ and $\Bbb R(X)$ isomorphic as fields?   $ $    $ $  For $1.$, here are some comments:   This is true for $\Bbb C$, see . The key point here is the fact that $\Bbb C$ is algebraically closed, of cardinality $2^{\aleph_0}$, and of characteristic $0$. The same argument doesn't work for $\Bbb R$. We can actually find a subfield $F$ of $\Bbb R$ such that $F \cong F(X) \cong F(X,Y) \cong \cdots$, see .  that $\Bbb Q(X)$ does not embed in $\Bbb Q$ (as a field).  This is equivalent to: Is there an injective ring morphism $\Bbb R[X] \hookrightarrow \Bbb R$ ? (By taking the fraction fields).   As for 2. :   They are isomorphic as abelian groups, and actually as $\Bbb Q$-vector spaces. They are not isomorphic as $\Bbb R$-algebras, because they haven't the same transcendence degree over $\Bbb R$. Let's try to see how to construct such a field isomorphism. We could try$^{[1]}$ to find a subset $B$ of $\Bbb R$ such that $\Bbb R = \Bbb Q(B)$ and $C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)$.  The set $B$ must contain transcendental elements, pick such an $x_0 \in B$.  Then $\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)$ and we could try$^{[2]}$ to find a field isomorphism $\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$, so that $\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)$.  For ${[1]}$, I tried to use Zorn's lemma on $\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}$ and the partial order $B_1 ≤ B_2 \iff B_1 \supset B_2$. But this is not clear to me why this should be an inductive set.  For ${[2]}$, my idea was to define $f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$ by choosing a bijection $b : B \setminus \{x_0\} \to B$ (these sets should be uncountable), and then trying to extend $f\vert_{B \setminus \{x_0\}} = b$ to a field morphism.     Edit for $^{[1]}$: in  article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are   .  I think that $B$ is a minimal generating set of a subfield $K \subset \Bbb R$ over $\Bbb Q$ if and only if any element of $B$ cannot be written as a rational fraction of other elements of $B$. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function $f: B \to \Bbb C$ such that $b$ and $f(b)$ are conjugate over $\Bbb Q$ (i.e. have the same minimal polynomial) extends to a (unique) field morphism $\sigma : K = \Bbb Q(B) \to \Bbb C$.  Actually, the collection $\mathscr B$ defined above doesn't have to be an inductive set, at least if we replace $\Bbb R$ by $K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)$, because the descending chain $\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}$ satisfy $K = \Bbb Q(E_n)$, but $K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q$, so the chain has no upper bound w.r.t. $≤$ in $\mathscr B$. Prove that $\det(M-I)=0$ if $\det(M)=1$ and $MM^T=I$ $M$ is a $3 \times 3$ matrix such that $\det(M)=1$ and $MM^T=I$, where $I$ is the identity matrix. Prove that $\det(M-I)=0$  I tried to take $M$ $=$ $$ \begin{pmatrix}         a  &amp;b &amp; c \\         d &amp; e &amp; f \\         g &amp; h &amp; i         \end{pmatrix} $$   but its expansion is too long. I've recently read a  from a local paper that included the following:     The pair was drinking prior to the shooting.   To me, this appears wrong and I would say that the proper way to make the statement would be:     The pair were drinking prior to the shooting.   Is the original article correct? Let $X$ be a normed vector space.   Show that if a subsequence of a Cauchy sequence converges, then the whole sequence converges. Use the part 1 to show that $S = \{x\in X : \|x\| = 1\}$ is complete if and only if $X$ is Banach.     I have found the first part in a book, but I do not how to prove part 2. Thanx regards Can't open a folder in windows 7 64bit, by double clicking, it shows me that the default option for opening folder is "cmd" (it's highlighted when i'm right clicking a folder, and the "open" option comes next). I've tried running some registry fixing tools, but it didn't fix it. Can someone suggest me what should I do. Find the supremum of the following set Find the supremum of the following set:  $$A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}$$  So I have as my answer that $\sup(A) = \frac12$ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound.   For step $1$: if $a$ is a member of $A$ then $a = (n-1)/(2n+3)$ for some natural number $n$ and $\frac{n-1}{2n+3} \leq \frac{n}{2n} = \frac12$ so $\frac12$ is an upper bound for the set  Step 2 is where I am uncertain. I think I need to show that for any $y&lt;\frac12$ there exists a member of the set between $y$ and $\frac12$ but not sure how to do this. Thanks for all the answers so far! convert ceil to floor Shell with better editing capabilities Shells (at least what I know) have really bad command editing capabilities.  For example:    I can't point the cursor to any point of the command. I can't select and backspace/delete/replace text in the command.   Currently these are so distracting, I use gedit to edit commands before pasting them to the terminal.  Is there any solution that's good for these? How to install the latest Arduino IDE? I have an Arduino UNO, which I need to get working with Ubuntu.  Can somebody explain to me the steps required to install the IDE on Ubuntu? Can anyone come up with an explicit example of two functions $f$ and $g$ such that: $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective?  I tried the following: $$f:\mathbb{R}\rightarrow \mathbb{R^{+}} $$ $$f(x)=x^{2}$$  and  $$g:\mathbb{R^{+}}\rightarrow \mathbb{R}$$ $$g(x)=\sqrt{x}$$  $f$ is not injective, and $g$ is not surjective, but $f\circ g$ is bijective  Any other examples? Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible. Is it better to initialize class member variables on declaration  private List&lt;Thing&gt; _things = new List&lt;Thing&gt;(); private int _arb = 99;   or in the default constructor?  private List&lt;Thing&gt; _things; private int _arb;  public TheClass() {   _things = new List&lt;Thing&gt;();   _arb = 99; }   Is it simply a matter of style or are there performance trade-offs, one way or the other? Best way to obtain clarifying information when reputation doesn't allow comments? I know only the very basics of TikZ, having only used it to produce a pretty flow chart for one of my documents. I recently learned that you can use Inkscape and export to TikZ.  What are the limitations of the Inkscape->TikZ workflow? Is there some compelling reason to take the time to learn TikZ (which I'm told has a fairly steep learning curve) rather than simply exporting to TikZ from Inkscape?