As a land surveyor, I often bring cad data into a GIS and create polygon representing parcels.  Traditional tools I have for cleaning topology are inappropriate so I am looking for something different - even if it means teaming up with someone to build additional extensions.  (I scanned the list of existing extensions and some pieces are already there).  My idea is to fix minor topology errors by constraining key geometric elements. That is, straight polylines and circular arcs which share a common centre point.  A key requirement is support for true circular curves.  I could keep the data in SQL Server but prefer a GUI for a better user experience.  As I am not familiar with QGIS, I am looking for some indication if the data structure used by QGIS can support this.
I am trying to make a floor plan for my office, have managed to make all of the rectangular components (e.g. the rooms) however some of the desk layouts are circular and I haven't been able to work out how to make exact and measured circles. They need to have a radius of 1.45m.
I want it to know how to convert string field values to integer values in QGIS Dufour... there are numbers and strings in the same field, when I use toreal(string) it stops and it won´t create the field. How can a regression be significant yet all predictors be non-significant? My multiple regression analysis model has a statistically significant F value however all beta values are statistically non-significant.   All the regression assumptions are met. No multicollinearity was found. Correlations among all predictors are all less than 0.60.   What else might be the cause of the insignificant predictors? I am a platform diver. I am trying to figure out how deep in the water I go depending on which platform I dive from.  The platforms are 1m, 3m, 5m, 7.5m and 10m above the water surface.  By ignoring air resistance (we assume that I do not do any tricks) I can easily compute the energy that my 90 kg body has when it is hitting the water:  E = mgh = 0.5mv^2   My problem is to find the deceleration in water and thus how deep my feet will go into the water before coming to a halt. We can assume that I stand straight and fall straight down and have a height of 190cm.  How do I compute that? Problem: James has a pile of n stones for some positive integer n ≥ 2. At each step, he   chooses one pile of stones and splits it into two smaller piles and writes the product   of the new pile sizes on the board. He repeats this process until every pile is exactly   one stone.      For example, if James has n = 12 stones, he could split that pile into a pile of size   4 and another pile of size 8. James would then write the number 4 · 8 = 32 on the   board. He then decides to split the pile of 4 stones into a pile with 1 stone and a   pile with 3 stones and writes 1 · 3 = 3 on the board. He continues this way until he   has 12 piles with one stone each.      Prove that no matter how James splits the piles (starting with a single pile of n   stones), the sum of the numbers on the blackboard at the end of the procedure is   always the same.   Hint: First figure out what the formula for the final sum will be. Then prove it using strong induction.  The formula I came up with is $n(n-1) / $2. It could be totally wrong...  My (Partial) Solution:  1) Base Case: (n=2) James has one pile of 2 stones. Suppose he takes the top-most stone and puts in one pile A, which now has size 1. He takes the remaining stone and places it into another pile B, which now has size 1. Then the product of the sizes is 1. The sum of all the products on his board is 1. Now, if he were to start again, and take the bottom stone first and put it in pile A, he has a pile of size 1, and then takes the top stone and puts it in pile B, he has a pile of size 1, and the product is 1, with sum of all products on the board is 1.   2) Inductive Hypothesis (Strong Induction): Suppose for some $k \geq 2$, $n$ stones can be split in $ 2 \leq n \leq k $ stones and $k-n$ stones.  3) Inductive Step:  Consider $n=k+1$. What do I do now? , Laser Interferometer Gravitational-Wave Observatory, is a large-scale physics experiment aiming to directly detect gravitational waves. The device measures the phase shift laser beams.  If I understand this well, the measurement assumes that a suitable gravitational wave reduces and prolongs one of the tubes and the extension will take effect in the phase-shifting laser beams.  But I think if there's a gravitational wave that passes through it, it will affect everything, even the laser beam. This means that when the wave lengthens or shortens the tube, it also lengthens or shortens the wavelength of the laser beam.  In my opinion, therefore, this device does not work. Or am I wrong? When people are asked to match monchromatic violet light with an additive mix of basic colours, they (paradoxically) mix in red.  In fact, the CIE 1931 color space chromaticity diagram shows this effect begins at about 510nm (greenish-cyan), where people mix in no red.  From that point on, the higher the frequency of the light source, the more red they mix in.    This effect is reflected by the red curve of the , which has an additional bump in the area of blue light.  However, that curve does not match the actual spectral sensitivity of red cones.  So where does this additional perception of red at higher frequencies come from? No cycle containing edges $e$ and $g$ implies there is a vertex $u$ so that every path sharing one end with $e$ and another with $g$ contains $u$ There is a proof in my textbook for the following claim, which doesn't make a whole lot of sense to me.  My annotations are in bold.  Could someone perhaps elaborate on what's going on?  Claim.  If there does not exist a cycle containing edges $e$ and $g$ then there exists a vertex $u \in V (G)$ such that every path in $G$ sharing one end with e and another with $g$ contains $u$.  Proof: The claim trivially holds if $e$ or $g$ is a loop OK, so we assume that neither is. Let $P$ with vertex set $v_1, v_2, . . . , v_k,$ in order, be a path with $e$ joining $v_1$ to $v_2$ and $g$ joining $v_{k−1}$ and $v_k$. Let $f_i \in E(P_i)$ be the edge with ends $v_i$ and $v_{i+1}$. Let $j$ be chosen minimum so that no cycle in $G$ contains $e$ and $f_j$ We can do this because we know that at least the edge $g$ will not create a cycle by assumption, right?. We will show that $u = v_j$ satisfies the claim.  Suppose not. Let $C$ be a cycle containing $e$ and $f_{j−1}$ What if $f_j = e$?  then what cycle? a single vertex? and let $P′$ be a path from an end of $e$ to an end of $f$ Not sure what the book meant by $f$ here, any guesses? avoiding $u$. Choose a subpath $Q$ of $P′$ with one end in $V (C)$ and another in ${v_{j+1}, v_{j+2}, . . . , v_k}$ as short as possible. Then $C \cup Q \cup P$ contains a cycle containing both $e$ and $f_j$, a contradiction. (The last statement requires some case checking.) This ending seems a bit abrupt and non-obvious to me  Thanks for the help I have a mesh and in the region around each triangle, I want to compute an estimate of the principal curvature directions. I have never done this sort of thing before and  does not help a lot. Can you describe or point me to a simple algorithm that can help me compute this estimate?  Assume that I know positions and normals of all vertices. Different PDF output with same source files (NeurIPS 2021 package) I initially thought that dark energy must in some way violate conservation of mass and energy since the component of the energy density of space that comes from dark energy is constant, and space is expanding. Therefore, as space expands the energy in the universe that comes from dark energy would increase. I presumed the source of this energy was not coming from the conversion of other types of energy to dark energy, so it must violate conservation.  I decided to Google this and came upon this article:    It says that dark energy does NOT violate conservation and quotes Carroll, Press, and Turner (1992):  "…the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically [i.e. without loss or gain of heat], one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρ c2. So the mathematics is consistent."  Is there a way to explain this in layman's terms? (The blog attempted to do this, but it was very unclear to me.) More specifically, can you explain where my initial train of thought described above fails when I erroneously concluded that dark energy violates conservation?  Thank you. Showing that $\lnot Q \lor (\lnot Q \land R) = \lnot Q$ without a truth table I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:  $$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$  But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:  $$\lnot Q \land (\lnot Q \lor R)$$  Is there a way to show this is true without using a truth table? What am I missing?!  Thanks in advance! Correct tracked camera orientation I have been playing around with camera tracking and I'm happy enough with the tracking and alignment of the markers with the footage when viewed through the camera.    The problem is, when I am not looking through the camera I can see that the grid isn't aligned to the camera.    How do you correctly align the grid to this tracked camera? I have set a floor and two points for scale, also clicked setup tracking scene, but this just added cube and plane which were on the grid. Given   $$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$  $Y = Z\sqrt{1-\rho^2} + \rho X$  And   $$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right) $$  Show that $P(X&gt;0,Y&gt;0)= \frac{1}{4}+\frac{1}{2\pi}(\arcsin \rho) $  I'm supposed to use the fact that X and Z are independent standard normal random variables, but I don't quite understand how. Any help would be greatly appreciated. Can I use Wi-Fi and cable (Ethernet) networking at same time? I am on an intranet that is not connected to the Internet and a Wi-Fi router with connection to the Internet. When I am connected to the wireless network, I can browse the Internet, but I cannot browse my LAN with my Ethernet cable connected.  How can I browse the Internet via my Wi-Fi router and at the same time browse my local LAN web application via my connected Ethernet cable? Explanation for "emails"? Why does LD keep outputting "no version information available" On every loading of a lib, I get the error:     no version information available   This lib has been compiled on another PC (ubuntu 10.04) than the one running it (mandriva 2010.2).  $ ldd /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so: /lib/libz.so.1: no version information available (required by /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so)  linux-gate.so.1 =&gt;  (0xffffe000)  libz.so.1 =&gt; /lib/libz.so.1 (0xb77a7000)  libc.so.6 =&gt; /lib/i686/libc.so.6 (0xb7655000)  /lib/ld-linux.so.2 (0xb787a000)   Edit: the workaround didn't work. I want to install lubuntu on my current (pre-existing) Ubuntu Install because of it's light-weight footprint.   Can I preserve currently installed apps after installation? I often use Cycles' built in Glass BSDF shader to achieve realistic refraction effects, but in many cases, it seems that Cycles' glass caustics creates an unfortunately large number of fireflies (pixels that are abnormally bright relative to their surroundings). In some models of mine, even when I render for 3000 or more samples/pixel, my final image is still covered in fireflies. While I can improve this in many cases by increasing the total amount of light available in the scene, this loses a lot of the potency of the refractive effects I'm looking for, especially in scenes with low light levels.  So my question is, how can I reduce the amount of fireflies while using Cycles' caustics without sacrificing the quality of the refractive effects?  Note: I am aware that Cycles' caustics will perform much better once Cycles has full support for bidirectional path-tracing, but it looks unlikely that that will become , unfortunately. A high school Olympiad problem Get users to use the download button instead of right-click save As part of our system, a web application, we have an image viewer. Previously we always displayed the full resolution image but scaled down with CSS to fit the page. For performance reasons we now display a properly scaled version of the image with the option to download the full resolution image with a button under the image.  As part of the users' workflow, they sometimes need to download this full resolution image to submit to another system. The problem is that many users just right-click the image and press "save image as". This used to work fine but now it will instead provide them with a low resolution image. The users don't notice this and this causes problems further down the line. How can we force users away from this behaviour? Sure, we could tell them not to do it but that doesn't feel very scalable. My least bad idea so far is to hijack the right-click in the browser but this is A) not super nice UI and B) not sure to cover all cases.  In short: How can we get users to use the download button instead of saving the low-res thumbnail?