The .asc LiDAR files provided by the environment agency have height values provided to a precision of the level used to measure atoms. How to quickly lose a few of the pointless digits on the end to get from attometers to centimetres?    There must be a simple solution but I can't see it.   I have tried raster calculator but the results of the calculation below are still in attometers.   (ToInt("DSM@1"*10))/10
121682
How can I use gdal_translate for AAIGrid so that the raster values are not more than 4 decimal places.
44745
I have been attempting to use the Spatial Analyst Curvature tool with mixed results.  I originally tried to use it with a raster that was projected using a 1983 Teale Albers projection and the results were black, indicating the lowest values.  I then transformed the raster into UTM projection and ended up with grid-like lines on the raster which have values that are obviously not accurate.  Does anyone know why these lines have appeared or what I'm doing wrong.  Thanks.
39610
I need a vertically dashed line in my matrix, but my current solution looks like the dashes are going out of the array environment (and they are - i checked in photoshop). Can I manually shorten it by say 1 mm or 0.5 mm?  Here is MWE:  \documentclass[a4paper]{article} \usepackage{amsmath} \usepackage{arydshln} \usepackage{mathtools} \begin{document}  \begin{equation*} \setlength{\dashlinegap}{2pt} \left[\begin{array}{cccc:c} a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} &amp; b_1 \\ a_{21} &amp; a_{22} &amp; \cdots &amp; a_{2n} &amp; b_2 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; a_{nn} &amp; b_n \end{array} \right] \end{equation*}  \end{document}
8113
gives a table of certain constants in different systems of units. I noticed that in Gaussian and electrostatic CGS the value of $\epsilon_0$ equals the dimensionless $1$.  I wondered whether anyone could provide justification for this? I was under the impression that $\epsilon_0$ is not even necessary in these units since it's essentially just a dimensional conversion factor in the SI, however I suppose it would still be possible to convert it into CGS units. How is this done?
105997
I've been looking at the following solution to a similar problem  However with my problem the 3 points are spread over a large distance and the distance is a distance-over-ground (ignoring altitude). So the curvature of the earth comes into play.  Example of one of the data points is...[lat,long,km]  51.505348978413,-0.11270052661132,75.639168 51.4845086249581,-3.16947466601561,173.809152 55.9568033803799,-3.20266968478392,465.100416  This can be visualised at  by pasting it into the section "CSV Upload - [latitude,longitude,radius(km)] per line"  Using the solution in  I get the result..  51.8247121391 -0.802763718095  which isn't bad but it's about 14 miles out. I'm really wanting this down to ~5 miles of error.  I changed the ECEF calculation to one from here  and used an altitude of 0.  However, this is way way off the mark with...  56.2518456429 -0.813853131422  Intermediate results for xA,yA,zA...etc. are  lla2ecef  xA = [3978.79751502] yA = [-7.82628596545] zA = [5342.89431232]  xB = [3974.53105834] yB = [-220.086727248] zB = [5342.88794881]  xC = [3573.7981748] yC = [-199.973379186] zC = [5344.22411569]  original method  xA = [3965.56758195] yA = [-7.8002627162] zA = [4986.36680448]  xB = [3961.32002974] yB = [-219.355176278] zB = [4984.92406448]  xC = [3561.02859617] yC = [-199.258852046] zC = [5279.1109334]  So... the Z values for lla2ecef method are quite different. I'm guessing this is what's causing the issue, something to do with fixing the altitude to 0 and the fact my distances are distance-over-ground.  Am I going to need to do some projections instead?  My brain is about melted for today, can anyone shed light on ways to tackle this problem?
69314
I have several big files with some measurements.   It looks this way:  N 12344;PE 9.9999999;... #S 0 0 31 44 75 130 165 196... #S_+ "2 5 2 3 3 1 1 2 3 1 2 2...  N 12345;PE 9.9999999;... #S 0 0 34 57 84 133 152... #S_+ "1 0 1 1 2 3 0 0 0...  N 12346;PE 9.9999999;... #S 0 0 31 44 73 140 169... #S_+ "3 3 4 0 0 2 1 2 4...  N 25104;PE 9.9999999;... #S 0 0 36 52 102 108 145... #S_+ "1 1 0 1 0 0 3 0 1...  N 25105;PE 9.9999999;... #S 0 0 32 58 88 130 143...   Sample is here:   The file is much bigger but I replaced part of the data with "..." to make it shorter.  I need to somehow replace the line ends before "#S" - in fact simply merge the "N" line with the following two ones into one line (or with the following three ones so I can get rid of the blank lines). Expect output like this:  N 12344;PE 9.9999999; #S 0 0 31 44 75 130 165 196 #S_+ "2 5 2 3 3 1 1 2 3 1 2 2... N 12345;PE 9.9999999; #S 0 0 34 57 84 133 152 #S_+ "1 0 1 1 2 3 0 0 0... N 12346;PE 9.9999999; #S 0 0 31 44 73 140 169 #S_+ "3 3 4 0 0 2 1 2 4... N 25104;PE 9.9999999; #S 0 0 36 52 102 108 145 #S_+ "1 1 0 1 0 0 3 0 1... N 25105;PE 9.9999999; #S 0 0 32 58 88 130 143...   Is this possible to achieve using some command-line utility in linux?   My knowledge is quite limited in this area so I would appreciate any help.  thanks
73179
On , $N_A = 6.022\;141\;29 \times 10^{23} \text{ mol}^{-1}$ has 9 significant figures and a standard uncertainty of $0.000\;000\;27 \times 10^{23} \text{ mol}^{-1}$.  First, can it be written as the following?  $$ N_A = 6.022\;141\;29 \times 10^{23} \pm 0.000\;000\;27 \times 10^{23} \text{ mol}^{-1} $$ If yes, consider the following quote from  (emphasis added):     Whenever you make a measurement, the number of meaningful digits that you write down implies the error in the measurement. For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. To record this measurement as either 0.4 or 0.42819667 would imply that you only know it to 0.1 m in the first case or to 0.00000001 m in the second. You should only report as many significant figures as are consistent with the estimated error. The quantity 0.428 m is said to have three significant figures, that is, three digits that make sense in terms of the measurement. Notice that this has nothing to do with the "number of decimal places". The same measurement in centimeters would be 42.8 cm and still be a three significant figure number. The accepted convention is that only one uncertain digit is to be reported for a measurement. In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m.   and the following quote from  (emphasis added):     The absolute error places bounds telling you that the actual value is expected to fall between x - D x and x + D x. The absolute error is an easy way to report the error and makes it easy to check that you are not reporting too many significant figures. For example, suppose you had a measurement that was reported to be 10.02 ± 0.3 m. Because the reported value is uncertain in the tenths place, it is wrong to report it to the hundredths place. It should be reported to the same decimal place as the absolute error: 10.0 ± 0.3 m. Another point is that the absolute error and the reported value must have the same units.   If there is such a convention (cf. Rice's lab guide) and it is wrong to report more significant figures than is warranted by the absolute error (cf. the joint project's guidelines), and NIST should have adhered to such convention and striven to be correct, is it true that NIST adheres and is correct because the standard uncertainty has two least significant digits (i.e., $27$)?  Had the standard uncertainty been $0.000\;000\;2\text{ mol}^{-1}$, would NIST have reported Avogadro's Number as $N_A = 6.022\;141\;3\times 10^{23}$ with 8 significant figures?
70602
I am going to ask a simple question, for sure.  The pressure with respect to the altitude is given by this formula    Where    sea level standard atmospheric pressure p0 = 101325 Pa sea level standard temperature T0 = 288.15 K Earth-surface gravitational acceleration g = 9.80665 m/s2. temperature lapse rate L = 0.0065 K/m universal gas constant R = 8.31447 J/(mol·K) molar mass of dry air M = 0.0289644 kg/mol   ( from  )  In addition to this, we have $L = \frac{g}{C_p}$ where $g = 9.80665\ m/s^2$ and $C_p$ is the constant pressure specific heat $ = 1007\ J/(Kg\ K)$  Understood and thence the above formula can be written in this simple way:  $$\boxed{P = P_0\left(1 - \alpha h\right)^{\beta}}$$  where  $\alpha = \frac{g}{C_p\ T_0} \approx 3.3796\cdot 10^{-5}\ m^{-1}$  $\beta = \frac{C_p\ M}{R} \approx 3.5081971$  On the other side we learnt from the elementary physics that the formula for the pressure is also given by  $$\boxed{P = P_0 + \rho g h}$$  Where $\rho$ is the air density $(1.23\ kg/m^3)$.  The question  First thing first: I assumed that  $$h = h_1 - h_0$$  is that correct? I mean it's the difference between two heights (maybe from a table and the ground, just to say).  Since the two formulae seems quite different, I tried with a numerical example in calculating the pressure in two points, with a difference of height about $0.18$ m and I got a really similar result.  Since the first formula is more technique, I think it's the correct formula but I would like to understand if one could pass from the first to the second or vice versa somehow.  Also I would like to know if there are cases in which I can use only the second formula or only the first formula!
84788
I got the following parameters:  Width and height of Map: 450px, 560px  units : 'meters'  maxExtent : new OpenLayers.Bounds(806677.9759,802420.9858,8876817.5647,853497.8186405064)  resolutions: [199.51887828322785, 99.75943914161392, 49.87971957080696,  24.93985978540348, 12.46992989270174, 6.23496494635087, 3.117482473175435,  1.5587412365877176, 0.7793706182938588, 0.3896853091469294, 0.1948426545734647,  0.09742132728673235, 0.048710663643366174, 0.024355331821683087, 0.012177665910841544, 0.006088832955420772, 0.003044416477710386, 0.001522208238855193]  numZoomLevels: 18  OpenLayers.DOTS_PER_INCH = 90.71428571428572;   Who inspire me how to calculate the scale of current Zoom Level?
5069
I am playing a bit around with Qgis 2.14.6 to understand how it deals with distance measurement.  I took a line-shapefile that I saved in 3 different coordinate systems (4326, 32642 and a national one that doesn't exist in Qgis yet).  Using 32642 and the national one, the length is quite similar (6442m vs. 6428m)  But using 4326 (unprojected WGS84), I get 8057m  In all cases, I just used the Field Calculator and introduced $length.  I was already positively surprised that the result using 4326 was in meters and not in degrees.  To calculate the distance on the dataset in 4326, I assume that Qgis does some kind of re-projection in the background. Does anyone know which one?   Shapefiles here if anyone wants to play around:   EDIT - adding more info after more tests  As you can see, now both my shapefile and the project are in EPSG:4326, so unprojected:    Here the rough measurement of my shapefile, with the Qgis measure tool:   When I use the field calculator, the preview gives me a value in degrees, while the result seems to be in meters:      This would mean that the function $length, like the measurement tool, always gives a result in meters, but using a very coarse approximation of the degrees => meters transformation:    Is this a bug? or should it really behave like this? I find it at best counter-intuitive (but maybe it's just me) and, at worse, completely misleading (it makes very little sense to use, everywhere on earth, the length of 1 degree at the Equator). Any input or opinion here? Should I open a bug report (at least for the inconsistency between the preview and the actual result)?
94547
I have a few LiDAR point data (.las files) with me. I would like to read the X,Y and Z from it using .Net. I found , which is in C++ and I am not sure if it can be used commercially after reading the licensing information. Has anyone found any good ones?
23881
I read on  ()    Shouldn't it be 1.03m or (if we really follow the conventions of the metric system) 1.03M instead?
39912
I am trying to figure out how I can transform my unit of length from decimal degree to km when I import the following simple ArcGIS polygon shape.  Ultimately, I would like to have this information available when I assign this spatial polygon to my window in spatstat [i.e., Window area = xx square unit &amp; Unit of length: xx km].   I did came across this post:   This is what I've done so far, but I don't understand why/if I need to do so. Doesn't my div0A SpatialPolygonsDataFrame already has my CRS argument defined? Why do I need to spTransform then?  library(spatstat) library(maptools) library(lattice) library(sp) library(foreign) require("rgdal") div0A &lt;- readOGR(dsn="Archive", layer="Projections") &gt; summary(div0A) Object of class SpatialPolygonsDataFrame Coordinates:      min       max x -80.00 -57.63430 y  66.25  78.16666 Is projected: FALSE proj4string : [+proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0] Data attributes:        Id      NPAzimutha         UTM20            UTM19           AlberEA        Min.   :0   Min.   :348877   Min.   :349232   Min.   :349162   Min.   :348656    1st Qu.:0   1st Qu.:348877   1st Qu.:349232   1st Qu.:349162   1st Qu.:348656    Median :0   Median :348877   Median :349232   Median :349162   Median :348656    Mean   :0   Mean   :348877   Mean   :349232   Mean   :349162   Mean   :348656    3rd Qu.:0   3rd Qu.:348877   3rd Qu.:349232   3rd Qu.:349162   3rd Qu.:348656    Max.   :0   Max.   :348877   Max.   :349232   Max.   :349162   Max.   :348656   proj4string(div0A) &lt;- CRS("+proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs") xxlcc &lt;- spTransform(div0A, CRS("+init=epsg:4326")) res1 &lt;- lapply(slot(xxlcc, "polygons"), function(x)   sapply(slot(x, "Polygons"), slot, "area"))  &gt; summary(unlist(res1))    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.   89.72   89.72   89.72   89.72   89.72   89.72  newW&lt;-as(xxlcc, "owin") &gt; summary(newW) Window: polygonal boundary single connected closed polygon with 29 vertices enclosing rectangle: [-80, -57.6343]x[66.25, 78.16666]units Window area =  89.7204 square units   Does this look correct? How can I specify/know my unit of length in km? Moreover, this window is ~Baffin Bay, west of Greenland and I am wondering if I should be using a different datum? E.g. EPSG Projection 4747 - GR96? Although, I will have to create ppp object, for which the coordinates have been taken with GPS satellite navigation system.  I would appreciate if someone can walk me through this or suggest an online example or further readings. I read chapter 4: Spatial Data Import and Export in Applied Spatial Data Analysis with R (2008), but I'm obviously very new to all of this.
42010
How can I compute a raster showing the number of return points per square meter using SAGA GIS or other open source software?    My goal is to obtain a mask which would show me the places (cells) where there were no hits and would help me in interpreting LiDAR-derived data.
60172
While working on my project, I got to a point where I don't know the answer. I have a 10-bit ADC (the one incorporated in the ATMEGA328P) and it returns values up to 1023 (210 - 1).   The official documentation states that the equation to convert the values from the ADC to the corresponding voltage is:   $$ V = \frac{V_{in} \cdot ADC}{2^{10}} $$  But, if the values go up to 1023, then wouldn't it be better to replace the 210 by 210&nbsp;-&nbsp;1?  Thanks.
57921
Using the siunitx package, if I specify an error with the \pm symbol then it automatically converts it to a form using parenthesis. I would like it to not do this and keep the \pm symbol, but still space the elements of the quantity and symbol in a good way.  e.g. \SI{10 \pm 2}{\kelvin} would produce 10(1) K  I would like it to produce: 10\pm1 K or something similar with spacing taken care of.
73581
For my study of physics I have to do various lab reports. Normally I just process all my data using python, gnuplot or Mathematica and copy it from there to my TeX document. Unfortunately all these applications use the . (dot) as a decimal seperator. Because I study in Germany the german standard says that I have to use , (comma) as the decimal seperator. In most cases the supervisors of the lab are forgiving, but one of them told me to redo my whole lab report because of this. Now I don't want to search all the 15 pages to replace all dots with commas. I could have used a  but this still implied editing the whole file.