https://community.wolfram.com RSS Feed for Wolfram Community showing questions tagged with Astronomy sorted by most replies. https://community.wolfram.com/groups/-/m/t/293403 TL;DR --- I have three issues with getting sun positions in Mathematica V10: - It is extremely slow compared to V9 - It doesn't *seem* to yield correct results - a) It needs an Internet connection, which b) is not assured to always return results and c) if not used optimally can easily consume your monthly allowance of W|A calls Long story === With the advent of V10 `AstronomicalData` has been deprecated, as shown on its documentation page: ![enter image description here][1]. I'm not an astronomer, so my main use of this function has been restricted to its capability to get the sun position using functions calls like this: { AstronomicalData["Sun", {"Azimuth", {2013, 3, 1, #, 0, 0}, {52.37`, 4.89`}}, TimeZone -> 1], AstronomicalData["Sun", {"Altitude", {2013, 3, 1, #, 0, 0}, {52.37`, 4.89`}}, TimeZone -> 1] } & /@ Range[0, 23] > {{341.47732, -43.93930}, {2.33417, -45.21747}, {22.94232, -43.19133}, > {41.52167, -38.28400}, {57.55208, -31.30276}, {71.47253, -23.03159}, > {84.02194, -14.08294}, {95.91940, -4.92723}, {107.80166, 4.03418}, {120.23770, 12.40167}, {133.72303, 19.72563}, {148.59433, > 25.48377}, {164.84179, 29.12209}, {181.93103, 30.19428}, {198.91284, > 28.55117}, {214.89602, 24.41962}, {229.46400, > 18.28613}, {242.70064, 10.70703}, {254.98998, > 2.19073}, {266.84760, -6.82566}, {278.85594, -15.94505}, {291.66723, -24.75464}, {306.00911, -32.75641}, {322.57956, > -39.29877}} So, this gets me the sun positions in steps of an hour during a particular day in Amsterdam (TZ 1). The same call still works in V10, though it now returns numbers with units; degrees in this case. On its first call, it reads some paclet information from a Wolfram server, but on any successive call no Internet connection is needed. I will be going into detail about timing further on, but I'll say here that the V10 function takes about three times longer than its V9 namesake. I blame the addition of units for that. With `AstronomicalData` apparently deprecated we are supposed to use its successors. In this case I need `SunPosition`. A direct translation of the above would be: SunPosition[GeoPosition[{52.37`, 4.89`}], DateObject[{2013, 3, 1, #, 0, 0}, TimeZone -> 1]] & /@ Range[0, 23] > {{95.7, -5.1}, {107.6, 3.9}, {120.0, 12.3}, {133.5, 19.6}, {148.3, 25.4}, {164.5, 29.1}, {181.6, 30.2}, {198.6, 28.6}, {214.6, 24.5}, {229.2, 18.4}, {242.5, 10.9}, {254.8, 2.4}, {266.6, -6.7}, {278.6, -15.8}, {291.4, -24.6}, {305.7, -32.6}, {322.2, -39.2}, {341.3, -43.5}, {2.0, -44.8}, {22.5, -42.9}, {41.1, -38.0}, {57.1, -31.1}, {71.0, -22.9}, {83.6, -13.9}} As with the new `AstronomicalData` the output is actually in degrees which I have removed in the above output for the sake of clarity. There are a few things to note: - `SunPosition` uses position and date objects, the latter being new in V10 - `SunPosition` does not have a `TimeZone` option, but you can set it in `DateObject` - `SunPosition` can use the old lat/long list position indication. It also can use a date list to enter the date instead of a `DateObject`. In the latter case you are out of options with respect to time zones and you have to add the appropriate amount of time offset - It is extremely slow, and it may even time-out: ![enter image description here][2] - Last but not least: the results seem to be plain wrong. It suggests that sunrise is somewhat before 1 am, which is -of course- incorrect. I assume that this has something to do with a `$GeoLocation` setting for the observer of the sun positions, but I haven't managed to sort out what I am supposed to enter to get the correct sun positions for the location provided in the same call. As to timing: I noticed very inconsistent timings for `SunPosition` compared to `AstronomicalData`, so I used the following code to collect a somewhat more statistical sound sample: SetAttributes[timingTest, HoldFirst]; timingTest[code_, repeats_Integer] := Table[ ClearSystemCache[]; code // AbsoluteTiming // First, {repeats} ] Using this, I collected timing of 20 calls to the following code snippets: - `AstronomicalData` V9 and V10: As above - `SunPosition`: As above - `SunPosition` without `GeoPosition`, just the lat/long list. - `SunPosition` without `GeoPosition`, and also without the `DateObject` date, just a classical date list (with the hour set to +1 to accommodate TZ 1) - `SunPosition` V10 without `GeoPosition` and with the `Map` (`/@`) gone and replaced by a `DateRange` inside the call. In the last case, the returned value is a `TimeSeries` object from which I extract the positions using the `"Paths"` method: SunPosition[{52.37`, 4.89`}, DateRange[{2013, 3, 1, 1, 0, 0}, {2013, 3, 1, 24, 0, 0}, "Hour"]]["Paths"][[1, All, 2]] The results were as follows: ![enter image description here][3] Clearly, the `SunPosition` results are very disappointing. Getting the sun positions with `SunPosition` is almost 40 times slower than using the old V9 method (which, I should add, wasn't particularly quick either. I have an implementation in Mathematica code which is faster). The V10 implementation of `AstronomicalData` is also more than three times slower than the V9 version. The `DateRange` version of the call saves a lot of communication overhead. Still, it is almost *five times slower* than in V9. The cause of all this slowness is that `SunPosition` simply does a call to Wolfram|Alpha. Sniffing the communication one sees the following string passed to the server: "1:eJxTTMoPSuNgYGAoZgESPpnFJcHcQEZwaV5AfnFmSWZ+XhoTsmxR/6GvGjH9wg4Qhr6XQxobsnzmXXYGhkxmIC+TEUSIgwggZihigIJgoAIGj/yizKr8PJigA5yBZtubwB1yrdxeDkXVIuvcH1aJOBRzAqUcS0vycxNLMpMBSAArww==" which can be turned into readable form using `Uncompress`: {"SunPosition", {4.89, 52.37}, {2013, 3, 1, 23, 0, 0.}, "Horizon", 2., 2., {52.09, 5.12}, Automatic} Here, we can recognize the lat/long of the position I used (but with lat/long reversed - Is this somehow significant?). At the end is my own `$GeoLocation`, but I don't believe it is used at all (and it shouldn't: I'm asking for the sun position over Amsterdam, not where I live). Changing it with `Block` I get the same results: Block[{$GeoLocation = GeoPosition[{52.37`, 40.89`}]}, SunPosition[{52.37`, 4.89`}, {2013, 3, 1, 1, 0, 0}]] Apart from the slowness, there's the issue of the necessary Internet connectivity (Want to give a demonstration and you don't have Internet? Sorry, you're out of luck). And what of the use of W|A calls? Each of the `SunPosition` tests (except the last one) took me 20 * 24 = 480 calls. So this part of my testing only already took 1440 calls, and one should be reminded that a typical Home Use license allows for only 3,000 calls per month. Things like this can go pretty fast. In fact, I once wrote an application that calculates the impact of building changes on shadows around your house throughout the year. It does in the order of 17,000 `AstronomicalData` calls. I couldn't implement that naively using `SunPosition` and have it actually work. Clearly, one should now use the `DateRange` version of the call as much as possible. ---------- To wrap up: I have one real question, i.e., how to get `SunPosition` to return the same values as `AstronomicalData`, and a request to the WRI team: please put `SunPosition` in the kernel and don't use W|A calls, because the situation as it is now is rather annoying and IMHO a real step backwards. [1]: /c/portal/getImageAttachment?filename=AstronomicalData.png&userId=43903 [2]: /c/portal/getImageAttachment?filename=timeout.png&userId=43903 [3]: /c/portal/getImageAttachment?filename=results.png&userId=43903 Sjoerd de Vries 2014-07-13T16:39:48Z https://community.wolfram.com/groups/-/m/t/763123 I would ask for some help to implement a code to be able to find the center of mass of an annular image of light projection from a fiber optics. The center of mass should be the point where it originates the average radius. On the other hand, the average radius ends at the point, inside the annulus, where the light intensity is maximum. Basically, an initial center is chosen (may be the center of the frame), then 18 sectors are built from that center to the larger circle of the annulus, which is contained in the frame. The fix for the assumed center is calculated as the sum of the average radius times the cosine (in x) and sine (for y) sectors around the circle. The center is shifted to there and the process repeated. In another words, the new coordinates obtained, original center plus delta x and delta y, must to shift the original center closer to the correct center of mass. However the operation should be performed some times until convergence is complete or that the residual value is less than or equal to 1 pixel. This should probably happen after two or three interactions, depending of course the amount initially chosen for the original center. So far, the current code is able to find the value delta x and delta y to be added to the center originally preset. My difficulty is define a loop to redo the operation as many times as necessary. I would really appreciate some help to finish this code to obtain the final value of the center of mass coordinates. Thanks for any help... Antonio de Oliveira 2015-12-23T18:26:51Z https://community.wolfram.com/groups/-/m/t/440994 Dear friends, I have a lot of Raw images in .NEF format (NIKKON). So, I read the section : [http://reference.wolfram.com/language/LibraryLink/tutorial/ImageProcessing.html][1] But I did not understand how implement these libraries in order to work with .NEF extension images, for astrophotography. EDIT 1: a sample of image *.NEF: [https://www.dropbox.com/s/2n5y5xjhhqeh7k0/DSC_5133.NEF?dl=0][2] [1]: http://reference.wolfram.com/language/LibraryLink/tutorial/ImageProcessing.html [2]: https://www.dropbox.com/s/2n5y5xjhhqeh7k0/DSC_5133.NEF?dl=0 Marcelo De Cicco 2015-02-12T13:19:44Z https://community.wolfram.com/groups/-/m/t/1500089 I am using Mathematica 10.3. I want fo perform a computational analysis of the motion of the Earth around the sun based on Kepler’s laws. Here is my code so far. eulerStep[{t_, state_List}, h_, f_List] := {t + h, state + h Through[f[{t, state}]]} solveSystemEuler [{t0_state0 _}, h_, n_Integer, f_List] := NestList[eulerStep[#, h, f] &, {t0, state0}, n] midptStep[{t_, state_List}, h_, f_List] := {t + h, state + h Through[ f[{t + 1/2 h, state + 1/2 h Through[f[{t, state}]]}]]} solveSytemMidPt[{t0_, state0_}, h_, n_Integer, f_List] := NestList[midptStep[#, h, f] &, {t0, state0}, n] L = 1/2 m (x'[t]^2 + y'[t]^2) + GMm/Sqrt[x[t]^2 + y[t]^2]; D[D[L, x'[t]], t] - D[L, x[t]] == 0 D[D[L, y'[t]], t] - D[L, y[t]] == 0 xdot[{t_, {x_, vx_, y_, vy_}}] := vx vxdot[{t_, {x_, vx_, y_, vy_}}] := -x/(x^2 + y^2)^(3/2) ydot[{t_, {x_, vx_, y_, vy_}}] := vy vydot[{t_, {x_, vx_, y_, vy_}}] := -y/(x^2 + y^2)^(3/2) start = {1, 0, 0, 1}; fcns = {xdot, vxdot, ydot, vydot}; orbit = solveSystemEuler[{0, start}, 0.01, 800, fcns]; << Statistics`DataManipulation` xypts = Column[Column[orbit, 2], {1, 3}]; ListPlot[xypts, PlotJoined -> True]; Running the program gave the following error messages. ![enter image description here][1] Please help me to fix my code. [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture.JPG&userId=1499975 Senlau Minto 2018-10-08T03:45:10Z https://community.wolfram.com/groups/-/m/t/2040012 I Have two set of data separately, which one of them is pressure and the other one is energy density.(P: pressure, E: energy density) I want to use these data as a function, P(E) or E(P) by utilizing an Interpolation between them. Could you please help me that how I can make a function of these data? For instance in Matlab I wrote the codes below and they work well. E=interp1(Data(:,2),Data(:,1),P); P=interp1(Data(:,1),Data(:,2),E); Regarding the codes in Matlab, I have a Matrix named Data which the first column is related to energy density and the second column is for pressure. In fact, by the codes above in Matlab, I could give an arbitrary pressure and then the result is the corresponding energy density or vice versa. For my Mathematica code, I separated the columns of the matrix mentioned above into two data as two Excel files. Then these two files were imported in the notebook in order to make an interpolation between them to utilize it as a function. Indeed the interpolation between energy density and pressure could work like a function which gives pressure in terms of energy density or vice versa. My question is that how I can write such a code in Mathematica. Davood Rafiei 2020-07-19T13:13:11Z https://community.wolfram.com/groups/-/m/t/574455 Please find attached a small routine that plots the intensity profile between two points in a circular image. I'd like a routine that was able to make several radial routes, such as the example, from the center of the image to its exterior. Perhaps with an adjustable angular variation. I am very grateful for any help. Antonio ![enter image description here][1] [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=sdf34546ygrwefads.png&userId=11733 Antonio de Oliveira 2015-10-02T19:52:07Z https://community.wolfram.com/groups/-/m/t/533151 Here is my code: (* Take inputs for ECEF *) x = Input["What is the x coordinate?"]; y = Input["What is the y coordinate?"]; z = Input["What is the z coordinate?"]; (* Put in Coordinate Form *) GeoPositionXYZ[{x, y, z}, "ITRF00"]; (* Convert to LLA *) GeoPosition[%] (* Display Map *) GeoGraphics[GeoMarker[GeoPosition[%]],GeoRange -> "World", GeoProjection -> "Robinson"] It lets me input coordinates for one point in ECEF, then converts it to latitude/longitude/height and shows it on a map. I want to keep showing the 2d map... but I also want to show a 3d globe that locates the coordinates on the 3d cartesian plane also. I need to be able to put in more than one coordinate. It needs to keep asking for more x's, more y's, more z's... until I stop inputting them. Eventually it's going to be modified to pull them out of a text file, but for now these two changes need to be made. Any ideas/hints/help/guidance is appreciated. Thanks. Nathan Lundholm 2015-07-20T07:16:28Z https://community.wolfram.com/groups/-/m/t/251690 Dear community members, I'm currently try to use Wolfram Mathematica to some gravity research. But i can't find built-in methods do differential geometry calculations in Mathematica. For example, is there any way to compute Einstein or Ricci tensor by metric? Or something more complicate, like create manifold, some medium with fixed state equation and write Einstein equations for this system? Thanks, Boris Latosh Boris Latosh 2014-05-14T18:31:41Z https://community.wolfram.com/groups/-/m/t/2299843 Dear all, when evaluating "HankelH1[n,z]" in Mathematica the resulting real part significantly differs from the result I obtain when I use "scipy.special.hankel1(n,z)" in Python for large values of n. For example, "HankelH1[5,3]" and "scipy.special.hankel1(5,3)" are in sufficient agreement, however for "HankelH1[16,3]" and "scipy.special.hankel1(16,3)" the result significantly differs. Does anybody have an idea why that is the case? Thank you very much. Kind regards D D 2021-06-27T05:37:11Z https://community.wolfram.com/groups/-/m/t/772569 Hello, I’m a graphic artist working for an education company in Arizona. I was given the task of writing a twenty page reader about black holes for middle school students, and all of my research has gone well. But I’ve reached a dead end regarding time dilation. I’m writing a scenario where the reader visits a ten-solar mass black hole while his or her friend stays at a safe distance. I would like to write the following: > If you could stay just in front of the event horizon, you could watch > your ten year old friend turn 100 years old in just [*xxx* *amount of > time*]. Unfortunately, I can’t get an adequate answer to this. I was directed to the time dilation calculator here http://www.wolframalpha.com/input/?i=time+dilation+calculator , but I don’t know how to use it. Just playing with it I’ve gotten negative numbers, *i*, and “exceeds the speed of light’. I have no idea what any of this means. What’s the gravitational acceleration? What’s the rest frame? What’s the radius of what? I know the time should be very short, but “a blink of an eye” isn’t useful. Would some kind-hearted soul be willing to walk me through this in layman’s language? Or better yet, give me an accurate (but not necessarily precise) number. Any help is greatly appreciated. Thank you and best regards, Jack Jack M 2016-01-13T02:39:29Z https://community.wolfram.com/groups/-/m/t/803034 Can anybody tell me how to compute the ground position of the moon? Specifically, the point on earth where the moon is directly overhead for a specific time? I know there is MoonPosition, but how do I convert that to a Lat, Lon. Thanks Thomas Dobroth 2016-02-25T19:42:14Z https://community.wolfram.com/groups/-/m/t/563314 Hi everyone, As the title said, I want to plot this function using mathematica: ![Luminosity distance][1] [1]: /c/portal/getImageAttachment?filename=distance_luminosity.png&userId=563000 where the only variable is the redshift z, how can I do that with the function Plot in mathematica ? The goal si to obtain the dL(z) as function of redshift z. Thanks. Matteo De Leone 2015-09-11T13:12:02Z https://community.wolfram.com/groups/-/m/t/487004 I'm trying to use Mathematica to drive an autofocus system for a telescope and ccd camera. Three calculations are commonly used FWHM, HFD (half flux diameter) and sigma. So I need to be able to calculate those from a small (100x100) crop from an image produced by my tethering software. I found a Mathematica program to separate the stars out of the small image that works great so that is all sorted. As I understand it you do a gaussian distribution on the pixel data then calculate the two values to use to figure out the direction and when you are in focus. I'm new to Mathematica discovered it when I bought a Raspberry Pi and since I can buy a copy for about half what commercial software to do autofocus costs I thought it would be the way to go as I can use Mathematica for so much more. But I'm not having any luck so far doing the GD. So any clues tips or links are very much appreciated. Dan Dan Pollock 2015-04-27T00:09:28Z https://community.wolfram.com/groups/-/m/t/473497 Dear All, I want to know the distance of the moon from earth for some particular date. I found the following in the help: ![enter image description here][1] But if I evaluate it I get: ![enter image description here][2] Did the syntax change? Or should I use another function? On a side note: Is this the center-to-center distance or minimal surface to surface distance? P.S.: This example was found in the PlanetaryMoonData at the end of the 'Scope' section. [1]: /c/portal/getImageAttachment?filename=ScreenShot2015-04-04at16.27.24.png&userId=73716 [2]: /c/portal/getImageAttachment?filename=6584ScreenShot2015-04-04at16.28.06.png&userId=73716 Sander Huisman 2015-04-04T14:41:06Z https://community.wolfram.com/groups/-/m/t/1438449 Hello, I am currently trying to solve the reduced three body problem in mathematica. I have the equations of motion $\ddot{x}-2\dot{y}=-\frac{\partial\Omega}{\partial x } $ $\ddot{y}+2\dot{x}=-\frac{\partial\Omega}{\partial y } $ $\Omega=-\frac{1}{2}\mu r_1^2-\frac{1}{2}(1-\mu)r_2^2-\frac{\mu}{r_1}-\frac{1-\mu}{r_2}$ Where $r_{1,2}$ are the distances between the small body and the two massive bodies , and $\mu$ is such that the ratio of the masses is $\mu:1-\mu$. I have shown that $J=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2+\Omega(x,y)$ is conserved. Then to obtain the motion I wrote the code below, but I cannot seem to get $J$ to be conserved. Does anyone have a clue why? Plotting x(t) and y(t) gives some discontinuous derivatives at t=6s which is when there is the first major change in the value of J. I tried forcing NDSolve to use runge-kutta but that didn't help, forcing it to use Euler gave nonsense (outward spiralling circular orbit, as if there were no gravity) In[139]:= r1[x_, y_, u_] := Sqrt[ (x + 1 - u)^2 + y^2] u = 0.5 r2[x_, y_, u_] := Sqrt[(x - u)^2 + y^2] Om[x_, y_, u_] := -1/2*u*r1[x, y, u]^2 - 1/2*(1 - u)*r2[x, y, u]^2 - u/r1[x, y, u] - (1 - u)/r2[x, y, u] In[184]:= sol = NDSolve[ {x''[t] - 2 y'[t] == -D[Om[x[t], y[t], u], x[t]], y''[t] + 2 x'[t] == -D[Om[x[t], y[t], u], y[t]], x[0] == y[0] == 0, x'[0] == y'[0] == 0.5}, {x, y}, {t, 100} ] In[183]:= ParametricPlot[Evaluate[{x[t], y[t]} /. sol], {t, 0, 100}, PlotRange -> All] In[116]:= J[t_] := 1/2*dX[t]^2 + 1/2*dY[t]^2 + Om[X[t], Y[t], u] In[106]:= pX[t_] := x[t] /. sol X[t_] := pX[t][[1]] In[110]:= pdX[t_] := x'[t] /. sol In[111]:= dX[t_] := pdX[t][[1]] pY[t_] := y[t] /. sol Y[t_] := pY[t][[1]] pdY[t_] := y'[t] /. sol dY[t_] := pdY[t][[1]] (* dX[t] gives derivative of x[t], X[t] gives x[t]. We introduce X \ and Y to obtain reals, not one element lists, which correspond, e.g., \ to x[t]/.sol *) In[182]:= Plot[Evaluate[J[t]], {t, 1, 30}] thomas rialan 2018-09-04T12:09:43Z https://community.wolfram.com/groups/-/m/t/217048 Hello, I have constructed a function which is simply a three-dimensional gaussian: As you can see, these are both gaussian terms, so they should be easy to integrate. I have tried many methods of NIntegrate, which do not converge or give a consistent answer. This function is very simple and well-behaved, it is a ball at r=0 and a shell at r=3, both of gaussian shape. How can I produce the definite integral of this function? [mcode]Conv[x_, y_, z_, xp_, yp_, zp_] := (1/4)*1/Sqrt[Pi]*Exp[-((xp - x)^2 + (yp - y)^2 + (zp - z)^2)] + (3/4)*(1/Sqrt[Pi])*Exp[-4*(Sqrt[(xp - x)^2 + (yp - y)^2 + (zp - z)^2] - 3.5)^2] [/mcode] C. E. Coppola 2014-03-11T19:12:07Z https://community.wolfram.com/groups/-/m/t/310109 For horizontal coordinates of stars, I have noted that StarData in Mathematica 10 returns values I don't expect. Taking a simple example, the altitude of Polaris. Regardless of date or time of day it will be close to the observers latitude. My home is near the 64th parallel. For latitude 64N, v9 gives a reasonable value: ![enter image description here][1] But not so in v10: ![enter image description here][2] And a query directly in W|A gives the same, wrong, answer: ![enter image description here][3] Changing $GeoLocation has no effect [1]: /c/portal/getImageAttachment?filename=v9.jpg&userId=93385 [2]: /c/portal/getImageAttachment?filename=v10.jpg&userId=93385 [3]: /c/portal/getImageAttachment?filename=WA.jpg&userId=93385 Hans Milton 2014-07-31T21:26:34Z https://community.wolfram.com/groups/-/m/t/3067368 Hi there, I have some basic code for plotting the gravitional time dilation as one moves closer to the event horizon of a black hole of a given mass. See code: ClearAll["Global`*"]'; tp = 1; G = 6.67408*10^(-11); M = (6.5*10^9)*(1.989*10^30); c = 2.99*10^8; rs = (2*G* M)/(c^2); Print[]; Print[" Black hole mass:", M, "kg", " \ Schhwarzchild radius:", rs, "m"]; Print[]; Plot[ tp*Sqrt[1 - (rs/r)], {r, 0, 2*10^14}, AxesLabel -> {" ", "Time (s)"}, AxesLabel -> {Style[" (m)", Bold, 26], Style["Time (s)", Bold, 16]}, LabelStyle -> Directive[Black, 16], AxesOrigin -> {0, 0}, GridLines -> {{rs}, {}}, GridLinesStyle -> {Directive[{Dashed, Thick}, Red], Directive[Thick, Red]}, ColorFunction -> "NeonColors"] I want to create a 3D plot for different masses of black holes ranging from the one given in the code to one a thousand times more massive, with a 3D sheet. I am not great at this, and while my basic code works fine, when I go to do a 3D plot, nothing seems to work. Estelle Asmodelle 2023-11-20T03:35:09Z https://community.wolfram.com/groups/-/m/t/919574 We can easily get properties of individual stars with `StarData`, i.e. `StarData[star, property]`. But this is not very useful. I might as well look up the star in Wikipedia or elsewhere. Such a database becomes useful not as an encylopedia, but as a comprehensive dataset which we can use to calculate various statistics and relationships between properties. For example, plot a Hertzsprung–Russel diagram. **How can I retrieve a property of *all* stars in the database?** This seems to work in principle, but in practice it takes such a long time that it is plainly unusable. It would probably take more than an hour to finish. magn = StarData[EntityClass["Star", All], "AbsoluteMagnitude"]; In that time I could google up several easier to use databases, download the data, figure out how to import it to Mathematica, etc. And that's just *one* property. Is there a better way then? Is there something wrong with the syntax I am using that causes this to be so slow? If not, then what is the point of all these `*Data` functions, given that several of them are practically unusable unless we're satisfied with looking at items (stars) one by one? Is there anyone on this forum who is able to make real practical use of this function? If yes, how? `StarData` contains about 100,000 stars. That's not a lot. 100,000 floating point number take less than a megabyte of storage and arithmetic with such arrays typically takes less than a millisecond. Szabolcs Horvát 2016-09-06T09:16:45Z https://community.wolfram.com/groups/-/m/t/574679 I need help figuring out how to have a Mathematica script open a file. Each line of the script is in the format: C Name t_i t_f x y z x_dot y_dot z_dot x_sig y_sig z_sig x_dot_sig y_dot_sig z_dot_sig I have attached the file. I need my script to open the file, and for each satellite it needs to take the first row, take Name, X, Y, Z... so the second column, 5th column, 6th column, and 7th column, for the first row of the file for each satellite. (there are several satellites numbered GPS15, GPS24, GPS25, etc.... It then needs to map x, y, z for all of the satellites and label them on the map with their Name. I know it involves lists. I've tried, I can't figure out how to do this. Thank you so much! Nathan Lundholm 2015-10-03T19:31:11Z