From roberto.soleti at gmail.com Thu Nov 7 14:34:36 2019 From: roberto.soleti at gmail.com (Stefano Roberto Soleti) Date: Thu, 7 Nov 2019 11:34:36 -0800 Subject: [AstroPy] Convolution with astropy Message-ID: Dear astropy users, I am having a problem understanding how the convolution of two models is done in astropy. I want to convolve my function, centered around a certain value x, with a Gaussian with mean 0 and sigma > 0. However, when I try to plot the result of the convolution, the function is centered around ~x/2. I think I am missing something fundamental here but I couldn?t find anything apart from the basic example on astropy docs. Here it is my code snippet: import numpy as np import math import matplotlib.pyplot as plt from astropy.modeling.models import custom_model from astropy.modeling import models from astropy.convolution import convolve_models from scipy.constants import alpha @custom_model def spectrum(x): me = 0.511 eMax = 104.97 E = np.sqrt(x*x + me*me) result = (1./eMax)*(alpha/(2*math.pi))*(np.log(4*E*E/me/me)-2.)*((E*E+eMax*eMax)/eMax/(eMax-E)) result[result < 0] = 0 return result x = np.linspace(0,120,1000) s = spectrum() g1 = models.Gaussian1D(1, 0, 2) conv = convolve_models(g1, s) plt.plot(x, conv(x), 'k-') plt.plot(x, s(x)) Thank you very much for your help! From bsipocz at gmail.com Tue Nov 19 17:21:02 2019 From: bsipocz at gmail.com (Brigitta Sipocz) Date: Tue, 19 Nov 2019 14:21:02 -0800 Subject: [AstroPy] astropy v4.0rc1 release candidate Message-ID: Dear all, The first astropy v4.0 release candidate is now available and ready for testing. https://pypi.python.org/pypi/astropy/4.0rc1 We would appreciate if you could try out this RC and report back any success or failure via the following wiki page. If you maintain any downstream packages please check, and preferably whether they are compatible or there are any blocking issues: https://github.com/astropy/astropy/wiki/v4.0-RC-testing Please also open issues for each different type of failure (please check the issue tracker first to make sure no one has reported it earlier). In addition of running the astropy test suite itself, we especially encourage you to also test it with as many packages of yours as possible. There is a section to list packages at the bottom of the wiki page. You should be able to install the RC by doing: pip install astropy --pre or pip install astropy==4.0rc1 We also made some refactoring in our testing infrastructure, so please make sure you have the latest version of pytest-astropy, including pytest-astropy-header installed. If you use conda, be sure to either remove the existing conda astropy package or make a new environment. Happy testing, Brigitta -------------- next part -------------- An HTML attachment was scrubbed... URL: From franco.basti at gmail.com Sat Nov 23 14:00:56 2019 From: franco.basti at gmail.com (franco basti) Date: Sat, 23 Nov 2019 14:00:56 -0500 Subject: [AstroPy] Convert from AltAz to ICRS Message-ID: Can anyone help me converting from a alt az to ra,dec ? all I can find are examples that convert from ra,dec to alt,az Thx Franco -------------- next part -------------- An HTML attachment was scrubbed... URL: From ejensen1 at swarthmore.edu Sat Nov 23 15:20:59 2019 From: ejensen1 at swarthmore.edu (Eric Jensen) Date: Sat, 23 Nov 2019 15:20:59 -0500 Subject: [AstroPy] Convert from AltAz to ICRS In-Reply-To: References: Message-ID: <8569E4CC-6B5B-4F68-B23C-7DDE9705B91A@swarthmore.edu> Hi Franco, You can use the same transform_to method that is given in the example docs, just in the other direction, e.g. object_radec = object_altaz.transform_to(?icrs') Referring to the example here: https://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html#sphx-glr-generated-examples-coordinates-plot-obs-planning-py Here is some code based on that example, that extends it to show a transformation back to RA and Dec: import astropy.units as u from astropy.time import Time from astropy.coordinates import SkyCoord, EarthLocation, AltAz # Get M33's coordinates m33 = SkyCoord.from_name('M33') # Define a location for observation: bear_mountain = EarthLocation(lat=41.3*u.deg, lon=-74*u.deg, height=390*u.m) utcoffset = -4*u.hour # Eastern Daylight Time time = Time('2012-07-12 23:00:00') - utcoffset # Find the alt/az at that time: m33altaz = m33.transform_to(AltAz(obstime=time,location=bear_mountain)) print("M33's Altitude = {0.alt:.2f}, Azimuth = {0.az:.1f}".format(m33altaz)) # prints: M33's Altitude = 0.13 deg, Azimuth = 47.3 deg # Look at the same alt/az, # but half a sidereal day later: new_altaz = SkyCoord(AltAz(obstime=time + 0.5*u.sday, az=m33altaz.az, alt=m33altaz.alt, location=bear_mountain)) # Transform back to RA, Dec; # should get the same Dec as before but RA 12 hours different: newradec = new_altaz.transform_to('icrs') print(m33.to_string('hmsdms')) print(newradec.to_string('hmsdms?)) # Prints # 01h33m50.904s +30d39m35.79s # 13h33m59.1873s +30d47m03.8071s Interestingly, the declination isn?t *exactly* the same and the RA doesn?t differ by exactly 12 hours - I wonder why? But overall you can see the idea for doing what you want. Best, Eric > On Nov 23, 2019, at 2:00 PM, franco basti wrote: > > > Can anyone help me converting from a alt az to ra,dec ? > all I can find are examples that convert from ra,dec to alt,az > > Thx > Franco > > _______________________________________________ > AstroPy mailing list > AstroPy at python.org > https://mail.python.org/mailman/listinfo/astropy -------------- next part -------------- A non-text attachment was scrubbed... Name: smime.p7s Type: application/pkcs7-signature Size: 3952 bytes Desc: not available URL: From keithmclaurin at gmail.com Tue Nov 26 00:06:54 2019 From: keithmclaurin at gmail.com (keithmclaurin at gmail.com) Date: Tue, 26 Nov 2019 00:06:54 -0500 Subject: [AstroPy] Convert from AltAz to ICRS In-Reply-To: <8569E4CC-6B5B-4F68-B23C-7DDE9705B91A@swarthmore.edu> References: <8569E4CC-6B5B-4F68-B23C-7DDE9705B91A@swarthmore.edu> Message-ID: <595E44E9-2129-41AB-BD32-A94D6BF77F43@gmail.com> The RA and dec changes mostly due to precession is my understanding. Sent from my iPhone > On Nov 23, 2019, at 15:20, Eric Jensen wrote: > > Hi Franco, > > You can use the same transform_to method that is given in the example docs, just in the other direction, e.g. > > object_radec = object_altaz.transform_to(?icrs') > > Referring to the example here: > https://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html#sphx-glr-generated-examples-coordinates-plot-obs-planning-py > > Here is some code based on that example, that extends it to show a transformation back to RA and Dec: > > import astropy.units as u > from astropy.time import Time > from astropy.coordinates import SkyCoord, EarthLocation, AltAz > > # Get M33's coordinates > m33 = SkyCoord.from_name('M33') > > # Define a location for observation: > bear_mountain = EarthLocation(lat=41.3*u.deg, lon=-74*u.deg, height=390*u.m) > utcoffset = -4*u.hour # Eastern Daylight Time > time = Time('2012-07-12 23:00:00') - utcoffset > > # Find the alt/az at that time: > m33altaz = m33.transform_to(AltAz(obstime=time,location=bear_mountain)) > print("M33's Altitude = {0.alt:.2f}, Azimuth = {0.az:.1f}".format(m33altaz)) > > # prints: M33's Altitude = 0.13 deg, Azimuth = 47.3 deg > > # Look at the same alt/az, > # but half a sidereal day later: > new_altaz = SkyCoord(AltAz(obstime=time + 0.5*u.sday, > az=m33altaz.az, > alt=m33altaz.alt, > location=bear_mountain)) > > # Transform back to RA, Dec; > # should get the same Dec as before but RA 12 hours different: > newradec = new_altaz.transform_to('icrs') > print(m33.to_string('hmsdms')) > print(newradec.to_string('hmsdms?)) > > # Prints > # 01h33m50.904s +30d39m35.79s > # 13h33m59.1873s +30d47m03.8071s > > Interestingly, the declination isn?t *exactly* the same and the RA doesn?t differ by exactly 12 hours - I wonder why? But overall you can see the idea for doing what you want. > > Best, > > Eric > > >> On Nov 23, 2019, at 2:00 PM, franco basti wrote: >> >> >> Can anyone help me converting from a alt az to ra,dec ? >> all I can find are examples that convert from ra,dec to alt,az >> >> Thx >> Franco >> >> _______________________________________________ >> AstroPy mailing list >> AstroPy at python.org >> https://mail.python.org/mailman/listinfo/astropy > > _______________________________________________ > AstroPy mailing list > AstroPy at python.org > https://mail.python.org/mailman/listinfo/astropy From ejensen1 at swarthmore.edu Tue Nov 26 07:58:58 2019 From: ejensen1 at swarthmore.edu (Eric Jensen) Date: Tue, 26 Nov 2019 07:58:58 -0500 Subject: [AstroPy] Convert from AltAz to ICRS In-Reply-To: <595E44E9-2129-41AB-BD32-A94D6BF77F43@gmail.com> References: <8569E4CC-6B5B-4F68-B23C-7DDE9705B91A@swarthmore.edu> <595E44E9-2129-41AB-BD32-A94D6BF77F43@gmail.com> Message-ID: <6CE26C72-6BFE-437C-8DBB-B75A21C28B63@swarthmore.edu> > On Nov 26, 2019, at 12:06 AM, keithmclaurin at gmail.com wrote: > > The RA and dec changes mostly due to precession is my understanding. Thanks for responding. A quick back-of-the-envelope calculation shows that precession (360 degrees in 26,000 years) should be less than 0.1 arcsecond over 12 hours, whereas the difference here is quite a bit bigger, about 7.5 arcminutes. So it must be something else - maybe some internal rounding or (more likely) some more subtle misunderstanding by me of some conversion / reference frame issue. In any case, it?s not really germane to the original query, but still has my curiosity piqued. Eric > >> On Nov 23, 2019, at 15:20, Eric Jensen wrote: >> >> Hi Franco, >> >> You can use the same transform_to method that is given in the example docs, just in the other direction, e.g. >> >> object_radec = object_altaz.transform_to(?icrs') >> >> Referring to the example here: >> https://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html#sphx-glr-generated-examples-coordinates-plot-obs-planning-py >> >> Here is some code based on that example, that extends it to show a transformation back to RA and Dec: >> >> import astropy.units as u >> from astropy.time import Time >> from astropy.coordinates import SkyCoord, EarthLocation, AltAz >> >> # Get M33's coordinates >> m33 = SkyCoord.from_name('M33') >> >> # Define a location for observation: >> bear_mountain = EarthLocation(lat=41.3*u.deg, lon=-74*u.deg, height=390*u.m) >> utcoffset = -4*u.hour # Eastern Daylight Time >> time = Time('2012-07-12 23:00:00') - utcoffset >> >> # Find the alt/az at that time: >> m33altaz = m33.transform_to(AltAz(obstime=time,location=bear_mountain)) >> print("M33's Altitude = {0.alt:.2f}, Azimuth = {0.az:.1f}".format(m33altaz)) >> >> # prints: M33's Altitude = 0.13 deg, Azimuth = 47.3 deg >> >> # Look at the same alt/az, >> # but half a sidereal day later: >> new_altaz = SkyCoord(AltAz(obstime=time + 0.5*u.sday, >> az=m33altaz.az, >> alt=m33altaz.alt, >> location=bear_mountain)) >> >> # Transform back to RA, Dec; >> # should get the same Dec as before but RA 12 hours different: >> newradec = new_altaz.transform_to('icrs') >> print(m33.to_string('hmsdms')) >> print(newradec.to_string('hmsdms?)) >> >> # Prints >> # 01h33m50.904s +30d39m35.79s >> # 13h33m59.1873s +30d47m03.8071s >> >> Interestingly, the declination isn?t *exactly* the same and the RA doesn?t differ by exactly 12 hours - I wonder why? But overall you can see the idea for doing what you want. >> >> Best, >> >> Eric >> >> >>> On Nov 23, 2019, at 2:00 PM, franco basti wrote: >>> >>> >>> Can anyone help me converting from a alt az to ra,dec ? >>> all I can find are examples that convert from ra,dec to alt,az >>> >>> Thx >>> Franco >>> >>> _______________________________________________ >>> AstroPy mailing list >>> AstroPy at python.org >>> https://mail.python.org/mailman/listinfo/astropy >> >> _______________________________________________ >> AstroPy mailing list >> AstroPy at python.org >> https://mail.python.org/mailman/listinfo/astropy > _______________________________________________ > AstroPy mailing list > AstroPy at python.org > https://mail.python.org/mailman/listinfo/astropy From bjw at as.arizona.edu Tue Nov 26 13:34:48 2019 From: bjw at as.arizona.edu (Benjamin Weiner) Date: Tue, 26 Nov 2019 13:34:48 -0500 Subject: [AstroPy] Convert from AltAz to ICRS In-Reply-To: References: Message-ID: Hi Eric, The precession that is causing this issue is the precession of the earth relative to the axis of the ICRS coordinate system, which is within an arcsec or so of J2000. The example used July 2012 so the earth's axis is 12 years off the coordinate system axis. Try the example again but with an observation time in July 2000: time2 = Time('2000-07-12 23:00:00') - utcoffset m33altaz = m33.transform_to(AltAz(obstime=time2,location=bear_mountain)) new_altaz = SkyCoord(AltAz(obstime=time2 + 0.5*u.sday, az=m33altaz.az, alt=m33altaz.alt, location=bear_mountain)) newradec = new_altaz.transform_to('icrs') print(m33.to_string('hmsdms')) print(newradec.to_string('hmsdms')) # prints: # 01h33m50.904s +30d39m35.79s # 13h33m51.157s +30d39m20.5722s These are much closer to 12 hours RA difference. cheers, Ben On Tue, Nov 26, 2019 at 12:01 PM wrote: > Send AstroPy mailing list submissions to > astropy at python.org > > To subscribe or unsubscribe via the World Wide Web, visit > https://mail.python.org/mailman/listinfo/astropy > or, via email, send a message with subject or body 'help' to > astropy-request at python.org > > You can reach the person managing the list at > astropy-owner at python.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of AstroPy digest..." > > > Today's Topics: > > 1. Re: Convert from AltAz to ICRS (keithmclaurin at gmail.com) > 2. Re: Convert from AltAz to ICRS (Eric Jensen) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Tue, 26 Nov 2019 00:06:54 -0500 > From: keithmclaurin at gmail.com > To: Astronomical Python mailing list > Subject: Re: [AstroPy] Convert from AltAz to ICRS > Message-ID: <595E44E9-2129-41AB-BD32-A94D6BF77F43 at gmail.com> > Content-Type: text/plain; charset=utf-8 > > The RA and dec changes mostly due to precession is my understanding. > > Sent from my iPhone > > > On Nov 23, 2019, at 15:20, Eric Jensen wrote: > > > > Hi Franco, > > > > You can use the same transform_to method that is given in the example > docs, just in the other direction, e.g. > > > > object_radec = object_altaz.transform_to(?icrs') > > > > Referring to the example here: > > > https://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html#sphx-glr-generated-examples-coordinates-plot-obs-planning-py > > > > Here is some code based on that example, that extends it to show a > transformation back to RA and Dec: > > > > import astropy.units as u > > from astropy.time import Time > > from astropy.coordinates import SkyCoord, EarthLocation, AltAz > > > > # Get M33's coordinates > > m33 = SkyCoord.from_name('M33') > > > > # Define a location for observation: > > bear_mountain = EarthLocation(lat=41.3*u.deg, lon=-74*u.deg, > height=390*u.m) > > utcoffset = -4*u.hour # Eastern Daylight Time > > time = Time('2012-07-12 23:00:00') - utcoffset > > > > # Find the alt/az at that time: > > m33altaz = m33.transform_to(AltAz(obstime=time,location=bear_mountain)) > > print("M33's Altitude = {0.alt:.2f}, Azimuth = {0.az: > .1f}".format(m33altaz)) > > > > # prints: M33's Altitude = 0.13 deg, Azimuth = 47.3 deg > > > > # Look at the same alt/az, > > # but half a sidereal day later: > > new_altaz = SkyCoord(AltAz(obstime=time + 0.5*u.sday, > > az=m33altaz.az, > > alt=m33altaz.alt, > > location=bear_mountain)) > > > > # Transform back to RA, Dec; > > # should get the same Dec as before but RA 12 hours different: > > newradec = new_altaz.transform_to('icrs') > > print(m33.to_string('hmsdms')) > > print(newradec.to_string('hmsdms?)) > > > > # Prints > > # 01h33m50.904s +30d39m35.79s > > # 13h33m59.1873s +30d47m03.8071s > > > > Interestingly, the declination isn?t *exactly* the same and the RA > doesn?t differ by exactly 12 hours - I wonder why? But overall you can > see the idea for doing what you want. > > > > Best, > > > > Eric > > > > > >> On Nov 23, 2019, at 2:00 PM, franco basti > wrote: > >> > >> > >> Can anyone help me converting from a alt az to ra,dec ? > >> all I can find are examples that convert from ra,dec to alt,az > >> > >> Thx > >> Franco > >> > >> _______________________________________________ > >> AstroPy mailing list > >> AstroPy at python.org > >> https://mail.python.org/mailman/listinfo/astropy > > > > _______________________________________________ > > AstroPy mailing list > > AstroPy at python.org > > https://mail.python.org/mailman/listinfo/astropy > > > ------------------------------ > > Message: 2 > Date: Tue, 26 Nov 2019 07:58:58 -0500 > From: Eric Jensen > To: Astronomical Python mailing list > Subject: Re: [AstroPy] Convert from AltAz to ICRS > Message-ID: <6CE26C72-6BFE-437C-8DBB-B75A21C28B63 at swarthmore.edu> > Content-Type: text/plain; charset=utf-8 > > > > On Nov 26, 2019, at 12:06 AM, keithmclaurin at gmail.com wrote: > > > > The RA and dec changes mostly due to precession is my understanding. > > Thanks for responding. A quick back-of-the-envelope calculation shows > that precession (360 degrees in 26,000 years) should be less than 0.1 > arcsecond over 12 hours, whereas the difference here is quite a bit bigger, > about 7.5 arcminutes. So it must be something else - maybe some internal > rounding or (more likely) some more subtle misunderstanding by me of some > conversion / reference frame issue. In any case, it?s not really germane > to the original query, but still has my curiosity piqued. > > Eric > > > > >> On Nov 23, 2019, at 15:20, Eric Jensen wrote: > >> > >> Hi Franco, > >> > >> You can use the same transform_to method that is given in the example > docs, just in the other direction, e.g. > >> > >> object_radec = object_altaz.transform_to(?icrs') > >> > >> Referring to the example here: > >> > https://docs.astropy.org/en/stable/generated/examples/coordinates/plot_obs-planning.html#sphx-glr-generated-examples-coordinates-plot-obs-planning-py > >> > >> Here is some code based on that example, that extends it to show a > transformation back to RA and Dec: > >> > >> import astropy.units as u > >> from astropy.time import Time > >> from astropy.coordinates import SkyCoord, EarthLocation, AltAz > >> > >> # Get M33's coordinates > >> m33 = SkyCoord.from_name('M33') > >> > >> # Define a location for observation: > >> bear_mountain = EarthLocation(lat=41.3*u.deg, lon=-74*u.deg, > height=390*u.m) > >> utcoffset = -4*u.hour # Eastern Daylight Time > >> time = Time('2012-07-12 23:00:00') - utcoffset > >> > >> # Find the alt/az at that time: > >> m33altaz = m33.transform_to(AltAz(obstime=time,location=bear_mountain)) > >> print("M33's Altitude = {0.alt:.2f}, Azimuth = {0.az: > .1f}".format(m33altaz)) > >> > >> # prints: M33's Altitude = 0.13 deg, Azimuth = 47.3 deg > >> > >> # Look at the same alt/az, > >> # but half a sidereal day later: > >> new_altaz = SkyCoord(AltAz(obstime=time + 0.5*u.sday, > >> az=m33altaz.az, > >> alt=m33altaz.alt, > >> location=bear_mountain)) > >> > >> # Transform back to RA, Dec; > >> # should get the same Dec as before but RA 12 hours different: > >> newradec = new_altaz.transform_to('icrs') > >> print(m33.to_string('hmsdms')) > >> print(newradec.to_string('hmsdms?)) > >> > >> # Prints > >> # 01h33m50.904s +30d39m35.79s > >> # 13h33m59.1873s +30d47m03.8071s > >> > >> Interestingly, the declination isn?t *exactly* the same and the RA > doesn?t differ by exactly 12 hours - I wonder why? But overall you can > see the idea for doing what you want. > >> > >> Best, > >> > >> Eric > >> > >> > >>> On Nov 23, 2019, at 2:00 PM, franco basti > wrote: > >>> > >>> > >>> Can anyone help me converting from a alt az to ra,dec ? > >>> all I can find are examples that convert from ra,dec to alt,az > >>> > >>> Thx > >>> Franco > >>> > >>> _______________________________________________ > >>> AstroPy mailing list > >>> AstroPy at python.org > >>> https://mail.python.org/mailman/listinfo/astropy > >> > >> _______________________________________________ > >> AstroPy mailing list > >> AstroPy at python.org > >> https://mail.python.org/mailman/listinfo/astropy > > _______________________________________________ > > AstroPy mailing list > > AstroPy at python.org > > https://mail.python.org/mailman/listinfo/astropy > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > AstroPy mailing list > AstroPy at python.org > https://mail.python.org/mailman/listinfo/astropy > > > ------------------------------ > > End of AstroPy Digest, Vol 158, Issue 4 > *************************************** > -- Benjamin Weiner Staff Scientist / Associate Astronomer, MMT / Steward Observatory bjw at as.arizona.edu http://mingus.mmto.arizona.edu/~bjw/ -------------- next part -------------- An HTML attachment was scrubbed... URL: From Paolo.Tanga at oca.eu Wed Nov 27 02:16:26 2019 From: Paolo.Tanga at oca.eu (Paolo Tanga) Date: Wed, 27 Nov 2019 08:16:26 +0100 Subject: [AstroPy] Convert from AltAz to ICRS In-Reply-To: References: Message-ID: An HTML attachment was scrubbed... URL: From tpursimo at not.iac.es Fri Nov 29 07:38:45 2019 From: tpursimo at not.iac.es (Tapio Pursimo) Date: Fri, 29 Nov 2019 12:38:45 +0000 (WET) Subject: [AstroPy] get_moon ... and the phase Message-ID: Hello, Can anybody help me how to get the moon phase using astropy? I can get the coordinates using get_moon, but the phase is apparently missing. I would not like to use/install "ephem" nor "astroplan". Best regards, Tapio Pursimo From ejensen1 at swarthmore.edu Fri Nov 29 12:09:21 2019 From: ejensen1 at swarthmore.edu (Eric Jensen) Date: Fri, 29 Nov 2019 12:09:21 -0500 Subject: [AstroPy] get_moon ... and the phase In-Reply-To: References: Message-ID: Hi Tapio, Here?s how I handled this in some of my code: # Moon illumination formula from Meeus, ?Astronomical # Algorithms". Formulae 46.1 and 46.2 in the 1991 edition, # using the approximation cos(psi) \approx -cos(i). Error # should be no more than 0.0014 (p. 316). moon_illum = 0.5 * (1. - sin(dec_sun)*sin(dec_moon) - cos(dec_sun)*cos(dec_moon)* cos(ra_sun - ra_moon)) That gives the percent illumination of the moon, between 0 and 1. If you can get the RA and Dec of the sun and moon, and the above precision is good enough for your purposes, this is a simple way to do it. Eric > On Nov 29, 2019, at 7:38 AM, Tapio Pursimo wrote: > > > Hello, > > Can anybody help me how to get the moon phase using astropy? > I can get the coordinates using get_moon, but the phase is apparently > missing. > I would not like to use/install "ephem" nor "astroplan". > > Best regards, > > Tapio Pursimo > > _______________________________________________ > AstroPy mailing list > AstroPy at python.org > https://mail.python.org/mailman/listinfo/astropy -------------- next part -------------- A non-text attachment was scrubbed... Name: smime.p7s Type: application/pkcs7-signature Size: 3952 bytes Desc: not available URL: