Is comet Holmes still around?

aphh · Aug 30, 2008

Using binoculars I spotted a foggy object a few degrees down and to the right of Cassiopeia's rightmost lower star?

Carrickagh · Aug 30, 2008

I believe the closest comet to Cass is currently Broughton. But it's a magnitude 11.And it would be several degress away, nowhere near the location you describe.If you are looking toward Perseus there is a very nice double cluster. This may be near the area you describe.M52 and M103 are nearby, as well as some other nebulae. Are you sure of the location of your little fuzzy object? My guess is M52.CK <div class="Discussion_UserSignature"> </div>

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I believe the closest comet to Cass is currently Broughton. But it's a magnitude 11.And it would be several degress away, nowhere near the location you describe.If you are looking toward Perseus there is a very nice double cluster. This may be near the area you describe.M52 and M103 are nearby, as well as some other nebulae. Are you sure of the location of your little fuzzy object? My guess is M52.CK Posted by Carrickagh</DIV> Yes Comet Holmes is still around, but, while relatively bright is very hard to see since that brightness is spread over an area several times that of the full moon.""Great outburst occured on Oct. 24, 2007, and it bacame a naked eye comet of 2 mag. It kept so bright as 5.5 mag still on Apr. 30 (Carlos Labordena), but it was extremely faint and difficult to see. The size was so large, the diameter was larger than 60 arcmin. Now appearing in the morning sky, but still low. It will become observable in good condition in autumn again. The extremely faint large diffuse object may be detected with a best sky condition, around 5-6 mag with a diameter of 1 or 2 degrees. <pre>Date(TT) R.A. (2000) Decl. Delta r Elong. m1 Best Time(A, h) </pre><pre>Aug. 23 8 14.33 29 29.0 4.516 3.700 32 6.1 3:55 (244, 14) </pre><pre>Aug. 30 8 22.63 28 60.0 4.485 3.726 36 6.1 4:02 (247, 19) ""</pre><pre> This location is in the earl;y morning sky, between Gemini and Cancer</pre><pre>From http://www.aerith.net/comet/weekly/current.html</pre><pre> </pre><pre>MW</pre> <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Using binoculars I spotted a foggy object a few degrees down and to the right of Cassiopeia's rightmost lower star? Posted by aphh</DIV> What time? It sounds like the Double Cluster in Perseus <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

3488 · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Yes Comet Holmes is still around, but, while relatively bright is very hard to see since that brightness is spread over an area several times that of the full moon.""Great outburst occured on Oct. 24, 2007, and it bacame a naked eye comet of 2 mag. It kept so bright as 5.5 mag still on Apr. 30 (Carlos Labordena), but it was extremely faint and difficult to see. The size was so large, the diameter was larger than 60 arcmin. Now appearing in the morning sky, but still low. It will become observable in good condition in autumn again. The extremely faint large diffuse object may be detected with a best sky condition, around 5-6 mag with a diameter of 1 or 2 degrees. Date(TT) R.A. (2000) Decl. Delta r Elong. m1 Best Time(A, h) Aug. 23 8 14.33 29 29.0 4.516 3.700 32 6.1 3:55 (244, 14) Aug. 30 8 22.63 28 60.0 4.485 3.726 36 6.1 4:02 (247, 19) "" This location is in the earl;y morning sky, between Gemini and CancerFrom http://www.aerith.net/comet/weekly/current.htmlMW Posted by MeteorWayne</DIV> Thanks Wayne,I was trying to find the Ephermeris of Comet 17/P Holmes. So is crossing the ecliptic on the descending node.Andrew Brown. I will try & find Holmes again myself.Andrew Brown.  <div class="Discussion_UserSignature"> "I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before". Linda Morabito on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.http://www.launchphotography.com/ http://anthmartian.googlepages.com/thisislandearthhttp://web.me.com/meridianijournal </div>

adrenalynn · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> Thanks Wayne,I was trying to find the Ephermeris of Comet 17/P Holmes. So is crossing the ecliptic on the descending node.Andrew Brown. I will try & find Holmes again myself.Andrew Brown.  Posted by 3488</DIV> You won't find it with anything less than a 10" scope under very dark skies.  I had almost no luck a couple weeks ago with photographing it under long exposures in the 8".  A 14" scope with good skies is the entry point for Holmes these days, IMHO.   Certainly not binoculars... <div class="Discussion_UserSignature"> .bipartisan (bī-pär'tĭ-zən, -sən) [Adj.] Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.IMPEACH OBAMA! </div>

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>You won't find it with anything less than a 10" scope under very dark skies.  I had almost no luck a couple weeks ago with photographing it under long exposures in the 8".  A 14" scope with good skies is the entry point for Holmes these days, IMHO.   Certainly not binoculars... Posted by adrenalynn</DIV> You would have to use a very low power, wide field of view lens, though. It is 2 degrees across or more, so is 4X the diameter of the moon. ANd I've seen no reports of any central condensation, though I'll have to look into that a bit more. <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

aphh · Aug 31, 2008

There are no deep space objects in this image http://fi.wikipedia.org/wiki/Kuva:Andromeda,_fi.png in the region where I saw the round fog object last night. Closest ones are M110 and M31, but they are right down from α Cassiopeia, and I had to go to the right and down to find the object. Also, the magnitude of M110 is 8.5, so not a binocular object. I will try to see it again and locate better coordinates. <img src="http://sitelife.space.com/ver1.0/Content/images/store/10/5/5a1eb257-6c3b-4d80-85f0-f38512bea5bd.Medium.jpg" alt="" />  By comparison, if the magnitude of Jupiter's moons are 4.5, and I could barely see them with binoculars, then this fog has to be much brighter, because it jumped out from the background when located. I had been drinking, but not so much that I would start to "see things" <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" />.

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>There are no deep space objects in this image http://fi.wikipedia.org/wiki/Kuva:Andromeda,_fi.png in the region where I saw the round fog object last night. Closest ones are M110 and M31, but they are right down from α Cassiopeia, and I had to go to the right and down to find the object.Also, the magnitude of M110 is 8.5, so not a binocular object.I will try to see it again and locate better coordinates.  By comparison, if the magnitude of Jupiter's moons are 4.5, and I could barely see them with binoculars, then this fog has to be much brighter, because it jumped out from the background when located. I had been drinking, but not so much that I would start to "see things" .   Posted by aphh</DIV> Here's an APOD of the double cluster. It is visible to the naked eye as well under dark skies. I saw them easily when meteor observing this morning.http://apod.nasa.gov/apod/ap071207.htmlHere's another view:http://theskywasbruised.blogspot.com/2007/11/double-cluster-in-perseus.htmlANd another:http://www.astrophoto.net/double_cluster.htmlAnd here's something that shows the location. It's the two sets of Yellow numbers superimposed on top of each other between Cassiopia and Perseus.http://www.utahskies.org/deepsky/constellations/perseus.html <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

aphh · Aug 31, 2008

Thanks for the info, but I just saw the fog again and it appears to be deep space objects M110 and M31. But can I see them with binoculars? It is fainter than I remembered.

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks for the info, but I just saw the fog again and it appears to be deep space objects M110 and M31. But can I see them with binoculars? It is fainter than I remembered. Posted by aphh</DIV> M31 and M110 are in Andromeda, not Perseus or Cassiopia.M31 is also visible without binoculars, while M110 most likely requires a decent sized telescope, though I haven't looked at it lately. I'll get the binocs out for afew minutes during tonights meteor session.It's the wrong location. They are to the right (not below) Cassiopia.Did you look at the map that I gave you that showed the double cluster location between CAS and PER? That's the position you described....In addition, M31 is 10 or 20 times the size of M110. If the fuzzies you are looking at are near the same size, it ain't them. <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

aphh · Aug 31, 2008

Thanks Wayne,I went outside and could see Double Cluster now. But the object I described is in fact in Andromeda, and it's M31. It is directly down from α Cassiopeia (brightest star of Cassiopeia) towards the equator of the celestial sphere.It's a shame I'm such an inexperienced astronomer, so I need to use the few constellations I know as a reference. It is difficult to learn about new constellations here, because due to light pollution they do not appear the same as in books.Also, to give you better reference I'd need to know about the Hour Angle and star time, but I don't know how to define Hour Angle for my local position, I know it should be 0H when something (vernal equinox?) crosses South Meridian.But how do I locate *it*?

aphh · Aug 31, 2008

In this image meridian 0 goes to the right of Cassiopeia. Does this mean, that when this meridian crosses south on the observer's location, local sidereal time is 0H? <img src="http://sitelife.space.com/ver1.0/Content/images/store/7/11/77dbb3fa-fd4e-4672-abd4-1525bb216c51.Medium.jpg" alt="" />

MeteorWayne · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>In this image meridian 0 goes to the right of Cassiopeia. Does this mean, that when this meridian crosses south on the observer's location, local sidereal time is 0H? Posted by aphh</DIV> Did Cassiopia look like the "W" depicted in this image, or did it look more like the number 3?That's why I asked what time you made the observation, so I can align my astronomy program to you point of view at the Lat, Long, and Time.Just so I don't have to look it up, when you give me the time; give me your Lat and Long again. Doesn't have to be precise. Nearest degree would be fine, or give me the nearest large city.How light polluted are your skies?Can you see all 4 stars in the bowl of the little dipper (Ursa Minor)?Do you know where the north star is from your location? It's a good thing to know, since it's due north, and is always at that same spot any season of the year, any time of the day (even when it's light out, it's still there)!No other star, planet, moon, comet, asteroid can make that claim?What constellations do you know? If it's only a few, you can list them. <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

aphh · Aug 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Did Cassiopia look like the "W" depicted in this image, or did it look more like the number 3?That's why I asked what time you made the observation, so I can align my astronomy program to you point of view at the Lat, Long, and Time.Just so I don't have to look it up, when you give me the time; give me your Lat and Long again. Doesn't have to be precise. Nearest degree would be fine, or give me the nearest large city.How light polluted are your skies?Can you see all 4 stars in the bowl of the little dipper (Ursa Minor)?Do you know where the north star is from your location? It's a good thing to know, since it's due north, and is always at that same spot any season of the year, any time of the day (even when it's light out, it's still there)!No other star, planet, moon, comet, asteroid can make that claim?What constellations do you know? If it's only a few, you can list them. Posted by MeteorWayne</DIV>Cassiopeia looked a lot more like 3, that's why I thought I was going to the right, when in fact I was going straight down towards Andromeda.The observation was made at 2AM local time (+2 UTC). Coordinates are 60N, 25E.I can't locate Ursa Minor at all because of light pollution, that's why I don't even know Polaris at this point. My plan is to install a camera and take pictures of the stars, so I could eventually locate Polaris and Ursa Minor.That's how bad it is.Right now I can only positively locate Cassiopeia, Summer Triangle, Orion (when it's visible later in the year), Ursa Major, stars Capella and Arcturus.And ofcourse, tonight I learned about Andromeda, because 3 of the center stars, that are perpendicular to the other stars, form a curved line down from M31.

aphh · Aug 31, 2008

I think I came up with a potential solution to determine local Sidereal time.I will have to choose one star, that resides above the sea, so that I have the horizon unobstructed in sight. Using sextant I will measure the star's altitude every hour (or to be precise, each minute).When the star has reached it's highest elevation, or culmination, the local sidereal time is 0H. Now the problem is, that the sidereal time may start in the daytime, in which case I need to find the perigee, or lowest elevation to the star.Then the local Sidereal time would be +12H. Knowing local Sidereal time would allow me to determine the Right Ascension of a "unknown" object in the sky, and thus provide you with much more accurate universal coordinates. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" />I think I'm starting to sound like a astronomist, but I need to if I wanted to pass the current course in school (AstroNavi 101).

aphh · Aug 31, 2008

Ok, it seems I don't need to get a sextant to determine local sidereal time, as this website can calculate it: http://tycho.usno.navy.mil/sidereal.htmlNow that I know the hour angle, or the position of the sky locally at a given time, I should be able to calculate the Right Ascension of a sky object using a known point of sky as a reference.So,let's say I made the observation at LST 01:48:27 and the local azimuth for the object was +2H 30' 00''. Now there is enough reference to calculate the direction to the object at your local place of observation at any given time by adding or subtracting the difference in LST.So if you now observed the same object at your local coordinates at LST 02:48:27, then the object should reside at direction +3H 30' 00" at your local coordinates.Is this all correct? Basically all I needed to know is the direction to my local South Meridian and then a device to determine the angle, or azimuth, of the desired sky object from the local meridian, and that's it. I could now give universal coordinates for you.So, only one question remains: how do I determine the direction of local South Meridian? Do I still need the sextant to measure the highest elevation of the sun or is it enough to take notes of the direction of the sun at noon? This is rather complex, but I think I'll figure it out.

aphh · Aug 31, 2008

Okay, this sums it up in one line:"hour angle as it is usually used in astronomy, which is how far west an object is from one's local meridian." http://en.wikipedia.org/wiki/Right_ascensionThe key to provide universal directions is to know the Hour Angle of the object (how far to the west from local meridian in Hours, Minutes and Seconds) and the Local Sidereal Time when the observation was made.Now anybody could look at the same direction anywhere and at any time adding or subtracting to the given Hour Angle based on their LST...

aphh · Sep 1, 2008

I went outside at exactly 1200 hrs and from my usual vantage point recorded the position of sun. I now should have the local South Meridian figured out.To test this method I'd need to select a sky object I could verify, say Arcturus, and check it's Right Ascension in the celestial sphere. Then I should choose a point in time in Local Sidereal Time and calculate the bearing, or Hour Angle, of Arcturus from my local vantage point at that time, step outside at the correct LST and see whether Arcturus will be where it is supposed to be.If this is correct method, I should be able to give much more precise directions than before, and even calculate the direction of a sky object from your local longitude.Next up: Declination vs. local elevation...

aphh · Sep 1, 2008

Some problems,First off, summertime. Do I need to account for the summertime, when determining Noon?Secondly, looks like I still need the sextant, as local time is political time determined by local time zone. Not true noon.Some work still needed to determine the actual Local Meridian.

adrenalynn · Sep 1, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>You would have to use a very low power, wide field of view lens, though. It is 2 degrees across or more, so is 4X the diameter of the moon. ANd I've seen no reports of any central condensation, though I'll have to look into that a bit more. Posted by MeteorWayne</DIV> NO major central condensation to really speak of.  Low power and a field reducer are the rule for something like that.  Given the distance, I'm starting to wonder if a wide fast lens of reasonable aperture with stacked exposures makes more sense. <div class="Discussion_UserSignature"> .bipartisan (bī-pär'tĭ-zən, -sən) [Adj.] Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.IMPEACH OBAMA! </div>

aphh · Sep 1, 2008

I calculated the true noon to occur at about 13:18 today for my longitude. This is when the sun should be in the south meridian, and I could use the direction as reference for future observations in the sky.I am a noob, so I need to figure out everything step by step from the beginning.Let's say I spot and interesting object in the sky one hour to the west from my south meridian. Now I only need to record the local sidereal time when the observation was made, and with a little calculation I should be able to offer you the Right Ascension of the object.Easy as cake. Or is it pie.

MeteorWayne · Sep 2, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Some problems,First off, summertime. Do I need to account for the summertime, when determining Noon?Secondly, looks like I still need the sextant, as local time is political time determined by local time zone. Not true noon.Some work still needed to determine the actual Local Meridian.  Posted by aphh</DIV> You could look at a map

I think you need to understand how Right Ascension and Declination are constructed. It's like a big imaginary globe surrounding the earth. +90 degrees Declination is directly above the north pole, very close to the north star (which is actually at 89 degrees 18 minutes, so makes a tiny circle around the true pole). That point will be as far above the horizon as your latitude in degrees. It's always there, just like polaris, day and night, whatever time of the year it is.-90 degrees Dec is over the south pole, obviously you can't see that from where you are. 0 degrees Declinition is directly above the equator. That 360 degree circle (0 degrees Declination) above the equator is divided in 24 hours (each hour is an angle of 360/24=15 degrees). It will always be above your southern horizon at an elevation of 90 degrees minus your latitude, and curve down to the horizon on the left and right.The 0 hour reference is is Pisces, a Constellation far too faint for you to see. It is nearly directly below the edge of the Great Square of Pegasus however, on the western (left) side where Andromeda attaches at the top left.It will be due south at local midnight at the autumnal equinox, at a height equal to 90 degrees minus your latitude. it then curves down and hits the horizon to your left and right. 6 Hours RA will be just past Orion on your eastern Horizon; in fact 0 degrees declination passes through the top of Orion's belt.. 20 Hours RA will be about 15 degrees above your western (right) horizon if you are facing south, below the bright star ALtair... Hope that helps. I always recommend getting a globe properly mounted with the 23 1/2 degree axial tilt. It makes these 3D problems much easier to visualize.  <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

aphh · Sep 2, 2008

Thanks, I think I now understand celestial sphere and right ascension and also it's relation to local coordinates. But there will be other noobs like myself, so it does not hurt to go back to basics every once in awhile.The next step that I was figuring out was how to give coordinates of space objects so that others could instantly know what I'm talking about or even have a quick glimpse of it by themselves.The Right Ascension is now pretty clear, next one would be how to translate local elevation of a sky object from local horizon into universal declination on the celestial sphere. This one requires a bit of math and spherical trig, but not too much.I'll get into that a bit later. Well, actually it only requires spherical trig if you wanted to calculate distances between sky objects, like the distance from alpha Cassiopeia to M31. Like I tried and failed in the original post. The intention, ofcourse, is to have coordinates readily available the next time I'll spot an interesting (for me) object. And also pass the AstroNavi 101 course in the school <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" />

MeteorWayne · Sep 2, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks, I think I now understand celestial sphere and right ascension and also it's relation to local coordinates. But there will be other noobs like myself, so it does not hurt to go back to basics every once in awhile.The next step that I was figuring out was how to give coordinates of space objects so that others could instantly know what I'm talking about or even have a quick glimpse of it by themselves.The Right Ascension is now pretty clear, next one would be how to translate local elevation of a sky object from local horizon into universal declination on the celestial sphere. This one requires a bit of math and spherical trig, but not too much.I'll get into that a bit later. Well, actually it only requires spherical trig if you wanted to calculate distances between sky objects, like the distance from alpha Cassiopeia to M31. Like I tried and failed in the original post. The intention, ofcourse, is to have coordinates readily available the next time I'll spot an interesting (for me) object. And also pass the AstroNavi 101 course in the school Posted by aphh</DIV> There is also the Hour Angle, the concept you mentioned:hour angle <div class="pseg"><div class="ds-single">The angular distance, measured westward along the celestial equator, between the celestial meridian of the observer and the hour circle passing through a celestial body. A body's hour angle is measured in hours, minutes, and seconds, and corresponds to its right ascension as measured with respect to the observer's meridian (which changes with time) rather than the vernal equinox (which is fixed on the celestial equator). A celestial object that crossed the observer's meridian 3 hours and 20 minutes ago has an hour angle of +3 hours 20 minutes. An object that will not cross the meridian for another 3 hours and 20 minutes has an hour angle of -3 hours 20 minutes."</div><div class="ds-single">However, I know of few who use that, since it is measured to a specific point on earth. Right AscensionA is universal (haha) </div><div class="ds-single">MW</div></div> <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

Is comet Holmes still around?

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