Help identify object in the sky (planet visible in daytime)

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drabina

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Hi,<br /><br />I am new here. I just signed up to look for help. On 2006:05:06 17:17:30 (EXIF data from my camera) I was at the Niagara Falls and when looking at the sky, I have noticed small white round object. It was stationary so I figured it was a planet. Here is the image:<br /><br />image<br /><br />Could somebody tell me what planet it is? This image is enlarged because I already maxed out my 80-300mm zoom on my digital camera and the planet was way up in the sky.<br /><br />Thanks,<br />Alex
 
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origin

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You can't see planets in the day (well except for the earth) and it would certainly be really hard to photograph them. Lools like a balloon to me. <div class="Discussion_UserSignature"> </div>
 
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votefornimitz

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I plugged your numbers and determined that Venus would be to the South-West, but it was only 40 degrees away from the sun at the time, so I doubt it was visible...<br /><br />Do you happen to know in which direction this was taken in? <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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votefornimitz

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Actually, I think I found your culprit....<br /><br />http://www.iaw.com/~falls/angels.html <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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drabina

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Here is a link to the map and the direction I was looking at:<br /><br />map<br /><br />VoterforNimitz: I doubt it could be a balloon because it was way up in the sky and with naked eye, this was only about the size of a needle pin. The link you provided claims that the balloons go as high as 400 feet. I am sure the object was way higher. Also, wouldn't I be seeing a carriage or something that holds people in it if I took the picture from the ground?<br /><br />I have maxed out my zoom 80-300 with my digital camera so the actual converted length was 450mm (1.5x factor). I guess I can somehow figure out the size of the object based on the dimensions of the original image, the size of the object and the focal length of the lens.<br /><br />Thanks for all the replies.
 
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origin

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I don't know it is looking more and more like it might be the helium balloon. Here is a picture of the balloon. <br /><br />Edited to add: good catch, votefornimitz <div class="Discussion_UserSignature"> </div>
 
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bearack

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It really doe's appear to have been the balloon. I blew it up 450% and, hard to see, but appears to have a bsket below.<br /><br />http://img.photobucket.com/albums/v607/Bearack/balloon2.jpg <div class="Discussion_UserSignature"> <p><br /><img id="06322a8d-f18d-4ab1-8ea7-150275a4cb53" src="http://sitelife.space.com/ver1.0/Content/images/store/6/14/06322a8d-f18d-4ab1-8ea7-150275a4cb53.Large.jpg" alt="blog post photo" /></p> </div>
 
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brigandier

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I don't think it's a planet either. It's awfully white... and it has pretty defined edges. It's most likely that balloon
 
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votefornimitz

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Looking at you're original picture and the map of you're location, I can effectively rule out what you saw was a planet....<br />Your map indicates you as looking north-northeast.<br />Star charts for that day at that location show no planets in that directions...<br /><br />I'll see if I can determine the diameter of one of the balloons for you to compare to your estimates... <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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Mee_n_Mac

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<font color="yellow">I have maxed out my zoom 80-300 with my digital camera so the actual converted length was 450mm (1.5x factor). I guess I can somehow figure out the size of the object based on the dimensions of the original image, the size of the object and the focal length of the lens. </font><br /><br />I think we also need to know the sensor size (W x H) in pixels. I take it from the 1.5X that it's an APS-C sized sensor in a dSLR. With a 450mm (35mm equivalent) focal length, we can get the FOV and then with the above info, figure out how many deg/pixel. Counting the pixels in the image gives us it's angular size and then we can rule out various things. <br /><br />EDIT : If I've got the date and time correct, Venus would have been 68% illuminated and subtended ~16 arc seconds. Jupiter would have been a full disc and ~45 arc seconds in diameter. <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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votefornimitz

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Its 73' feet in diameter and is moored 1/2 a mile north of Goat Island (1)<br />Using goggle maps that location is somewhere in the ball park of being between Rainbow Blvd South and Rainbow Blvd North. (2)<br />Putting it right in your field of view, being NNE of your location given by the map....<br />Being about 1700 ft NE of you and 400 ft above you, it would be about 1746 feet away from you...<br /> <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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Mee_n_Mac

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Check my math but 73 ft @ 1746 ft distance would mean an angular width of ~2.4 deg. With a 35 mm lens at 450 mm FL, that would fill about half the horizontal FOV. The object seems too small to be that balloon that close.<br /><br /><br />http://en.wikipedia.org/wiki/Angle_of_view<br /><br /> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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votefornimitz

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Plugging in the numbers you get that such a balloon at such a distance would take up little over 0.02% of the sky's area....<br />Seems reasonable given your photographs... <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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votefornimitz

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Hmmm...<br />That implies a serious math error on my part... <br /><br />I found the surface area of a sphere with radius 1746 ft and got 38,308,773.9 ft^2...<br />Neglecting curvature of the Earth, only half of that area is above the horizon....<br />Leaving you with 19,154,386.9 ft^2....<br />The surface area of the balloon (being a circle since its seen head on) would be (pi)73^2, giving you 4185.38ft^2<br />Dividing 4185.38 by 19,154,386.9 gives you a number a little over 0.02%....<br />Where'd I go wrong?<br />I've yet to have a calculus course and my current precalc teachers doesn't know what he's doing, so I'm betting I forgot to factor something in... <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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drabina

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Wow, I never expected forum members to be that much involved. I really appreciate this.<br /><br />I have also contacted my friend who is very good with math (engineer) and he gave me a formula that will calculate the approx. size of the object. I just have to go back home and over the weekend plug in some numbers from the lens, EXIF and camera specs into the formula. I will post my findings.
 
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votefornimitz

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I've got a Math honor society competition tomorrow, and what better way to get warmed up...Despite the fact I managed to screw this calc up....<br /><br />No problem though.... <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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Mee_n_Mac

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Well I could be wrong but here's my calcs ...<br /><br />The angular distance subtended by the object is found from it's length (diameter, d) / 2 and the distance to the object (D). These form 2 legs of an right triangle. The angle is then arc tan [(d/2)/D]. The object subtends twice this "angular distance" so it's 2 * arc tan [(d/2)/D].<br /><br /><br />So 2*arc tan[(73/2)/1746] = 2*arc tan(0.0209) =~ 2.4 deg or so I thought.<br /><br /><br />See the wiki :<br />http://en.wikipedia.org/wiki/Angle_of_view<br /><br /><br />EDIT: By FOV I mean the camera's FOV which I took from the table in the wiki.<br /><br /><br /> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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drabina

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Another thing that I remember, the balloon from your links that day was setup not 1 mile north of Goat Island, but somewhere near the center of the town (on the grass). I think it was close to the Hard Rock Cafe. So unless there are two or more balloons, the one from the photo could not be the one on the photograph. I think I have another picture of the balloon in the center of the town. Have to check at home. <br /><br />Plus if the balloon from one of your links is about 187px wide and the gondola is about 45px wide then the gondola is approx. 1/4th of the balloon's diameter. That in my opinion would be easily visible in my picture. So maybe that is a weather balloon as one of my friends suggested?
 
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votefornimitz

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Those may very well be the case, I wasn't there and have to rely on the often inaccurate internet for my values, so its possible...<br /><br />I do see what you mean about the gondola, it should be visible...Al I can say was that with the information given, its not a planet...<br />It may in fact be a weather balloon... <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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Mee_n_Mac

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I'm going to <b>guess</b> that the dSLR was a 6MP version given the date above. That makes it ~ 3000 pixels wide and 2050 or so tall. So with 4.75 deg spread across 3000 pixels, you get about 5.7 arc secs per pixel. How many pixels wide was the object ... 187 ?<br /><br />That gives us ~18 arc mins of width or about 25' at 1 mile distance. <br /><br /><br />EDIT: Wait the balloon in a pic was 187 px not <i>the object the pic</i>. So the above is not the size of the object but you can scale accordingly (if my assumption was correct).<br /><br />EDIT2 : Looking at the image of the object provided above, it's 3008 x 2000 pixels so I'll continue to say it's a 6mp picture. The object is ~ 25 pixels wide out of the 3008 or 0.0083 deg "wide". That makes it 8" in diameter at 1 mile (or 4" at 1/2 mile or 16" at 2 mi, etc, etc).<br /><br /><br />Maybe it's just a regular balloon some kid let go of ????<br /><br /><br /><br /> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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votefornimitz

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I don't know anything about photography, so I'm clueless...<br /> <div class="Discussion_UserSignature"> <span style="color:#993366">In the event of a full scale nuclear war or NEO impact event, there are two categories of underground shelters available to the public, distinguished by depth underground: bunkers and graves...</span> </div>
 
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adrenalynn

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Time to warm-up the forensics software. I'm going to fetch the original image now. I'll be back in later this afternoon/evening if I can add anything to the discussion. I do love a mystery! <div class="Discussion_UserSignature"> <p>.</p><p><font size="3">bipartisan</font>  (<span style="color:blue" class="pointer"><span class="pron"><font face="Lucida Sans Unicode" size="2">bī-pär'tĭ-zən, -sən</font></span></span>) [Adj.]  Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.</p><p><strong><font color="#ff0000"><font color="#ff0000">IMPE</font><font color="#c0c0c0">ACH</font> <font color="#0000ff"><font color="#c0c0c0">O</font>BAMA</font>!</font></strong></p> </div>
 
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Mee_n_Mac

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Kewl. I don't have my image SW on this 'puter so I'm suspicious of the pic size I see reported. I think the 3000 x 2000 is the original size but I think the pict as posted has been resized (to 33%). So my estimate on size might be off by a factor of 3.<br /><br /><br />But I like the lost balloon idea .... it just makes sense.<br /><br /> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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adrenalynn

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It does make sense.<br /><br />This isn't the "original" original image. As you suspected, it's been previously touched. There's no EXIF data attached. It's been *heavily* jpeg'd. The image size is 6.016mpixel and the crop factor makes that difficult to derive. As, I believe, Mac' suggested - we need to know what kind of camera this is so we can get the actual sensor size before any real forensics can be performed.<br /><br />Due to the post-jpeg, I can't absolutely authenticate it either.<br /><br />I'm hoping that the OP can chime back in and hopefully get us the original file with EXIF intact.<br /><br />In the meantime, I've done a point-source refocus on the image, and mathematically corrected the aberations for CA and SA. (Reindeer Graphics Fovea Pro 4 absolutely rocks for this stuff. It's what I use when I testify as an expert witness for video and stills and have to do measurements of post-crime-scene details) <div class="Discussion_UserSignature"> <p>.</p><p><font size="3">bipartisan</font>  (<span style="color:blue" class="pointer"><span class="pron"><font face="Lucida Sans Unicode" size="2">bī-pär'tĭ-zən, -sən</font></span></span>) [Adj.]  Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.</p><p><strong><font color="#ff0000"><font color="#ff0000">IMPE</font><font color="#c0c0c0">ACH</font> <font color="#0000ff"><font color="#c0c0c0">O</font>BAMA</font>!</font></strong></p> </div>
 
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drabina

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Guys, I will post the original picture (as downloaded from camera) in about 2 hours or so (9pm EST). There is going to be EXIF data. My camera is Pentax istDL (6mp).
 
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