Telescope

[astropro2009]

Should I buy the Celestron Dobsonian 12" or 10", and if I bought an 10" advanced series, the images I could see through them, which can be seen on this link:<br />http://www.celestron.com/c2/product.php?ProdID=50<br /><br />Would I be able to see these images in clearer, worse, the same or better detail with the 12" Dobsonian? and how does the 10" Dob compare to the 12" Dob in your opinion?<br />

 

[betelgeuze]

I plan to buy a telescope, my main interest are planets and the moon so I guess I should go for a good refractor. I want and automatic object searcher and a pc connector.<br />My budged is around 1500euro (max 2000euro). <br /><br />Can anyone suggest me a telescope?

 

[doubletruncation]

<font color="yellow">The other method and more common with deep-space objects is called Prime focus. Or basicly connecting the camera directly to the scope without an eyepiece. Though couldn't tell you how to determine the field of view. I know with my scope is close to .5º ( a half moon) .</font><br /><br />For prime focus the field of view depends on size of your camera chip (or film). If you know the physical size of your light sensitive element (e.g. the size of your CCD chip in mm) then, in degrees it is:<br />180 * A / (pi * f) = 57.3 * A/f<br />where A is the physical size of your CCD and f is the focal length of your scope. Easiest thing to do though is probably just to take an image of a double star like alberio. <div class="Discussion_UserSignature"> </div>

 

[doubletruncation]

It depends, I think, on what you want to do with it. As SVMsmiles mentioned, don't rely on the pictures that you see taken with the telescope to judge what you'll actually see through an eyepiece. If you want to take pictures with the scope, then I would consider getting something with an equitorial mount instead of a dobsonian.<br /><br />If you're doing visual observing of the planets, then the number that is probably the most relevant is the highest useful magnification. Take the focal length of the telescope and divide by that magnification to get the focal length of the eyepiece you would need to achieve that magnification. To figure out how large a planet would look through that eyepiece just take its angular diameter and multiply by the magnification.<br /><br />e.g., at opposition Jupiter has an angular diameter of 0.78', the 10" dobsonian has a maximum useful magnification of 600x, so Jupiter would be 7.8 degrees big if you used the right eyepiece (in this case it would have to be a 1270 mm / 600 ~ 2mm focal length eyepiece). The 12" has a maximum useful magnification of 721x, (you'd need a 1500 mm / 721 ~ 2mm focal length eyepiece to achieve that magnification), with that magnification Jupiter would appear to be about 9.4 degrees in diameter. (Whether or not this magnification is actually useful will depend in part on the seeing).<br /><br />If you want to look at nebulae, then you really want to go for high surface brightness. In this case, look at the minimum useful magnification and the light gathering power (or the effective area of the primary (that is primary surface area minus the surface area of any obstruction)). Take the light gathering power and divide by the minimum useful magnification squared, this will give you a number proportional to the apparent surface brightness of an object (to figure out exactly what it means you'd have to calibrate it somehow to your own observing conditions). So, for example, with the 10" dobsonian the factor wou <div class="Discussion_UserSignature"> </div>

 

[toothferry]

<font color="yellow">"I plan to buy a telescope, my main interest are planets and the moon so I guess I should go for a good refractor. I want and automatic object searcher and a pc connector. My budged is around 1500euro (max 2000euro). "</font><br /><br />hmmm... well if I were shopping for one I'd have this high on my list, the 120ED apo optical tube.. then take it too a star party and borrow a mount and eyepieces until you can slowly buy the extras. Otherwise, the 100ED is available for the very same price, including equatorial mount and 26mm plossl eyepiece. anyhow they look VERY interesting.. and competitive to Televue, the world leader in apo refractors I believe (am I correct folks?)<hr /><br />http://www.telescope.com/shopping/product/detailmain.jsp?itemID=201578&itemType=PRODUCT&iMainCat=4&iSubCat=13&iProductID=201578<br /><font color="orange"><br />This big-aperture, premium f/7.5 APO takes imaging to a new level of excitement and affordability<br />The 120ED is a breakthrough in performance at this low price, offering optical quality, workmanship, and attention to detail you'd expect to find on scopes costing thousands more. If you've shied away from high-end refractors because of the cost, you should consider the 120ED. <br /><br />Superb apochromatic optics set the 120ED apart. The 900mm focal length (f/7.5) is great for planetary work and fast enough for faint galaxies and nebulas. FPL-53 extra-low dispersion (ED) glass in one of the two objective lens elements and multi-coating on all air-to-glass surfaces means light passes to the eyepiece or camera without color fringing. Besides providing more light-grasp for deep-sky imaging &emdash 43% more than a 100mm scope &emdash the optics easily handle high magnification. <br /><br />Included are a 2" Crayfor</font>

 

[toothferry]

[yelow]"Another thing....have you met Brad (NEVERS) yet? The wife and I are heading to LV the end of Sept (our only vacation that we don't take telescopes) We are staying at the LV Hilton starting Sept 28th to Oct 3rd. Brad is going to join us for dinner Saturday evening Sept 30th there. If ya want join us there. "<br /><br />I've never met any of you guys. I certainly have a lot of respect for NEVERS, though I don't see his posts as often as I use to. A star party in Arizona certainly does sounds tempting especially to see what difference high altitude and arid weather makes compared to the limited number of oberservations Ive personally done here... heck ..but I'm like East of the Mississippi... and no mountain tops.

 

[toothferry]

<font color="yellow">"The only info I have is a small chart that shows a specific range, where light is reduced significantly within the range of 400 - 700 nm.Weither this concidered broad or narrow, I couldn't tell you. ( LPR, btw stands for Light Pollution Reduction for those that dont know ) "</font>hr /><br />Thanks for the light interference chart... I've got the Narrowband ordered.. along with a couple of other goodies

 

[toothferry]

<font color="yellow">"If you want to look at nebulae, then you really want to go for high surface brightness. In this case, look at the minimum useful magnification and the light gathering power (or the effective area of the primary (that is primary surface area minus the surface area of any obstruction)). Take the light gathering power and divide by the minimum useful magnification squared, this will give you a number proportional to the apparent surface brightness of an object (to figure out exactly what it means you'd have to calibrate it somehow to your own observing conditions). So, for example, with the 10" dobsonian the factor would be 1317 / (36^2) = 1.016 whereas the 12" dobsonian would have 1898 / (44^2) = 0.98. So in other words, in practice the telescopes would be nearly identical in your ability to make out faint nebulae, however at the same surface brightness the object would look slightly larger in the 12" than in the 10" (about 5/4 times as big). "</font><br /><br />nice post! question: how do you come up with "minimum useful magnification?" did you just arbitrarily use 3.6X per inch of aperture (as in the "36^2" for 10" telescope in the above sample) and is that proportional to the exit pupil size?

 

[doubletruncation]

<font color="yellow">how do you come up with "minimum useful magnification?</font><br /><br />I took the value listed by celestron (e.g. http://www.celestron.com/c2/product.php?CatID=7&ProdID=23 for the 12" dobsonian). In general it is the magnification for which the exit pupil is the same size as the pupil of your eye. In fact the surface brightness in general is proportional to the size of the exit pupil squared as long as the exit pupil is smaller than the pupil of the eye.<br /><br />The exit pupil size is = eyepiece focal length / focal ratio, the magnification is equal to the telescope focal length / eyepiece focal length, if the pupil size of your eye is 7mm in a dark site, then the minimum useful magnification is equal to:<br />diameter of telescope / 7mm. <br /><br />I think one of the morals is that when you look at an extended source the smallest telescope in which you can resolve the source (i.e. you can tell that it isn't a point source of light) will be just as good as any larger telescope for being able to detect it. But with a larger telescope it can look bigger with the same surface brightness. <br /><br />If you're trying to take an image of the nebula at prime focus though, rather than looking at it through an eyepiece, then for a given CCD and exposure time, the surface brightness will be proportional to the inverse of the focal ratio squared. The magnification of the image will be higher for the longer focal length telescope as usual. <div class="Discussion_UserSignature"> </div>

 

[doubletruncation]

If you have an optical element in the tube then your field of view will change and you should replace the focal length of the telescope with the effective focal length of the entire optical system. The expression for the effective focal length when using eyepiece projection is given on this website. This assumes that you're using an eyepiece in the eyepiece projection tube to magnify the image, if you just have the tube attached to your telescope without an optical element in it, then the field of view should be the same as without the tube (your ccd will have to be placed at the focal point of the telescope, if you have a tube attached without a lens in it then you would just need to bring your focuser in the distance of the tube to focus your image). <br /><br />By the way, the above expression for the field of view assumes that your ccd chip is smaller than the light beam - i.e., that your entire ccd chip is exposed to light. <div class="Discussion_UserSignature"> </div>

 

[doubletruncation]

<font color="yellow">For looking at point sources, the only relevant number is the light gathering power (again, effectively the aperture squared). The 12" has 1898 / 1317 = 1.4x the light gathering power of the 10", which means that you should be able to see point sources (stars, asteroids, nuclei of galaxies) that are 1.4 times fainter, or about 0.4 magnitudes fainter, with the 12" than with the 10".</font><br /><br />One caveat I probably should have mentioned before is that the limiting flux to which you can see scales inversely as the collecting area of the telescope only when the stars are much brighter than the sky background. If you're at a dark site then stars brighter than 18th-19th magnitude should be brighter than the sky, so the limiting flux does scale inversely as the collecting area for most telescopes within the reach of amateur astronomers. If you're near a city, however, then the sky can be substantially brighter and the limiting flux may end up scaling as the inverse of the square root of the collecting area (or as the diameter only rather than the diameter squared). This at least is true for doing photometry with a camera, I don't know if your eye performs this well or if it is even worse. So if you're near a city the benefit of a large telescope to a small telescope in terms of the faintest stars (or point sources) that you can see is not as great as it is if you have a dark site - it still is better to have the bigger telescope, it's just that the degree of improvement is not as great as you might expect. <div class="Discussion_UserSignature"> </div>