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I was wondering if it's possible to measure star distance using parallax at home with a small telescope (Something like a 3-inch reflector)? Or is this not really practical?

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I think what you need to do is have a CCD imager on a telescope with a large f-ratio, such that each pixel on the detector covers a small angle on the sky - I would say at most 0.25 arcseconds. The field of view also needs to be wide enough that you can get many faint stars in the same CCD image. You must have a telescope able to track the sky.

Then what you do is you take pictures on the nights of best seeing, making sure that you do not saturate the neaby star you are interested in, but getting reasonable signal to noise on the many faint stars.

Then you need to astrometrically calibrate your images so that you can estimate the apparent position of the star of interest with respect to the (faint, and hence assumed) background stars.

If you have seeing of 2 arcseconds, then the best centroiding precision you might reasonably expect to achieve is a precision of around 1/10 of the seeing disc or 0.2 arcseconds. If you have that sort of data repeated a number of times over the course of say 2 years (because you need to fit a solution which includes the annual proper motion as well as the annual parallax), then I reckon you might be able to measure parallaxes for stars out to a few parsecs (maybe ten?) with no problem.

The problem with a three inch reflector is that you will need to do very long exposures in order to get lots of faint background stars in your image and then you may have problems with tracking accuracy.

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  • $\begingroup$ So you think it's being able to get long enough exposures and the tracking for this that will be the limiting factor? $\endgroup$ – Phil_12d3 Mar 17 '16 at 8:59
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    $\begingroup$ @Phil_12d3 Tracking error will reduce the precision with which you can measure star centroids. If your motor tracks at a given percentage accuracy then this will be less serious in short exposures. An alternative approach might be to shift and stack numerous short exposures. $\endgroup$ – Rob Jeffries Mar 17 '16 at 10:04
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Short answer - not really, parallax for the closest stars is right on the limits of resolution for a good ground-based amateur equipment.

The nearest star would show a parallax angle of under an arcsecond. (The Parsec even takes its definition based on one arcsecond of parallax.)

Ground based amateur observations are probably limited to an arcsecond at most, so it's probably not possible to measure them.

A special (and totally fictional) case - if a nearby star like Alpha Centauri were exactly between us and a much more distant star, and it were possible to see them both without Centauri overpowering that distant star, then maybe it would be possible to observe them as a close double. That's not true in practice however.

Update:

Prompted by Rob's much better answer, the earliest parallax determination I've been able to find was using

the 6.2-inch (157.5 mm) aperture Fraunhofer heliometer at Königsberg

(Source: Heliometer article on Wikipedia.)

In purely aperture terms, that's an amateur-sized instrument (with some admittedly very precise measuring gear built in.) So my suggestion now is to ignore my initially doubtful response and try measurements, perhaps using a camera at fairly high magnification as Rob suggests...

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  • $\begingroup$ How do you think parallaxes were obtained in the 19th century? The size of the seeing disc, while related to the precision that can be achieved, is not the limiting precision achievable. $\endgroup$ – Rob Jeffries Mar 16 '16 at 16:42
  • $\begingroup$ Hmmm I assume they'd measure star separations between the centres of airy discs. I doubt that's very precise with the amateur gear mentioned though. (Perhaps you're right my answer is concentrating too much on resolving power, repeatability is the real problem.) $\endgroup$ – Andy Mar 16 '16 at 17:05
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    $\begingroup$ The very first parallax did indeed use the double star technique. $\endgroup$ – Rob Jeffries Mar 16 '16 at 17:13
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    $\begingroup$ I've removed this as the answer just because I think there is probably some more discussion to be had. Will choose an answer in a few days if nothing more is said. Thanks though! $\endgroup$ – Phil_12d3 Mar 16 '16 at 22:02

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