Difference between revisions of "Overview of measuring distances"

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(Created page with " Most FITS image viewers provide a way to measure distances within the tool. Look for ruler icons or names in menus. You can also measure distances by hand by comparing pixe...")
 
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Technically, to be absolutely correct, because you are calculating distances on a sphere, in order to do this, you need to do spherical trigonometry. This matters because the angle subtended by 1 hour of RA on the celestial equator is much larger than that subtended by 1 hour of RA near the celestial pole. For quick and dirty purposes, it should be mostly fine to simply subtract the RA and Dec to get a reasonable estimate of the distance BUT WATCH YOUR UNITS because RA by default is in hours:min of time:sec of time, not deg:arcmin:arcsec.  
 
Technically, to be absolutely correct, because you are calculating distances on a sphere, in order to do this, you need to do spherical trigonometry. This matters because the angle subtended by 1 hour of RA on the celestial equator is much larger than that subtended by 1 hour of RA near the celestial pole. For quick and dirty purposes, it should be mostly fine to simply subtract the RA and Dec to get a reasonable estimate of the distance BUT WATCH YOUR UNITS because RA by default is in hours:min of time:sec of time, not deg:arcmin:arcsec.  
  
The spherical trig does make a difference, though. See [http://spiff.rit.edu/classes/phys301/lectures/precession/precession.html#sep this excerpt from someone's class notes] with some really nice graphics and explanations of why you need to do this, and how to do it right. (hint: For large-ish distances, you need a cosine of the declination. I won't make you do the full spherical trig in most cases.) For the ambitious, anticipating skills you'll need downstream from this worksheet, try programming a spreadsheet to do this for you, given two RA/Dec position pairs. NB: Be sure to watch your units on the Dec-- some cosine functions want radians, and some take degrees. (Bonus: how much of a difference in your target region does it make if you leave out the cos(dec) term? Is that going to get worse or better if we move closer to the north celestial pole?)
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The spherical trig does make a difference, though. See [http://spiff.rit.edu/classes/phys301/lectures/precession/precession.html#sep this excerpt from someone's class notes] with some really nice graphics and explanations of why you need to do this, and how to do it right. (hint: For large-ish distances, you need a cosine of the declination.) For the ambitious, anticipating skills you'll need downstream from this worksheet, try programming a spreadsheet to do this for you, given two RA/Dec position pairs. NB: Be sure to watch your units on the Dec-- some cosine functions want radians, and some take degrees. (Bonus: how much of a difference in your target region does it make if you leave out the cos(dec) term? Is that going to get worse or better if we move closer to the north celestial pole?)

Latest revision as of 01:27, 20 January 2021

Most FITS image viewers provide a way to measure distances within the tool. Look for ruler icons or names in menus.

You can also measure distances by hand by comparing pixel coordinates. Note that as you move your mouse around on the image in any of these FITS viewers, it will give you an updated readout of the RA and Dec near the top. You can change this from hh:mm:ss ddd:mm:ss format to decimal degrees for both RA and Dec -- for ds9, you do this by picking from the "wcs" menu at the top, either 'degrees' or 'sexagesimal'. Make a note of the RA/Dec from which you want to measure a distance, and the RA/Dec of the end point of the distance measure.

No matter how, exactly, you do this, WATCH YOUR UNITS. RA by default is in hours, not degrees. Dec by default IS in degrees. How do you convert between hours and degrees? (Hint: there are 24 hours of RA ...and 360 degrees.)

Technically, to be absolutely correct, because you are calculating distances on a sphere, in order to do this, you need to do spherical trigonometry. This matters because the angle subtended by 1 hour of RA on the celestial equator is much larger than that subtended by 1 hour of RA near the celestial pole. For quick and dirty purposes, it should be mostly fine to simply subtract the RA and Dec to get a reasonable estimate of the distance BUT WATCH YOUR UNITS because RA by default is in hours:min of time:sec of time, not deg:arcmin:arcsec.

The spherical trig does make a difference, though. See this excerpt from someone's class notes with some really nice graphics and explanations of why you need to do this, and how to do it right. (hint: For large-ish distances, you need a cosine of the declination.) For the ambitious, anticipating skills you'll need downstream from this worksheet, try programming a spreadsheet to do this for you, given two RA/Dec position pairs. NB: Be sure to watch your units on the Dec-- some cosine functions want radians, and some take degrees. (Bonus: how much of a difference in your target region does it make if you leave out the cos(dec) term? Is that going to get worse or better if we move closer to the north celestial pole?)