How Can You Determine the Right Ascension of a Star with a Meridian Circle?

  • Thread starter Thread starter JeffOCA
  • Start date Start date
  • Tags Tags
    Absolute
AI Thread Summary
A meridian circle can determine the right ascension (R.A.) of a star by measuring the transit times of both a reference star with known R.A. and the target star. By calculating the difference in transit times, the right ascension of the target star can be derived using the formula α = T when H=0. The discussion also raises the possibility of using the Sun's transit as a reference for determining R.A. near the equinox, given its known position. Additionally, the Flamsteed method is mentioned as a potential approach for determining absolute R.A., though resources for easy understanding are lacking. The conversation highlights the complexity of the topic and the need for further clarification on specific methods.
JeffOCA
Messages
48
Reaction score
0
Hi,

With a meridian circle, you can determine the latitude of your location (Horrebow-Talcott method) but also deriving time if you know the right ascension of a star.
However, I'd like to know how to determine right ascension of a star with a meridian circle.

If you know the right ascension of a reference star, you measure its transit time. Then, you measure the transit time of the second star (the star you want to know the right ascension) and then you add the difference of transit time to the right ascension of the reference star. So, you obtain the right ascension of the second star since \alpha = T when H=0 (at transit).

How to derive right ascension of a star when you have no other star with known R.A ? Can we use the transits of the Sun as a reference star since, near equinox, we can assume the Sun R.A to be known ?
I heard about the Flamsteed method to determine absolute R.A but I don't documents which explained it "quite easily".


Jeff
 
Astronomy news on Phys.org
Anyone ?
 
Jeff, you've asked a couple of very specific questions in this thread and some of your others that not many people will know. Don't be suprised if you don't get any hits.
 
Drakkith said:
Jeff, you've asked a couple of very specific questions in this thread and some of your others that not many people will know. Don't be suprised if you don't get any hits.

Ok, sorry.
 
This thread is dedicated to the beauty and awesomeness of our Universe. If you feel like it, please share video clips and photos (or nice animations) of space and objects in space in this thread. Your posts, clips and photos may by all means include scientific information; that does not make it less beautiful to me (n.b. the posts must of course comply with the PF guidelines, i.e. regarding science, only mainstream science is allowed, fringe/pseudoscience is not allowed). n.b. I start this...
Today at about 4:30 am I saw the conjunction of Venus and Jupiter, where they were about the width of the full moon, or one half degree apart. Did anyone else see it? Edit: The moon is 2,200 miles in diameter and at a distance of 240,000 miles. Thereby it subtends an angle in radians of 2,200/240,000=.01 (approximately). With pi radians being 180 degrees, one radian is 57.3 degrees, so that .01 radians is about .50 degrees (angle subtended by the moon). (.57 to be more exact, but with...
Back
Top