# RA or Right ascension of a Star at GMT in London?

If I’m quoted on a star Chart RA= 2.5hrs and its 10pm at night and I live in London. What setting would I use on my telescope to find RA relative to my location? If I already know how to set the Declination. Does this value change during the year and how do I take Daylight savings into account… Basically I want to know how to find a star!

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please read up on sidereal time, and how your local sidereal time changes during the year. you will find terms like “hour angle”. learn what they mean.
a planisphere is an excellent way to explore the relationships and see for yourself.

The RA is fixed to the sky, not the Earth. So you need to calibrate the RA every night. Typically, you get the coordinates of a bright star you know, point your scope at it, and set the RA on the scope to this known value. Then, you can use RA and Dec to find other stuff. I’m assuming you have a motor drive.
I have digital setting circles on my scope, which is alt-az. So i need to tell my scope where two stars are, and which way is up. The computer can then tell me where stuff is when i ask. And, it can suggest objects if i haven’t thought to do an observing plan. One advantage of my push-to computer is that a cheap 9 volt battery lasts a couple months, since i do the pushing.

First, you need to obtain your Local Sidereal Time. From this, you subtract the Right Ascension of the object you’re intending to view, which will give its Hour Angle. Hour Angle is measured positive to the west.
Greenwich Sidereal Time (in degrees) is given by the formula:
GST=280.461+360.98564737D
where D = number of days (of Universal Time) from Greenwich Mean Noon on January 1st 2000, equivalent to Julian Day 2451545.0
At 0h UT today, the Julian Date was 2454809.5, hence the elapsed time from January 1st 2000 is 3264.5 days. Add another 0.9167 days for 10 p.m. (22h) and the value for D is 3265.4167. Put this into the above equation and GST = 49.022 degrees (subtracting whole numbers of 360 degrees) which converts to 3h 16m 05s (1 hour is 15 degrees). The object is therefore at an HA of 3h 16m 05s – 2h 30m = 0h 46m 05s (i.e. west of the meridian).
For locations not on the Greenwich meridian, subtract the longitude (positive to the west), to obtain the Local Sidereal Time.
You need to convert your local time to UT before you start. For daylight time in the U.K., simply subtract one hour.

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