Sidereal Time Calculation

[jACK101]

I am confused.,Particularly about the Julian Calendar involvement in Sidereal time.

 

I have read that to set up a telescope to the South it is necessary to have the sidereal time when the sun crosses the south meridian. This presumably must be available as the Corresponding UTC.

 

All that is then necessary is to line up the sun at that precise time and to transfer that line to the ground using a plumb bob. This line should then be used to set up the telescope's tripod and thereafter everything is hunky dory.

 

Is it safe to assume that at midnight, sidereal time and UTC are the same?

 

If so, surely the problem is relatively simple, i.e. to calculate the time it will take to cross the South Meridian when travelling at 23 hours and roughly 56 minutes per revolution until it gets to the longitude required.

 

Am I simply displaying my total ignorance or would this be an acceptably accurate way to align a telescope for all practical purposes?

 

If not, where can I get an understandable method to calculate it more accurately?

 

I live at roughly 4 degrees west so my calculation from midnight would be 23 hours 56 minutes ( or whatever the accurate figure is) multiplied by 184 degrees divided by 360. This would be roughly 2 minutes earlier than using the 24 hour revolution time of the sun.

 

Have I now got everyone confused?

 

Thanks to any one who can throw some light on this.

 

Jack

[acey]

Why do you need to know sidereal time? If you do need to know it, there are various apps etc that will give you your current local sidereal time if you input your longitude (Google "sidereal time", "sidereal clock" etc). If you want to align your telescope due south then all you need is a compass. If you lack a compass then observe where the Sun (or any star) rises to its highest. If you are trying to locate objects knowing their right ascension then you don't need to know sidereal time.

[ronin]

Part of the problem is that there is a time diference.

Sidereal time is I believe measure as 1 complete revolution of the earth, unfortunately if you use the sun then you are doing 1 and a bit revolutions.

An "exact" sidereal day is measured by when a star would cross your meridian line, not the sun.

You would measure this by using a star like Alderamin in Cepheus, Seph or Caph in Casseiopia or one of the stars in the handle of the plough - they are reasonably high, brightish and above the horizon all year.

The problem is that although the although the earth rotates about it's axis and it also rotates around the sun.

So a Solar day is one rotation plus whatever angle we have travelled around the sun in that one rotation, about 1 degree. (1/365.??).

So the sun does not return to your meridian until it has done a bit less then a 361 degree rotation. (I think it is added on not taken off)

We tend to just refer to a 24 hours solar day, then every so often modify it all to get back to a nearer match - Feb 29 gets thrown in.

Does raise the question which is actually 24 hours, the one that uses the sun (and all our clocks refer to) which is actually a bit more then a 360 degree revolution or the one that uses a star and is actually one 360 degree revolution.

[nameunknown]

One way to do this is look up the time of sunrise and sunset in the local paper - half way between the two is "solar noon", so a stick in the ground will cast a shadow proper north at that time. Aligning a scope that way will get close enough that a proper polar alignment can be done using a polarscope and some correction for the fact that polaris is close to, but not at, the pole.

Otherwise, look up https://en.wikipedia.org/wiki/Equation_of_time

P

[Scooot]

This might help as well.

http://www.astro.cornell.edu/academics/courses/astro201/sidereal.htm

[jACK101]

My question was raised  because Chris Woodhouse in his "Astrophotography Manual on page 256 gives his method of alignment. First of all he says that a compass is not reliable enough. He suggests using a planetarium to find the local time for the sun when it will be on the meridian. I have interpreted that, perhaps wrongly, as being the sidereal time for the sun. I understood that when any body crossed the meridian its sidereal time was zero. I don't know what to do with a planetarium, hence the question.

Thanks for your responses, I clearly have a lot to learn.

Jack

[Scooot]

The North , South line is the meridian. You can polar align your scope by setting it up on the meridian, if you use the sun to plot the meridian in the first place. At solar noon the Suns shadow will point from south to north. There are easier ways to do it these days.

Current local sidereal time is always the same as the right ascension of a star that is crossing the meridian. So if you look at the screen shot below when sidereal time was roughly 00.00 and look up the right ascension of the star Deneb Kaitos which is just on the green meridian line, you'd see its right ascension is 00.44 (it's not quite on it). The right ascension of Uranus is 01.06 so it will cross the meridian in about an hour, when local sidereal time wil be 01.06. Hope this helps.

[acey]

My question was raised  because Chris Woodhouse in his "Astrophotography Manual on page 256 gives his method of alignment. First of all he says that a compass is not reliable enough. He suggests using a planetarium to find the local time for the sun when it will be on the meridian. I have interpreted that, perhaps wrongly, as being the sidereal time for the sun. I understood that when any body crossed the meridian its sidereal time was zero.

To find the time of local noon (i.e. time when Sun is on meridian) you can, for example, get it here:

http://www.timeanddate.com/sun/uk/london

(Enter your location and scroll down to see table column headed "solar noon")

If you need it more accurately then try googling for other sites (or maybe Stellarium does it) so you can enter your longitude rather than town. The method is potentially more accurate than a compass because magnetic north is not the same as geographical north. The degree of accuracy you need will depend on your intended task. But what you really seem to be concerned with here is finding south, so sidereal time is not important.