How to align a stick with earth's axis?

[aphh]

Let's say I wanted to put a camera or telescope on a tripod and follow an object in the sky as effortlessly as possible.

If I had the tripod standing and pointing to the star Polaris, I should be pretty close to having my base axis in line with earth's (~0.5 degrees). When the earth turns, the vertical axis of my tripod turns similarly and I only need to adjust one axis, the horizontal axis or pan, to keep the object in view.

Is the required tilt for the tripod's vertical axis the same as your latitude? I think it is; if you were to reside on the North pole, your vertical axis would point straight up i.e. 90 degrees (0 degree tilt). If you were to reside on the equator, your vertical axis would lay on the ground and point to North (90 degree tilt). So the needed tilt for the vertical axis for your tripod would be 90 degrees minus your current latitude and pointing north.

Is this correct thinking?

 

[MeteorWayne]

If I'm correctly understanding what you are asking, yes, If you are at the pole, the axis points straight up all the time. If you are not at the pole, your axis for a scope, when aimed at polaris, will match your latitude.

This is the concept for an equatorial mount. And as you said, then you only have to worry about one rotation, the 24 hour/360 degree rotation of the sky around the pole (which is very near Polaris).

 

[aphh]

This is what I want to try next, put the binoculars on a tripod that is tilted and directed appropriately and see if I could observe an object simply panning the tripod head to one direction as the world turns.

 

[MeteorWayne]

I don't think that will work without an equatorial mount. The path of the stars during a night is circles centered around the pole star.

http://antwrp.gsfc.nasa.gov/apod/ap091128.html

 

[aphh]

I don't think that will work without an equatorial mount. The path of the stars during a night is circles centered around the pole star.

http://antwrp.gsfc.nasa.gov/apod/ap091128.html

But if your vertical axis already points to the center of the circle, shouldn't you be able to follow a selected object simply turning the base axis i.e. panning the head of the tripod? The only problem would be to make the tripod's axis point to star Polaris as accurately as possible.

If you take the image you linked and point a pen to the center of the circle (vertical axis of the tripod) and then imagine a telescope on the pen fixed to point a star, they would both rotate the same when you rotate the pen.

 

[MeteorWayne]

I don't think that will work without an equatorial mount. The path of the stars during a night is circles centered around the pole star.

http://antwrp.gsfc.nasa.gov/apod/ap091128.html

But if your vertical axis already points to the center of the circle, shouldn't you be able to follow a selected object simply turning the base axis i.e. panning the head of the tripod? The only problem would be to make the tripod's axis point to star Polaris as accurately as possible.

I guess that could work, but I suspect it would be mighty inconvenint with binoculars :shock:

 

[aphh]

I guess that could work, but I suspect it would be mighty inconvenint with binoculars :shock:

I just want to understand the concept. With this method I could keep the binoculars easily pointed to a nice star cluster and show somebody diamonds in the sky.

 

[MeteorWayne]

In binoculars, things don't move that fast anyway. There are very nice parallogram binocular mounts that allow you to move the binocs easily and keep pointing at the same spot as you change the height.

http://www.youtube.com/watch?v=P2uluHEe_GI

 

[aphh]

If we think about this a bit more (this is all very basic astronomy), with this method you should be able to locate a star's elevation in the sky pretty easily.

Let's say you know the declination or elevation on the celestial sphere of star Arcturus, which would be ~19 degrees. Let's assume your latitude would be 45 degrees North. Knowing your latitude you should now know the elevation of star Arcturus at your location. Hence to you Arcturus would appear at elevation of 90 degrees - 45 degrees + 19 degrees = 64 degrees (and 12 hrs later at 64 degrees + 26 degrees + 26 degrees = 116 degrees).

My question is: when is this true? At what time will Arcturus appear at 64 degrees above horizon at latitude 45 N?

 

[MeteorWayne]

It depends on what time of the year it is.

For example tonight, Arcturus rises at 6:11 PM reaches it's highest elevation (64 degrees) at 1:38 AM, and sets at 9:02 AM.

In two months (June 16) the rise time is 2:11 PM, transit (64 degrees) 9:35 PM and sets at 5:02 AM.

It reaches it's minumim elevation (26 degrees below the horizon) approximately 12 hours after the transit.

 

[aphh]

It depends on what time of the year it is.

For example tonight, Arcturus rises at 6:11 PM reaches it's highest elevation (64 degrees) at 1:38 AM, and sets at 9:02 AM.

In two months (June 16) the rise time is 2:11 PM, transit (64 degrees) 9:35 PM and sets at 5:02 AM.

It reaches it's minumim elevation (26 degrees below the horizon) approximately 12 hours after the transit.

Thanks! Yes, it does depend on the time of the year, but the base time I belive is the star time. This is different from the sun time (or average sun time), because one star day is only 23 hr 56 minutes long.

The star has a position on the celestial sphere which is called Right Ascension. This is the amount in hours, minutes and seconds of how much the star trails the fixed 0 position of the celestial sphere. So if a star had Right Ascension of 14h 15 m 39.7s (Arcturus), it would culminate, or reach it's highest point, 14 hrs, 15 minutes and 39 seconds after the fixed 0 point of the celestial sphere has passed south of your location.

Now all we need to know is the correlation between star time and sun time, plus the zero point in the sky, and we should be able to calculate a star's position on any local coordinates at any given time. The star time and sun time are the same once (or twice?) each year at some location and the fixed 0 position, or Vernal Equinox, is in the south at some location at that moment.

 

[MeteorWayne]

Too much thinkin' for me

I use my astronomy prgram SkyMapPro

 

[aphh]

Star time would actually be more precise, as it does not vary with earth's velocity on the ecliptica, as does the sun time (hence we use average sun time, when clock displays midday sun may be slightly behind or already passed south). Only problem is that star days lead 24 hr days progressively by ~four minutes each day.

When the cycle is full, the star time and sun time are the same again for a brief moment and a year has passed. So the star time and sun time are the same in spring during the equinox. The base point of the celestial coordinate system points to Vernal Equinox.

http://en.wikipedia.org/wiki/Equinox#Ce ... te_systems
http://en.wikipedia.org/wiki/Equatorial ... ate_system

I have heard that some astronomers carry both clocks, one that shows average sun time and one that displays current star time or actually it is called sidereal time. Stars precisely follow the sidereal time, so knowing that and a star's coordinates, you would always know where a particular star resides in the sky.

http://docs.kde.org/development/en/kdee ... ereal.html

Edit: the last link has a very good and simple explanation of the relationship of the star time with the solar time. Basically what you need to know is your Local Sidereal Time and your Local Meridian and you should always know when a star culminates (=passes local meridian in the south).