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A doubt about declination coordinates


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Hello:

These days I'm beginning to enjoy "Guía del firmamento", by Comellas. On page 30 the author explains how to locate objects using his mixed recommeded method (locating an object easily identifiable by eye and then to reach its location with the coordinates). But my question is when I move the planisphere to follow graphically the explanation. For example, the star Antares (Alpha Scorpii), has a declination of -26 ° 19 ':

- Why, if it has a negative declination, I can see it from the Northern Hemisphere (36° North)?

- Why is reflected in the planisphere with aprox. + 12° declination?

Sure are two easy questions but I can't resolve myself! Any information would be wellcomed.

Kind regards.

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Hello Juan Carlos,

0° is on the Equatorial line and the horizon is at -35° here in Paris. That is why you have negative numbers. If you'd use an Alt grid then 0° would be the horizon.

So I would suspect that the +12° is on Alt-Az coordinate and the -26° is on Equatorial coordinate.

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I think that's right. The declination given is relative to the celestial equator and this is tilted relative to the earth's equator owing to the tilt of the earth's axis. Because of this tilt, all the stars appear to rotate around the Pole Star every 24 hours rather than around the zenith directly overhead and therefore some with moderately negative declinations will at their highest points be visible from northern locations.

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The circles on the planisphere are circles of declination. The planisphere will be designed for a particular latitude, so it will show stars down to a some limit of declination. In your case this limit will be below the celestial equator, i.e. a negative value of declination.

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The circles on the planisphere are circles of declination. The planisphere will be designed for a particular latitude, so it will show stars down to a some limit of declination. In your case this limit will be below the celestial equator, i.e. a negative value of declination.

Ok, thank you. But the scale of this circles is not graduated but equidistant, and the celestial equator not represented, hence my confussion.
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The planisphere is the inside surface of a sphere projected onto a flat circle. Hence the lines become distorted. Whenever a 3D image is represented on a 2D surface either the scale becomes distorted or the shape, normally the scale is distorted so that the image remains recognisable but some projections are the opposite such as gnomonic where the image is distorted to allow the grid to remain the same.

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The planisphere is the inside surface of a sphere projected onto a flat circle. Hence the lines become distorted. Whenever a 3D image is represented on a 2D surface either the scale becomes distorted or the shape, normally the scale is distorted so that the image remains recognisable but some projections are the opposite such as gnomonic where the image is distorted to allow the grid to remain the same.

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Than you. Then, a planisphere represent the real right ascension but not the real declination, this last only approximate. :o

I think that I need a star map for a precise coordinate information.

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The scale is accurate it's just been distorted to allow the constellations to be recognisable.

Look at it like you would an atlas based on the Earth which is an oblate spheroid. To display this on a two dimensional surface you would have to stretch the longitude lines greatly at the poles or compress the image at the equator. Or both. The actual angular distances will still be correct but the bearing and measured distances could be misleading.

If you look at a standard Mercator map of the Northern hemisphere and draw a straight line from London to New York it will cross the Atlantic not far from the Azores.

Now do the same thing on a globe, the straight line now goes through Iceland and Greenland.

The straight line on the 2D map is actually a curved line over the surface of the Earth.

However if you draw a line from Durban to London they will be very similar on both as the map is not being distorted in the North/South direction.

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The scale is accurate it's just been distorted to allow the constellations to be recognisable.

Look at it like you would an atlas based on the Earth which is an oblate spheroid. To display this on a two dimensional surface you would have to stretch the longitude lines greatly at the poles or compress the image at the equator. Or both. The actual angular distances will still be correct but the bearing and measured distances could be misleading.

If you look at a standard Mercator map of the Northern hemisphere and draw a straight line from London to New York it will cross the Atlantic not far from the Azores.

Now do the same thing on a globe, the straight line now goes through Iceland and Greenland.

The straight line on the 2D map is actually a curved line over the surface of the Earth.

However if you draw a line from Durban to London they will be very similar on both as the map is not being distorted in the North/South direction.

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Thank you for your clear post!

In the star atlas is the same?

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Yes the only way you will get both identical scale and image is on a sphere. You can get planispheres but they aren't very practical for star hopping.

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Yes the only way you will get both identical scale and image is on a sphere. You can get planispheres but they aren't very practical for star hopping.

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Thank you. But in a star atlas the declination coordinates are exactly scaled in grades I think. The projection logically distort the image but the coordinates are exact

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You're right, a star chart projects the sphere onto a plane and shows co-ordinates of the sphere. So what exactly is your problem? Are you looking for a representation of the sphere that doesn't introduce distortion? In that case your map needs to be spherical (or hemispherical), i.e. use a globe or sit inside a planetarium.

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You're right, a star chart projects the sphere onto a plane and shows co-ordinates of the sphere. So what exactly is your problem? Are you looking for a representation of the sphere that doesn't introduce distortion? In that case your map needs to be spherical (or hemispherical), i.e. use a globe or sit inside a planetarium.

:D I understand that projection distort the object, I use maps from years ago. This is not the problem. I will try explain it in my very poor English.

If you divide a semisphere in nine declination zones (10° every part) paralels and you project this eight equidistant zones in a plane surface, the resultant circles are not equidistant. And in my planisfhere are equidistant, the equatorial line is not marked and the negative declination objects are into the paralel n° 8. I don't understand how can calculate the aprox. declination in the planisphere.

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What planisphere are you using? If it's a Philips then declinations are marked on one or more of the right ascension lines (I don't have mine handy to inspect in detail).

If you need accurate declination co-ordinates then you will in any case be better using a large-scale map, not a planisphere, which is only really useful as a way of finding naked-eye targets.

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Ok try to remember that declination is basically latitude. Now on earth latitude does distort slightly towards the poles due to flattening but that isn't the case for dec so let's ignore that ;)

1 degree of latitude is 1852m, ignoring flattening this remains the same all over the globe.

1 degree of longitude is 1852m at the equator, at the pole it is 0m.

Now your 2D planisphere works in the same way but with RA and DEC.

Unless it is poorly made the coordinates will be accurate but the scale looks wrong as it has been distorted to make the image recognisable. The scale may also be difficult to use as the half way point between two lines may not be the real half way point.

As suggested by others you could be much better off with a star chart or computer based system.

Try carts du ceil, this produces very good charts and is free.

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