One of the first
tasks of any budding astronomer is to first master the layout of the night sky and that includes the angular distances or separations between objects. Critical in this process is knowing what the altitude of an astronomical object means - in relation to one's location- as well as how to find it. Next, the angular distances to other nearby objects - whether stars, planets or more exotic fare, e.g. nebulae. It is useless to use a linear measure such as meters, feet or inches because these have no meaning when referred to the sky, or celestial objects in it and distances between them. So we use * degrees*. Just a glance at the graphic in the article above- and blown up version below- shows the sense of this, given the sky presents a spherical extent.

To that end, when Janice and I began our series '*Discovering the Stars*' we realized we had to first deal with the basic angular system to measure celestial separations and relationships. Hence, the article clip shown above, for which the magnified image is given below.

This shows an observer at the center of his horizon ( 'horizon' system) and determining the altitude and azimuth of a star, S in it. A lateral view is also possible to use to more easily see angular relations, e.g.

For example, if the altitude of the star S is 45^{o }then the *zenith distance* z, would be found from:

z
= 90^{o} - A = 90^{o} - 45^{o}
= 45^{o}

In order to measure such an angular distance it is necessary to build a suitable instrument. One such is called a cross staff, designed specifically to measure angles in the sky. This ancient instrument was probably first used as long ago as 400 B.C. by the Chaldeans to make basic angular observations and computations. E,g.

One such ancient measurement was to obtain the zenith distance z of the pole star, and thereby to obtain latitude φ. Thus, if z = (90 - φ) then: φ = 90 - z . The device is relatively simple to construct, as indicated in the link below for any who might be interested:

__http://www.phy6.org/stargaze/Scrostaf.htm__But before one can use it one
must become familiar with the system of angular measure. Say that one wishes to
get the distance between the Moon and Saturn such as depicted in the star map
(from my Cybersky planetarium program) below:

But given degrees are only one unit, and many objects (e.g.
double stars) are much nearer, say fractions of a degree, one must be able to
use smaller measures too.

As for the cross staff shown in the image above,
that is mainly used to measure degrees. For a basic cross staff one
can generally measure from 15 degrees to 60 degrees with a fair amount of
accuracy. For simple estimates of angular measures up to 15 degrees, and as low as one degree
angular measure, the stars in the Big
Dipper provide a useful aid, e.g.

__+__ 1 degree due to natural disparity in
the size of hands.

Consider the angular
width of the full Moon, and we seek smaller angular units based upon it. An
arcminute is 1/30 the width of the full Moon. Hence, we conclude that 1 deg =
60 arcmin. The arcminute is further divided into 60 arcseconds, which is
typically used to measure the distance between components of a binary star, e.g.

Where the separation is in seconds of arc or arcsec. Conveniently, astronomers have learned that if the semi-major axis of the true relative orbit (the one displayed if the system were seen face-on) has an angular distance of a" (seconds of arc) then the semi-major axis in astronomical units would be:

a = (a" x d)

It should come as no surprise that planetary
widths - given they are tiny - would also be registered in arcseconds or
". Thus, the giant planet Jupiter can be seen up to 50" in diameter.
Mars will reach 24" in 2018 at a close opposition. Neptune is 2" and
Uranus is 4". (Pluto's angular width is barely 0.1"). To
fix ideas, to magnify Pluto's disk to one arcminute width (i.e. 1/30 of the
full Moon's diameter) would require a magnification of:

1 arcmin/ 0.1 arcsec = 60
arcsec/ 0.1 arcsec = 600 x

For completeness another angular measure used is
the* radian*.

For example the Sun has an *angular radius* of a = 959.63 "

But this must be in* radians* before one can compute the solar constant, for example.

One radian (1 rd) can first be converted into
arcsec as follows, given there are 3600 arcsec per degree.:

1 rd = 57.3 degrees = 57.3 deg/rad x (3600"/
deg)= 206 280 "

Then: a (rd) = 959.63"/ 206 280"/
rad = 0.00465 rad

*Suggested Problems*:

1) Two observers using cross
staffs obtain zenith distances from their respective locations of z = 45 degrees, and z = 35 degrees. How far apart in latitude are
their locations?

2) Consider the system *Epsilon Ursae Majoris* which semi-major axis subtends an angle
of 2½" and for which the parallax of the system is
0."127. Find the semi-major axis in astronomical units. (Hint:
p" = 1/d)

3) What telescope magnification would be
required to observe the planet Uranus as a disk *2 arcminutes* in diameter?

4) Find the Moon's angular diameter in radians.

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