. .
.
Observer Location Determination (Under Construction)
.
.

 

Introduction :

 

 

                      Knowledge of the co-ordinates of a few stars, and their zenithal distances at the time of observation are enough to determine the latitude and longitude of the observer. A simple concept of a 'substellar' point shall be used for the observer location determination.

 

 

Theory:

                                                                      

          Fig 1: The observer at O will observe the star to have a zenith distance of zs, and hence«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8736;«/mo»«mi mathvariant=¨bold-italic¨»O«/mi»«mi mathvariant=¨bold-italic¨»C«/mi»«mi mathvariant=¨bold-italic¨»S«/mi»«/math»will also be zs

 

As can be seen easily from the figure, if star S is at a zenithal distance of zs, then the observer must be located around a circle centered at L, with radius zs.

This point, L, is called the substellar point. Note that if we can plot 3 such circles then the observers location can be determines without doubt, since three circles can have atmost one common point, unless two of them are identical. First we shall see how to determine the substellar point of a star. Then we shall give procedural instructions as to how to use this substellar point to determint the position of an observer.

 

Determining substellar point:

 

                 To calculate the substellar point we need to know the RA and Declination of the star, the date of the year, and the current time in UT(Universal Time).

 

First we realize that the latitude of the substellar point of the star shall be equal to the declination of the star. This must be true since the celestial equator always maintains itself exactly over the earths equator. Now we are only left to align the earths line of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»0«/mn»«mo»§#176;«/mo»«/math» (Prime meridian), and the celestial line of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msup»«mn»0«/mn»«mi mathvariant=¨bold-italic¨»h«/mi»«/msup»«/math» (The point of the vernal equinox, or the first point of aries). We shall consider an observer on the equator, since we have already seen that it is easy to account for the latitiude. The vernal equinox is the point of the rising node of the earths orbit around the sun (the ecliptic), with respect to the celestial equator. This point is exactly on the zenith for an equatorial observer on the date of the vernal equinox (21st March) at his local noon. I.e. for an observer at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»0«/mn»«mo»§#176;«/mo»«/math» latitude, and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»0«/mn»«mo»§#176;«/mo»«/math» longitude, the first point of aries will be on the zenith at 1200 UT on 21st march.

 

Now, at this time, at this place, the substellar point of any object with«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»RA«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«msup»«mi mathvariant=¨bold-italic¨»H«/mi»«mi mathvariant=¨bold-italic¨»h«/mi»«/msup»«/math», and declination«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»§#948;«/mi»«/math»will have longitude«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»l«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«msup»«mi mathvariant=¨bold-italic¨»H«/mi»«mi mathvariant=¨bold-italic¨»h«/mi»«/msup»«mo»§#215;«/mo»«mfrac»«mrow»«mn»360«/mn»«mo»§#176;«/mo»«/mrow»«msup»«mn»24«/mn»«mi mathvariant=¨bold-italic¨»h«/mi»«/msup»«/mfrac»«mo»,«/mo»«/math»and latitude«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»§#946;«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi mathvariant=¨bold-italic¨»§#948;«/mi»«/math». Now, if we fix the date in the year, then as the earth rotates about the sun, once in 24h),the substellar point of any object will take on complete rotation on the earth surface, travelling east to west, maintaining its latitude. Similarly, if we keep the time of the day fixed, and we vary the date, then also the substellar point will take a complete rotation of the earth along it's surface, moving west to east, keeping the latitude constant. This happens since through one entire year teh earth takes one complete revolution of the sun, and effectively, from the frame of reference of the earth, it appears as if in the duration of one year the sky takes a revolution of the earth. Since both of these motions are almost entirely uniform, we can assume a linear movement of the substellar point with time in both cases.

If we wish to find the substellar point of any object at any time, now we only have to superimpose the above two described movements to be able to tell the substellar point of any object at any time of the year.

 

 

 

 

Cite this Simulator:

.....
..... .....
Copyright @ 2018 Under the NME ICT initiative of MHRD (Licensing Terms)
 Powered by AmritaVirtual Lab Collaborative Platform [ Ver 00.12. ]