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Matrix Method for Coordinates Transformation

Toshimi Taki

January 10, 2002

Revision A: February 17, 2002

Revision B: December 21, 2002

Revision C: January 11, 2003

Revision D: November 22, 2003

Table of Contents

1. Introduction....................................................................................................................... 3

2. References......................................................................................................................... 3

3. Notations ........................................................................................................................... 3

3.1 Note .............................................................................................................................. 3

3.2 Symbols ....................................................................................................................... 3

4. Basic Equations of Coordinates Transformation in Matrix Method............................ 5

4.1 Polar Coordinates and Rectangular Coordinates ................................................... 5

4.2 Coordinate Transformation........................................................................................ 8

4.2.1 New Coordinate System Rotated around Z-axis............................................... 8

4.2.2 New Coordinate System Rotated around X-axis .............................................. 9

4.2.3 New Coordinate System Rotated around Y-axis ............................................ 10

4.3 Obtaining Polar Coordinates from Direction Cosine............................................ 11

4.4 Notes on Approximation .......................................................................................... 12

4.4.1 Approximation of Trigonometric Functions.................................................... 12

4.4.2 Approximation of Other Functions................................................................... 12

5. Applications .................................................................................................................... 13

5.1 Transformation from Equatorial Coordinates to Altazimuth Coordinates ......... 13

5.1.1 Transformation Equations................................................................................. 13

5.1.2 Example Calculation .......................................................................................... 15

5.2 Angular Separation................................................................................................... 18

5.2.1 Equations ............................................................................................................ 18


5.3 Compensation of Mounting Fabrication Errors..................................................... 21

5.3.1 Telescope Coordinates ...................................................................................... 21

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5.3.2 Fabrication Errors of Mount.............................................................................. 21

5.3.3 Derivation of Equations..................................................................................... 22

5.3.4 Apparent Telescope Coordinate without Approximation .............................. 24

5.3.5 Example Calculations ........................................................................................ 28

5.4 Equations for Pointing Telescope........................................................................... 32

5.4.1 Introduction......................................................................................................... 32

5.4.2 Transformation Matrix........................................................................................ 33

5.4.3 Derivation of Transformation Matrix ................................................................ 34


5.5 Polar Axis Misalignment Determination................................................................. 42

5.5.1 Derivation of Equations..................................................................................... 42

5.5.2 Example Calculations ........................................................................................ 52

5.6 Dome Slit Synchronization ...................................................................................... 62

5.6.1 Object in First Quadrant .................................................................................... 63

5.6.2 Object in Second Quadrant............................................................................... 65

5.6.3 Object in Third Quadrant................................................................................... 67

5.6.4 Object in Fourth Quadrant ................................................................................ 69

5.6.5 Intersection ......................................................................................................... 71

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1. Introduction

Coordinates transformation is a basic part of astronomical calculation and spherical

trigonometry has been long used for astronomical calculation in amateur astronomy.

Spherical trigonometry equations can be a little bit difficult for amateurs to understand.

In the last two decades, development of personal computers has brought about a change in

the way astronomical calculations are carried out. In my opinion, spherical trigonometry is

not appropriate to astronomical calculation using personal computers. I recommend the

matrix method for coordinates transformation, because of its simplicity and ease of

generalization in writing computer programs.

In this monograph, I describe coordinates transformation using the matrix method. I also

extend the method to some specific applications, such as polar axis misalignment

determination of equatorial mount (Challis’ method) and a telescope pointing algorithm.

2. References

[1] Jean Meeus, “Astronomical Formulae for Calculators,” 1985, Willmann-Bell, Inc.

[2] Jean Meeus, “Astronomical Algorithms,” 1991, Willmann-Bell, Inc.

[3] Ko Nagasawa, “Calculation of Position of Astronomical Objects,” 1985, Chijin-Shokan

Co., in Japanese

[4] W. R. Vezin, “Polar Axis Alignment of Equatorial Instrument”

[5] Rev. James Challis, “Lectures on Practical Astronomy and Astronomical Instruments,”

1879.

[6] Toshimi Taki, “A New Concept in Computer-Aided Telescopes," Sky & Telescope,

February 1989, pp.194-196.

3. Notations

3.1 Note

In this monograph, angles are expressed in radian, because all computer languages for

personal computers use radian for trigonometric functions.

3.2 Symbols

Following symbols are used in this monograph.

X-Y-Z : General rectangular coordinate system

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Xe-Ye-Ze : Rectangular equatorial coordinate system

Xh-Yh-Zh : Rectangular altazimuth coordinate system

Xt-Yt-Zt : Rectangular telescope coordinate system

α : Right Ascension (in radian)

δ : Declination (in radian)

A : Azimuth, measured westward from the South. (in radian)

h : Altitude (in radian)

ξ : General polar coordinate measured counterclockwise from X-axis in XY-plane (in

radian)

ζ : General polar coordinate, measured upward from XY-plane. (in radian)

ϕ, θ : Telescope polar coordinates (in radian)

∆, ∆', ∆" : Telescope mount fabrication errors (in radian)

L : X-component of direction cosine of celestial object in X-Y-Z coordinates

M : Y-component of direction cosine of celestial object in X-Y-Z coordinates

N : Z-component of direction cosine of celestial object in X-Y-Z coordinates

H : Hour angle

φ : Observer’s latitude

θ0 : Sidereal time at Greenwich

t : Time

u, v : Telescope polar axis misalignment (in radian)

JD : Julian day number

d : Angular distance between two objects

[T] : Transformation matrix between coordinate systems

R : Atmospheric refraction (in radian)

5/72

4. Basic Equations of Coordinates Transformation in Matrix Method

4.1 Polar Coordinates and Rectangular Coordinates

In astronomical calculations, polar coordinate systems are usually used. See figure 4-1.

Point O is the observation point. Vector OR shows unit vector directing to a celestial object.

The position of the celestial object is express in polar coordinates (ξ, ζ). Normally, angle ξ is

measured counterclockwise from X-axis (viewing from positive Z) and angle ζ is measured

upward (toward Z-axis) from XY-plane.

Figure 4-1 Polar Coordinates and Rectangular Coordinates

An example of polar coordinates is right ascension and declination, (α, δ). See figure 4-2.

The other example is azimuth and altitude, (A, h). But azimuth is measured westward

(clockwise) from the South which is the opposite direction to the normal polar coordinate

system. See figure 4-3.

YX

Z Celestial Object

O

R (L,M,N)

Sphere with radius = 1

ζ

ξ

YX

Z Celestial Object

O

R (L,M,N)

Sphere with radius = 1

ζ

ξ

6/72

Figure 4-2 Equatorial Coordinates

Figure 4-3 Altazimuth Coordinates

Ye

Xe (Vernal Equinox)

Ze (Celestial North Pole)Celestial Object

O

R (Le,Me,Ne)

δ

α

Celestial Equator

Ye

Xe (Vernal Equinox)

Ze (Celestial North Pole)Celestial Object

O

R (Le,Me,Ne)

δ

α

Celestial Equator

Yh (East)

Xh (South)

Zh (Zenith) Celestial Object

O

R (Lh,Mh,Nh)

hA

Horizon

Yh (East)

Xh (South)

Zh (Zenith) Celestial Object

O

R (Lh,Mh,Nh)

hA

Horizon

7/72

The vector OR is also expressed in rectangular coordinates, (L, M, N). (L, M, N) is called

direction cosine. In the matrix method, direction cosines are used to express coordinate

transformation.

Relationship between rectangular coordinates and polar coordinates can be expressed in

matrix form as follows.

=

ζξζξζ

sin

sincos

coscos

N

M

L

…. Equation (4-1)

For equatorial coordinates,

=

δαδαδ

sin

sincos

coscos

e

e

e

N

M

L

…. Equation (4-2)

For horizontal coordinates,

−−

=

h

Ah

Ah

N

M

L

h

h

h

sin

)sin(cos

)cos(cos

…. Equation (4-3)

Note that (-A) is used in the equation (4-3) instead of A, because azimuth A is measured

clockwise.

8/72

4.2 Coordinate Transformation

4.2.1 New Coordinate System Rotated around Z-axis

New coordinate system, X’-Y’-Z’ is generated rotating X-Y-Z coordinates around Z-axis as

shown in figure 4-4.

The polar coordinates in X’-Y’-Z’ coordinate system is (ξ’, ζ’) and the direction cosine in

X’-Y’-Z’ coordinate system is (L’, M’, N’). The relationship between the direction cosines in

both coordinate systems is expressed as follows.

=

'sin

'sin'cos

'cos'cos

'

'

'

ζξζξζ

N

M

L

…. Equation (4-4)

−=

N

M

L

N

M

L

zz

zz

100

0cossin

0sincos

'

'

'

θθθθ

…. Equation (4-5)

−=

'

'

'

100

0cossin

0sincos

N

M

L

N

M

L

zz

zz

θθθθ

…. Equation (4-6)

Figure 4-4 Coordinates Rotation around Z-axis

X

X’

Y

Y’

OR : unit vector

L

M

L’M’

O

θz (rotate counterclockwise around Z-axis)

R

Looking Normal to XY-plane

X

X’

Y

Y’

OR : unit vector

L

M

L’M’

O

θz (rotate counterclockwise around Z-axis)

R

Looking Normal to XY-plane

9/72

4.2.2 New Coordinate System Rotated around X-axis

New coordinate system, X’’-Y’’-Z’’ is generated rotating X-Y-Z coordinates around X-axis as

shown in figure 4-5.

The polar coordinates in X’’-Y’’-Z’’ coordinate system is (ξ’’, ζ’’) and the direction cosine in

X’’-Y’’-Z’’ coordinate system is (L’’, M’’, N’’). Then the relationship between the direction

cosines in both coordinate systems is expressed as follows.

=

''sin

''sin''cos

''cos''cos

''

''

''

ζξζξζ

N

M

L

…. Equation (4-7)

−=

N

M

L

N

M

L

xx

xx

θθθθ

cossin0

sincos0

001

''

''

''

…. Equation (4-8)

−=

''

''

''

cossin0

sincos0

001

N

M

L

N

M

L

xx

xx

θθθθ …. Equation (4-9)

Figure 4-5 Coordinates Rotation around X-axis

Y

Y’’

Z

Z’’

OR : unit vector

M

N

M’’N’’

O

θx (rotate counterclockwise around X-axis)

R

Looking Normal to YZ-plane

Y

Y’’

Z

Z’’

OR : unit vector

M

N

M’’N’’

O

θx (rotate counterclockwise around X-axis)

R

Looking Normal to YZ-plane

10/72

4.2.3 New Coordinate System Rotated around Y-axis

New coordinate system, X’’’-Y’’’-Z’’’ is generated rotating X-Y-Z coordinates around Y-axis

as shown in figure 4-6.

The polar coordinates in X’’’-Y’’’-Z’’’ coordinate system is (ξ’’’, ζ’’’) and the direction cosine

in X’’’-Y’’’-Z’’’ coordinate system is (L’’’, M’’’, N’’’). Then the relationship between the

direction cosines in both coordinate systems is expressed as follows.

=

''''sin

''''sin''''cos

''''cos'''cos

'''

'''

'''

ζξζξζ

N

M

L

…. Equation (4-10)

−=

N

M

L

N

M

L

yy

yy

θθ

θθ

cos0sin

010

sin0cos

'''

'''

'''

…. Equation (4-11)

−=

'''

'''

'''

cos0sin

010

sin0cos

N

M

L

N

M

L

xy

yy

θθ

θθ …. Equation (4-12)

Figure 4-6 Coordinates Rotation around Y-axis

Z

Z’’’

X

X’’’

OR : unit vector

N

L

N’’’

L’’’

O

θy (rotate counterclockwise around Y-axis)

R

Looking Normal to ZX-plane

Z

Z’’’

X

X’’’

OR : unit vector

N

L

N’’’

L’’’

O

θy (rotate counterclockwise around Y-axis)

R

Looking Normal to ZX-plane

11/72

4.3 Obtaining Polar Coordinates from Direction Cosine

After coordinate transformation using the matrix method it is necessary to obtain the polar

coordinates (ξ’, ζ’) from the direction cosines.

Using equation (4-4), ξ’ and ζ’ are obtained from direction cosines as shown below.

''

'tanLM

=ξ …. Equation (4-13)

When L’ >= 0, ξ’ is in the 1st quadrant or the 4th quadrant.

When L’ < 0, ξ’ is in the 2nd quadrant or the 3rd quadrant.

''sin N=ζ …. Equation (4-14)

-π/2 (-90o) <= ζ’ <= +π/2 (+90o)

12/72

4.4 Notes on Approximation

4.4.1 Approximation of Trigonometric Functions

When we process small angles in trigonometry, approximation of trigonometric functions is

often used.

In the following approximations, θ is very small angle and expressed in radian.

θθ ≅sin …. Equation (4-15)

1cos ≅θ …. Equation (4-16)

For higher order approximation,

21cos

2θθ −≅ …. Equation (4-17)

4.4.2 Approximation of Other Functions

For other functions, following approximation can be used when x is very small compared to

1.

211

xx +≅+ …. Equation (4-18)

( ) xx 211 2 +≅+ …. Equation (4-19)

13/72

5. Applications

5.1 Transformation from Equatorial Coordinates to Altazimuth Coordinates

5.1.1 Transformation Equations

Altazimuth coordinate system, Xh-Yh-Zh is rotated –(π/2 - φ) around Yh-axis to equatorial

coordinate system, Xe’-Ye’-Ze’. φ is observer’s latitude. See figure 5.1-1.

The direction cosines are expressed in angles as follows.

−−

=

h

Ah

Ah

N

M

L

h

h

h

sin

)sin(cos

)cos(cos

…. Equation (5.1-1)

Where A is azimuth measured westward from the South and h is altitude.

−−

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

e

e

e


Where H is local hour angle measure westward from the South and δ is declination.

Relationship between the coordinates is expressed in matrix form as shown below.

−

−

−−

−

=

h

h

h

e

e

e

N

M

L

N

M

L

2cos0

2sin

0102

sin02

cos

'

'

'

πφ

πφ

πφ

πφ


−

−−

−

−

=

'

'

'

2cos0

2sin

0102

sin02

cos

e

e

e

h

h

h

N

M

L

N

M

L

πφ

πφ

πφ

πφ


h

h

L

MA =− )tan( …. Equation (5.1-5)

When Lh >= 0, (-A) is in the 1st quadrant or the 4th quadrant.

When Lh < 0, (-A) is in the 2nd quadrant or the 3rd quadrant.

hNh =sin …. Equation (5.1-6)

14/72

-π/2 (-90o) <= h <= +π/2 (+90o)

Figure 5.1-1 Altazimuth Coordinates and Equatorial Coordinates

Comparison with spherical trigonometric equations (ref. [1]) is performed below.

From equations (5.1-2), (5.1-4) and (5.1-5), we obtain the following equations.

δφφ

δπ

φπ

φ

δπ

φδπ

φ

δ

πφ

πφ

tancoscossin)sin(

tan2

sin)cos(2

cos

)sin(

sin2

sincos)cos(2

cos

cos)sin(

'2

sin'2

cos

')tan(

−−

=

−+−

−

−=

−+−

−

−=

−+

−

=−

HH

H

H

H

H

NL

MA

ee

e


Yh, Ye’ (East)

Xh (South)

Zh (Zenith)Celestial Object

O

Ze’ (North Pole)

Xe’ (Meridian)

φ-H

Equator

Horizon

Meridian

Yh, Ye’ (East)

Xh (South)

Zh (Zenith)Celestial Object

O

Ze’ (North Pole)

Xe’ (Meridian)

φ-H

Equator

Horizon

Meridian

15/72

δφδφ

δπ

φδπ

φ

sinsincoscoscos

sin2

coscos)cos(2

sinsin

+=

−+−

−−=

H

Hh …. Equation (5.1-8)

These equations are the same as equations (8.5) and (8.6) in ref. [1].

5.1.2 Example Calculation

Example 8.b in ref. [1]:

Find the azimuth and the altitude of Saturn on 1978 November 13 at 4h34m00s UT at the

Uccle Observatory (longitude –0h17m25.94s, latitude +50o47’55.0” = 0.88660302 (radian)); the

planet’s apparent equatorial coordinates, interpolated from the A.E., being

α = 10h57m35.681s = 10.9599114h = 2.86929809 (radian)

δ = +8o25’58.10” = 8.432806o = 0.14718022 (radian)

The apparent sidereal time at Greenwich, θ0 = 8h01m46.135s.

Local hour angle, H is,

H = θ0 - L - α

= 8h01m46.135s + 0h17m25.94s – 10h57m35.681s

= -2h38m23.606s

= -2.6398906h

= -2.6398906 x 15 / (180/π) … (radian)

= -0.69112174 (radian)

From equation (5.1-2),

=

−−

=

14664943.0

63051067.0

76220092.0

sin

cos)sin(

cos)cos(

'

'

'

δδδ

H

H

N

M

L

e

e

e


16/72

=

−=

−−−

−−=

−

−−

−

−

=

59539056.0

63051067.0

49796223.0

14664943.0

63051067.0

76220092.0

77492917.0063204809.0

010

63204809.0077492917.0

14664943.0

63051067.0

76220092.0

)68419331.0cos(0)68419331.0sin(

010

)68419331.0sin(0)68419331.0cos(

'

'

'

2cos0

2sin

0102

sin02

cos

e

e

e

h

h

h

N

M

L

N

M

L

πφ

πφ

πφ

πφ

From equations (5.1-5) and (5.1-6),

2661817.149796223.063051067.0

)tan( ===−h

h

L

MA

-A = 0.90232066 (radian) à A = -0.9032066 (radian) = -51.6992o

sin h = 0.59539056

h = 0.63775167 (radian) = 36.5405o

17/72

Figure 5.1-2 Hour Angle and Sidereal Time

Looking from North Pole Normal to Equatorial Plane

Xe (Vernal Equinox)

Ye

Ye’

O

Xe’ (Meridian)

α: Right Ascension

H: Hour Angle

θ : Local Sidereal Time

Celestial Object

Looking from North Pole Normal to Equatorial Plane

Xe (Vernal Equinox)

Ye

Ye’

O

Xe’ (Meridian)

α: Right Ascension

H: Hour Angle

θ : Local Sidereal Time

Celestial Object

18/72

5.2 Angular Separation

5.2.1 Equations

The angular distance d between two celestial objects, P1 and P2 is derived using the matrix

method.

Position of object 1, P1: (ξ1, ζ1)

Position of object 2, P2: (ξ2, ζ2)

Direction cosines of the two objects are,

=

1

11

11

1

1

1

sin

sincos

coscos

ζξζξζ

N

M

L


=

2

22

22

2

2

2

sin

sincos

coscos

ζξζξζ

N

M

L


Using scalar product of the two unit vectors, 1OP and 2OP , angular separation d is

obtained as follows. See figure 5.2-1.

)cos(coscossinsin

sinsincoscossinsincoscoscoscos

cos

212121

2121212121

212121

ξξςςςςςςςςξξςςξξ

−+=++=

++= NNMMLLd


This equation is identical to equation (9.1) in ref. [1].

Figure 5.2-1 Angular Separation

OP1

P2

d

OP1

P2

d

19/72

When angular separation is very small, (ξ1-ξ2) and (ζ1-ζ2) are nearly zero and equation

(5.2-2) can not be used. Equation (5.2-2) is transformed to a new equation as follows.

( )

( )

( )

( ) ( )ς

ξς

ςξ

ς

ςςξ

ςςςς

ξςςςς

ξςςςς

222

22

21

2

2121

2

2121

2121

cos22

1

cos2

cos

coscos2

coscossinsin

21coscossinsin

coscoscossinsincos

∆−

∆−≅

∆−∆=

∆−+=

∆−+≅

∆+=d

Where 2

21 ςςς

−=

( ) ( ) ( )ς

ξς 2222

cos22

12

1

cos

∆−

∆−≅

∆−≅

d

d

( ) ( ) ( ) ςξς 2222 cos∆−∆≅∆d

( ) ( )22cos ςςξ ∆+⋅∆=d …. Equation (5.2-4)

Note: When θ is very small, the following approximation can be used.

21cos

2θθ −≅ (in radian)


Example 9.a in ref. [1]:

Calculate the angular distance, d between Arcturus and Spica.

The 1950 coordinates of these stars are,

Arcturus : α1 = 14h13m22.8s = 213.3450o = 3.72357269 δ1 = +19o26’31” = 0.33932594

Spica : α2 = 13h22m33.3s = 200.6388o = 3.50180767 δ2 = -10o54’03” = -0.19025543

20/72

−−

=

=

33285154.0

51833597.0

78774214.

sin

sincos

coscos

1

11

11

1

1

1

δαδαδ

N

M

L

−−−

=

=

18910972.0

34611538.0

91893507.0

sin

sincos

coscos

2

22

22

2

2

2

δαδαδ

N

M

L

84034247.0

)18910972.0(33285154.0

)34611538.0()51833597.0()91893507.0(78774214.0

cos 212121

=−×+

−×−+−×−=++= NNMMLLd

d = 0.57288162 (radian) = 32.8237o

21/72

5.3 Compensation of Mounting Fabrication Errors

5.3.1 Telescope Coordinates

A telescope has telescope coordinate system as shown in figure 5.3-1. True telescope polar

coordinates is (ϕ, θ). “True” means that we consider hypothetical perfect telescope mount

without fabrication error.

If the Xt-axis points to the South and Zt-axis points to zenith, this mount is an alt-azimuth

mount.

A−=ϕ

h=θ

If the Zt-axis points to celestial north pole and Xt-axis points to meridian, this mount is an

equatorial mount.

H−=ϕ

δθ =

Figure 5.3-1 Telescope Coordinates

5.3.2 Fabrication Errors of Mount

In the real world all mountings have fabrication errors. There are three different fabrication

errors to be considered as shown in figure 5.3-2.

(1) ∆: Error in perpendicularity between horizontal axis and vertical axis, or polar axis

Xt

Yt

Zt

«

ϕ: Horizontal Angle

θ: Elevation Angle

Xt

Yt

Zt

«


θ: Elevation Angle

22/72

and declination axis

(2) ∆’: Collimation error between vertical or polar axis and telescope optical axis

(3) ∆’’: Shift of zero point in apparent elevation angle or declination angle

Figure 5.3-2 Telescope Mount Fabrication Error

5.3.3 Derivation of Equations

The apparent telescope coordinates (ϕ’, θ’) is the coordinates measured with setting circles

of the telescope mount. Relationship between the true telescope coordinates and the

apparent telescope coordinates are derived as follows. See figures 5.2-3 to 5.2-5.

(1) Telescope optical axis, R’’’-axis points to a celestial object of true telescope

coordinates (ϕ, θ). R’’’S’’’-plane is the plane defined by telescope optical axis and

telescope vertical axis. This means that direction cosines of the celestial object in

R’’’-S’’’-T’’’ coordinates is

0

0

1

.

∆

∆’

θ’’ + ∆”

Telescope Optical Axis

Telescope Vertical Axis

Telescope Horizontal Axis

∆

∆’

θ’’ + ∆”

∆

∆’

θ’’ + ∆”




23/72

(2) The coordinate system R’’’-S’’’-T’’’ is rotated –∆’ counterclockwise around T’’’-axis and

becomes a new coordinate system R’’-S’’-T’’.

(3) The coordinate system R’’-S’’-T’’ is rotated (θ’+∆’’) counterclockwise around S’’-axis

and becomes R’-S’-T’ coordinate system.

(4) The coordinate system R’-S’-T’ is rotated -∆ counterclockwise around R’-axis and

becomes R-S-T coordinate system.

(5) Finally, the coordinate system R-S-T is rotated –ϕ’ counterclockwise around T-axis and

becomes X-Y-Z coordinate system which is the true telescope coordinates.

∆∆+∆∆∆+∆∆∆+−∆∆+∆∆+∆∆∆++∆∆−∆∆+

=

∆∆∆−∆

∆+∆+

∆+−∆+×

∆∆∆−∆

−=

'sinsin'coscos)'''sin(

'cossin'cos)'''sin('sincos'cos'cos'sin)'''cos(

'cossin'sin)'''sin('sincos'sin'cos'cos)'''cos(

0

0

1

100

0'cos'sin

0'sin'cos

)'''cos(0)'''sin(

010

)'''sin(0)'''cos(

cossin0

sincos0

001

100

0'cos'sin

0'sin'cos

sin

sincos

coscos

θϕθϕϕθϕθϕϕθ

θθ

θθ

ϕϕϕϕ

θϕθϕθ


Equation (5.3-1) is an exact solution to obtain true telescope coordinate from apparent

telescope coordinate.

Using the following approximation,

'cossincossin'cossin'cos)'''sin(,'sincoscos'sincos'cos

'cossinsinsin'cossin'sin)'''sin(,'sincossin'sincos'sin

∆∆≈∆∆∆+∆∆≈∆∆∆∆≈∆∆∆+∆∆≈∆∆

ϕθϕθϕϕϕθϕθϕϕ

Equation (5.3-2) is derived from equation (5.3-1).

∆∆∆∆−∆∆∆+∆∆−∆∆∆−∆∆+

=

∆+∆+∆+

'coscos/)'sinsin(sin

'cos/)'cossincossin'sincoscossin(cos

'cos/)'cossinsinsin'sincossincos(cos

)'''sin(

'sin)'''cos(

'cos)'''cos(


θϕθϕθ


24/72

Using equation (5.3-2), an exact solution of θ ’ and an approximate solution of ϕ’ are

obtained.

Further approximation can be made as follows.

Assuming the errors are very small, equations (5.3-1) and (5.3-2) are simplified as follows.

∆+∆+∆−∆+∆+∆+∆+∆−∆+

=

)'''sin(

'cos)'''sin('cos''sin)'''cos(

'sin)'''sin('sin''cos)'''cos(

sin

sincos

coscos


θϕθϕθ


∆+∆−∆−∆+

=

∆+∆+∆+


θϕθϕθ

sin

cossincos'sincos

sinsinsin'coscos

)'''sin(

'sin)'''cos(

'cos)'''cos(


5.3.4 Apparent Telescope Coordinate without Approximation

In order to obtain an exact solution of apparent telescope coordinate ϕ’, iteration will be

made as follows.

Rewriting equation (5.3-1), we get the following equation.

∆∆∆∆−∆∆∆∆∆−+∆∆−∆∆∆∆∆−−∆∆+

=

∆∆∆∆−∆∆∆∆++∆∆−∆∆∆∆+−∆∆+

=

∆+∆+∆+


'cos/)cos/sin'cos)'sinsin(sin'sincos'cossin(cos

'cos/)cos/sin'sin)'sinsin(sin'sincos'sincos(cos


'cos/)'cossin'cos)'''sin('sincos'cossin(cos

'cos/)'cossin'sin)'''sin('sincos'sincos(cos

)'''sin(

'sin)'''cos(

'cos)'''cos(



θϕθϕθ


Using equation (5.3-2), the first approximate solution (ϕ’1, θ ’) is obtained. The first

approximate solution is input into equation (5.3-5),

25/72

∆∆∆∆−∆∆∆∆∆−+∆∆−∆∆∆∆∆−−∆∆+

=

∆+∆+∆+


'cos/)cos/sin'cos)'sinsin(sin'sincos'cossin(cos

'cos/)cos/sin'sin)'sinsin(sin'sincos'sincos(cos

)'''sin(

'sin)'''cos(

'cos)'''cos(

11

11

2

2


θϕθϕθ


Solving this equation, the second approximate solution (ϕ’2, θ ’) is obtained.

This iteration will be performed until the solution converges. If the mount fabrication errors

are about 1 degree, two iterations are enough.

26/72

Figure 5.3-3 True Telescope Coordinates

Figure 5.3-4 Apparent Telescope Coordinate

ϕ’

θ’ + ∆’’

«

R’’’ R’’

R, R’

XY

Z, T

S

S’, S’’

T’’, T’’’T’ ∆’

∆

ϕ’

θ’ + ∆’’

«

R’’’ R’’

R, R’

XY

Z, T

S

S’, S’’

T’’, T’’’T’ ∆’

∆

ϕ

θ

«

XY

Z

ϕ

θ

«

XY

Z

27/72

Figure 5.3-5 Apparent Telescope Coordinates with Mount Error

∆

∆’

θ’’ + ∆”

T’’’, T’’

R’’

S’’, S’

R’

T’T’

S’

S

T, Z

R, R’




∆

∆’

θ’’ + ∆”

T’’’, T’’

R’’

S’’, S’

R’

T’T’

S’

S

T, Z

R, R’

∆

∆’

θ’’ + ∆”

∆

∆’

θ’’ + ∆”

T’’’, T’’

R’’

S’’, S’

R’

T’T’

S’

S

T, Z

R, R’




28/72

5.3.5 Example Calculations

5.3.5.1 Apparent Telescope Coordinates à True Telescope Coordinates

Find true telescope coordinates from apparent telescope coordinates.

(1) Data

Mount errors are given as shown below.

∆ = 0.15o = 0.15 / 180 x π = 0.0026179939 (radian)

∆’ = -0.08o = -0.08 / 180 x π = -0.0013962634 (radian)

∆’’ = 0.2o = 0.2 / 180 x π = 0.0034906585 (radian)

Measured position (apparent telescope coordinates) of a celestial object, (θ’, ϕ’) is,

θ’ = 62.3000o = 62.3 / 180 x π = 1.08734012 (radian)

ϕ’ = 53.5000o = 53.5 / 180 x π = 0.93375115 (radian)

(2) Calculation

θ’ + ∆’’ = 1.08734012 + 0.0034906585 = 1.09083078 (radian)

Exact solution is obtained from equation (5.3-1),

=

−+−

−−−+

−

−+−−

−

=

88700327.0

36896762.0

27764743.0

)0013962634.0sin(0026197739.0sin

)0013962634.0cos(0026179939.0cos09083078.1sin

)0013962634.0cos(0026179939.0sin93375115.0cos09083078.1sin

93375115.0cos0026179939.0cos)0013962634.0sin(

)0013962634.0cos(93375115.0sin09083078.1cos

)0013962634.0cos(0026179939.0sin93375115.0sin09083078.1sin

93375115.0sin0026179939.0cos)0013962634.0sin(

)0013962634.0cos(93375115.0cos09083078.1cos

sin

sincos

coscos

θϕθϕθ

From equations (4-13) and (4-14),

29/72

32890702.127764743.036896762.0

tan ===LM

ϕ

ϕ = 0.92569835 (radian) = 53.0386o

sin θ = 0.88700327

θ = 1.09081440 (radian) = 62.4991o

Approximate solution is obtained from equation (5.3-4) as follows.

=

−−+

+−−

=

88701083.0

36896797.0

27764770.0

09083078.1sin

93375115.0cos09083078.1sin0026179939.0

93375115.0cos)0013962634.0(93375115.0sin09083078.1cos

93375115.0sin09083078.1sin0026179939.0

93375115.0sin)0013962634.0(93375115.0cos09083078.1cos

sin

sincos

coscos

θϕθϕθ


32890699.127764770.036896797.0

tan ===LM

ϕ

ϕ = 0.92569834 (radian) = 53.0386o

sin θ = 0.88701083

θ = 1.09083078 (radian) = 62.5000o

30/72

5.3.5.2 True Telescope Coordinates à Apparent Telescope Coordinates

Find apparent telescope coordinates from true telescope coordinates

(1) Data

Mount errors are the same as 5.3.4.2.

∆ = 0.15o = 0.15 / 180 x π = 0.0026179939 (radian)

∆’ = -0.08o = -0.08 / 180 x π = -0.0013962634 (radian)

∆’’ = 0.2o = 0.2 / 180 x π = 0.0034906585 (radian)

True telescope coordinates of a celestial object, (θ, ϕ) is,

ϕ = 0.92569835 (radian) = 53.0386o

θ = 1.09081440 (radian) = 62.4991o

(2) Calculation


=

−−−

−−+−−

−−−−+

=

∆+∆+∆+

88701083.0

37120378.0

27467653.0

)0013962634.0cos(0026179939.0cos/

))0013962634.0sin(0026179939.0sin09081440.1(sin

)0013962634.0cos(/))0013962634.0cos(0026179939.0sin92569835.0cos09081440.1sin

)0013962634.0sin(0026179939.0cos92569835.0cos92569835.0sin09081440.1(cos

)0013962634.0cos(/))0013962634.0cos(0026179939.0sin92569835.0sin09081440.1sin

)0013962634.0sin(0026179939.0cos92569835.0sin92569835.0cos09081440.1(cos

)'''sin(

'sin)'''cos(

'cos)'''cos(

θϕθϕθ


35142154.127467653.037120378.0

''

'tan ===LM

ϕ

ϕ ' = 0.93375083 (radian) = 53.5000o

sin (θ‘ + ∆'') = 0.88701083

31/72

θ' + ∆'' = 1.09083078 (radian) = 62.5000o

θ' = 62.5000o – 0.2o = 62.3000o

Approximate solution is obtained from equation (5.3-5).

=

+−−

−−+

=

∆+∆+∆+

88700327.0

37120343.0

27467625.0

09081440.1sin

92569835.0cos09081440.1sin)0026179939.0(

92569835.0cos)0013962634.0(92569835.0sin09081440.1cos

92569835.0sin09081440.1sin)0026179939.0(

92569835.0sin)0013962634.0(92569835.0cos09081440.1cos

)'''sin(

'sin)'''cos(

'cos)'''cos(

θϕθϕθ


35142165.127467625.037120343.0

''

'tan ===LM

ϕ

ϕ ' = 0.93375087 (radian) = 53.5000o

sin (θ‘ + ∆ '') = 0.88700327

θ' + ∆'' = 1.09081440 (radian) = 62.4991o

θ' = 62.5000o – 0.2o = 62.2991o

32/72

5.4 Equations for Pointing Telescope

5.4.1 Introduction

Using setting circles in telescope mount, you can point a telescope to a target object whose

equatorial coordinates is known. You don’t need align the telescope mount. You just have to

point your telescope to two reference stars and measure the setting circle readings of the

stars. Input the data to your computer, and the computer will create transformation

equations. After that, you just input equatorial coordinates of a target into the computer and

the computer will return the setting circle numbers for the target (ref. [6]).

The telescope coordinate system is defined as shown in figure 5.4-1. The position of a star

will be specified in horizontal angle, ϕ and elevation, θ. Note that the horizontal angle is

measured from right to left. This is the opposite direction to azimuth. The telescope is not

necessarily leveled or aligned with any directions. Equatorial mounts and altazimuth mounts

are the special cases. For equatorial mounts, ϕ corresponds to right ascension, α and θ

corresponds to declination, δ. For altazimuth mounts, ϕ corresponds to –(azimuth angle)

and θ corresponds to altitude.

Figure 5.4-1 Telescope Coordinates and Equatorial Coordinates

XY

Z«


θ: Elevation Angle

Xe (Vernal Equinox)

at t0

Ze (Celestial North Pole)

Ye at t0

XY

Z«


θ: Elevation Angle

Xe (Vernal Equinox)

at t0

Ze (Celestial North Pole)

Ye at t0

33/72

5.4.2 Transformation Matrix

The relationship between telescope coordinates and equatorial coordinates is derived in this

section.

Transformation from equatorial coordinates to telescope coordinates is expressed in matrix

form as follows.

[ ]

=

=

N

M

L

T

N

M

L

TTT

TTT

TTT

n

m

l

332331

232221

131211

.... Equation (5.4-1)

Transformation from telescope coordinates to equatorial coordinates is expressed as follows.

This is the inverse form of equation (5.4-1).

[ ]

=

−

n

m

l

T

N

M

L1 .... Equation (5.4-2)

Where,

=

θϕθϕθ

sin

sincos

coscos

n

m

l

.... Equation (5.4-3)

: Direction cosine of an object in telescope coordinate system

−−−−

=

δαδαδ

sin

)(sin(cos

)(cos(cos

0

0

ttk

ttk

N

M

L


: Direction cosine of an object in equatorial coordinate system

[ ]T , [ ] 1−T : Transformation matrix and its inverse matrix

t : Time

t0 : Initial time

ϕ : Horizontal angle of an object

θ : Elevation angle of an object

34/72

α : Right Ascension of an object

δ : Declination of an object

k = 1.002737908

5.4.3 Derivation of Transformation Matrix

Suppose that data set of equatorial coordinates and telescope coordinates for three

reference stars are obtained as follows.

Equatorial Coordinates Telescope Coordinates Reference

Star

Observation

Time Right

Ascension Declination

Horizontal

Angle

Elevation

Angle

Star 1 t1 α1 δ1 ϕ1 θ1



Using the data above, direction cosine of each star is expressed in both telescope

coordinates and equatorial coordinates.

=

1

11

11

1

1

1

sin

sincos

coscos

θϕθϕθ

n

m

l

.... Equation (5.4-5)

−−−−

=

1

0111

0111

1

1

1

sin

))(sin(cos

))(cos(cos

δαδαδ

ttk

ttk

N

M

L


=

2

22

22

2

2

2

sin

sincos

coscos

θϕθϕθ

n

m

l


−−−−

=

2

0222

0222

2

2

2

sin

))(sin(cos

))(cos(cos

δαδαδ

ttk

ttk

N

M

L


35/72

=

3

33

33

3

3

3

sin

sincos

coscos

θϕθϕθ

n

m

l


−−−−

=

3

0333

0333

3

3

3

sin

))(sin(cos

))(cos(cos

δαδαδ

ttk

ttk

N

M

L


Relationship between telescope coordinates and equatorial coordinates are,

[ ]

=

1

1

1

1

1

1

N

M

L

T

n

m

l

[ ]

=

2

2

2

2

2

2

N

M

L

T

n

m

l

[ ]

=

3

3

3

3

3

3

N

M

L

T

n

m

l

Combining the three equations above, we obtain the following equation.

[ ]

=

321

321

321

321

321

321

NNN

MMM

LLL

T

nnn

mmm

lll

Multiplying

1

321

321

321

−

NNN

MMM

LLL

to the both side of the equation above, the transformation

matrix is derived as follows.

[ ]

1

321

321

321

321

321

321

−

=

NNN

MMM

LLL

nnn

mmm

lll

T …. Equation (5.4-11)

36/72

Although we use three reference stars in equation (5.4-11), two stars are enough. An

independent vector (direction cosine) is created from reference star 1 and star 2 using

vector product. The new direction cosines will replace direction cosines for reference star 3.

The definition of vector product is shown in figure 5.4-2 and equation (5.4-12).

−−−

=×=

2121

2121

2121

213

lmml

nlln

mnnm

OPOPOP …. Equation (5.4-12)

Where,

=

1

1

1

1

n

m

l

OP ,

=

2

2

2

2

n

m

l

OP

Figure 5.4-2 Vector Product

New direction cosines are created from the coordinates of the reference star 1 and the

reference star 2 using equation (5.4.12). Note that the vector products are divided by the

length of the vector because direction cosines should be unit length.

−−−

×−+−+−

=

2121

2121

2121

22121

22121

22121

3

3

3

)()()(

1

lmml

nlln

mnnm

lmmlnllnmnnmn

m

l


O P1

P2

P3

O P1

P2

P3

37/72

−−−

×−+−+−

=

2121

2121

2121

22121

22121

22121

3

3

3

)()()(

1

LMML

NLLN

MNNM

LMMLNLLNMNNMN

M

L


Use equations (5.4-13) and (5.4-14) in equation (5.4-11) instead of (5.4-9) and (5.4-10).


The following data was measured using my 12.5 inch Dobsonian with setting circles.

Calculate the transformation matrix from telescope coordinates and equatorial coordinates

assuming the mount does not have fabrication errors.

Equatorial Coordinates Telescope Coordinates Reference

Star

Observation

Time Right

Ascension Declination

Horizontal

Angle

Elevation

Angle

Initial Time

t0

= 21h00m00s

= 5.497787

-- -- -- --

Star 1:

α And

t1

= 21h27m56s

= 5.619669

α1

= 0h07m54s

= 0.034470

δ1

= 29.038o

= 0.506809

ϕ1

= 99.25o

= 1.732239

θ1 =

= 83.87o

= 1.463808

Star 2:

α Umi

t2

= 21h37m02s

= 5.659376

α2

= 2h21m45s

= 0.618501

δ2

= 89.222o

= 1.557218

ϕ2

= 310.98o

= 5.427625

θ2

= 35.04o

= 0.611563


−=

=

994282.0

105396.0

0171648.0

463808.1sin

732239.1sin463808.1cos

732239.1cos463808.1cos

1

1

1

n

m

l


38/72

−=

−−−−

=

485390.0

0766175.0

870934.0

506809.0sin

))497787.5619669.5(81.00273790034470.0sin(506809.0cos

))497787.5619669.5(81.00273790034470.0cos(506809.0cos

1

1

1

N

M

L


−=

=

574148.0

618107.0

536934.0

611563.0sin

427625.5sin611563.0cos

427625.5cos611563.0cos

2

2

2

n

m

l


=

−−−−

=

999908.0

00598490.0

0121877.0

557218.1sin

))497787.5659376.5(002737908.1618501.0sin(557218.1cos

))497787.5659376.5(002737908.1618501.0cos(557218.1cos

2

2

2

N

M

L


39/72

−=

×−−×−×−−×

−×−××

×−−−+

×−−×+

−×−×=

0529714.0

626379.0

777717.0

)536934.0(105396.0)618107.0()0171648.0(

574148.0)0171648.0(536934.0994282.0

)618107.0(994282.0574148.0105396.0

)536934.0105396.0)618107.0)(0171648.0((

)574148.0)0171648.0(536934.0994282.0(

))618107.0(994282.0574148.0105396.0(

1

2

2

2

3

3

3

n

m

l


−

−=

×−−××−××−×−

×

×−−×+

×−×+

×−×−=

00707598.0

995776.0

0915436.0

0121877.0)0766175.0(00598490.0870934.0

999908.0870934.00121877.0485390.0

00598490.0485390.0999908.0)0766175.0(

)0121877.0)0766175.0(00598490.0870934.0(

)999908.0870934.00121877.0485390.0(

)00598490.0485390.0999908.0)0766175.0((

1

2

2

2

3

3

3

N

M

L

The inverse matrix is,

40/72

−−−

−−=

−−

−=

−

00707598.0995776.00915436.0

006518.10582508.0555830.0

0133413.0105481.0146349.1

00707598.0999908.0485390.0

995776.000598490.00766175.0

0915436.00121877.0870934.01

321

321

321

NNN

MMM

LLL

From equation (5.4-11), we obtain the following transform matrix.

[ ]

−−−

−−=

−−−

−−

−−

−=

=

−

56425.0018686.082552.0

61911.067086.040704.0

54617.074134.038932.0

00707598.0995776.00915436.0

006518.10592508.0555830.0

0133413.0105481.0146349.1

0529714.0574148.0994282.0

626379.0618107.0105396.0

777717.0536934.00171648.0

1

321

321

321

321

321

321

NNN

MMM

LLL

nnn

mmm

lll

T

If you want to aim the telescope at β Cet (α = 0h43m07s, δ = -18.038o) at 21h52m12s, from

equation (5.4-1),

radiansmh 188132.007430 ==α

radian314822.0038.18 −=−= oδ

radiansmht 725553.5125221 ==


41/72

−−=

−−×−×−−×−×−

=

−−−−

=

309647.0

0382687.0

950081.0

)314822.0sin(

))497787.5725553.5(002737908.1188132.0sin()314822.0cos(

))497787.5725553.5(002737908.1188132.0cos()314822.0cos(

sin

)(sin(cos

)(cos(cos

0

0

δαδαδ

ttk

ttk

N

M

L


[ ]

−=

−−

−−−

−−=

=

610308.0

604099.0

510635.0

309647.0

0382687.0

950081.0

56425.0018686.082552.0

61911.067086.040704.0

54617.074134.038932.0

N

M

L

T

n

m

l

From equations (4-13) and (4-14), telescope coordinates are calculated as follows.

183035.1510635.0

604099.0tan −=

−=ϕ

o21.130272546.2 == radianϕ

610308.0sin =θ o61.37656449.0 == radianθ

This calculated telescope coordinates is very close to the measured telescope coordinates,

ϕ = 130.46o, θ = 37.67o.

42/72

5.5 Polar Axis Misalignment Determination

The matrix method is applied to derive the equations of the declination drift method for polar

axis misalignment determination in this section.

The declination drift method was proposed by Challis [5] to determine the polar axis

misalignment of equatorial mount. The advantage of the declination drift method is its

simplicity of the measurement.

5.5.1 Derivation of Equations

5.5.1.1 Relationship between Coordinate Systems

Following Coordinate systems are used. See figure 5.5-1.

Equatorial coordinate system is Xe’-Ye’-Ze’. Ze’-axis directs to the north pole. Ye’-axis is in the

horizontal plane and directs to the east. Xe’-axis is on the meridian. Polar coordinates of a

celestial object in the equatorial coordinate system is (-H, δ), where H is hour angle and δ is

declination.

Telescope coordinate system is X-Y-Z and its polar coordinates is (ξ, ζ).

Misalignment of the telescope polar axis Z from the celestial polar axis Ze’ is defined as

follows.

First, equatorial coordinate system Xe’-Ye’-Ze’ is rotated θ clockwise around Ze’-axis (polar

axis), then the new coordinate system is rotated γ clockwise around the new X-axis.

43/72

Figure 5.5-1 Misalignment of Telescope Polar Axis

Direction cosine of a star in equatorial coordinates is,

−−

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

e

e

e

… Equation (5.5.1-1)

Direction cosine of the same star in telescope coordinates is,

=

ςξςξς

sin

sincos

coscos

N

M

L


Relationship between the coordinate systems is derived as follows using equations shown

in section 4.2.

Yh , Ye’ (East)

Xh (South)

Zh (Zenith)

O

Ze’ (North Pole)

Xe’ (Meridian)

EquatorHorizon

Meridian

Z (Polar Axis of Telescope)

Y

θ (rotate clockwise around Ze’-axis)

X

uv

Yh , Ye’ (East)

Xh (South)

Zh (Zenith)

O

Ze’ (North Pole)

Xe’ (Meridian)

EquatorHorizon

Meridian

Z (Polar Axis of Telescope)

Y

θ (rotate clockwise around Ze’-axis)

X

uv

44/72

−

−=

−

−=

'

'

'

coscossinsinsin

sincoscossincos

0sincos

'

'

'

100

0cossin

0sincos

cossin0

sincos0

001

e

e

e

e

e

e

N

M

L

N

M

L

N

M

L

γθγθγγθγθγ

θθ

θθθθ

γγγγ


−−

−

−=

δδδ


θθ

ςξςξς

sin

)sin(cos

)cos(cos

coscossinsinsin

sincoscossincos

0sincos

sin

sincos

coscos

H

H


5.5.1.2 Basic Equations for Declination Drift Method

Two stars are selected for measurement. Point the telescope to the first star and drive the

equatorial mount around polar axis only. Measure the drift of the star in declination in a

certain time interval. Same measurement is done for the second star.

Assume that declination drifts of two stars are obtained from observation as shown in table

5.5.1-1. Atmospheric refraction is neglected in this section. Effect of atmospheric refraction

will be discussed in section 5.5.1.4.

Table 5.5.1-1 Observed Data

Position of Star Time

Star Right

Ascension, α

Declination,

δ Start End

Drift of Declination

(Refraction is

neglected.)

Star 1 α1 δ1 t1a t1b ζ1b - ζ1a

Star 2 α2 δ2 t2a t2b ζ2b - ζ2a

From equation (5.5.1-4),

δγδθγδθγς sincos)sin(coscossin)cos(cossinsinsin +−+−= HH


45/72

Using the data from star 1 in table 5.5.1-1,

111111 sincos)sin(coscossin)cos(cossinsinsin δγδθγδθγς +−+−= aaa HH


Where,

H1a: Hour angle of the first star at time t1a

H1b: Hour angle of the first star at time t1b

ζ1a: Declination in Telescope Coordinates at time t1a

δ1: Declination of the first star in Equatorial Coordinates

Assuming that γ is very small, equation (5.5.1-6) is expressed as,

))sin(cos)cos((sincossin

sin)sin(coscos)cos(cossinsin

1111

111111

aa

aaa

HH

HH

−+−+=+−+−≅

θθδγδδδθγδθγς


Considering the following relationship (assuming that ∆ is very small),

11111 cossinsincoscossin)sin( δδδδδ ∆+≅∆+∆=∆+

We get the following equation from equation (5.5.1-7).

)sin(cos)cos(sin

))sin(cos)cos((sin

111

1111

aa

aaa

HH

HH

−+−+=−+−+≅

θγθγδθθγδς


Same equation is derived for time t1b.

)sin(cos)cos(sin 1111 bbb HH −+−+≅ θγθγδς … Equation (5.5.1-9)

Retracting equation (5.5.1-8) from (5.5.1-9),

))sin()(sin(cos))cos()(cos(sin 111111 ababab HHHH −−−+−−−=− θγθγςς

… Equation (5.5.1-10)

Putting θγ sin=u , θγ cos=v in equation (5.5.1-10),

))sin()(sin())cos()(cos( 111111 ababab HHvHHu −−−+−−−=− ςς

… Equation (5.5.1-11)

Using the data of star 2 in table 5.5.1-1,

))sin()(sin())cos()(cos( 222222 ababab HHvHHu −−−+−−−=− ςς

… Equation (5.5.1-12)

Equations (5.5.1-11) and (5.5.1-12) are the basic equations of declination drift method.

46/72

When you obtain the data in table 5.5.1-1 from observation, you can calculate the polar axis

misalignment u and v from equations (5.5.1-11) and (5.5.1-12).

Two important comments can be made based on the equations.

(1) In order to maximize the accuracy of the method, it is desirable to take two stars which

locations at observation are nearly 90 degree apart.

(2) It is not necessary to select stars near equator. Stars far from celestial equator can work.

This conclusion is derived from the fact that the declinations of the stars do not appear in

equations (5.5.1-11) and (5.5.1-12).

5.5.1.3 Challis’ Method

Challis’ original declination drift method requires three measurements with one star. See

table 5.5.1-2 for the required data.

Table 5.5.1-2 Observed Data – Challis’ Method


Star Right

Ascension, α

Declination,

δ Start End


(Refraction is

neglected.)

ta tb ζb - ζa Star 1 α1 δ1

ta tc ζc - ζa

Based on equations (5.5.1-11) and (5.5.1-12),

))sin()(sin())cos()(cos( ababab HHvHHu −−−+−−−=− ςς

… Equation (5.5.1-13)

))sin()(sin())cos()(cos( acacac HHvHHu −−−+−−−=− ςς

… Equation (5.5.1-14)

47/72

5.5.1.4 Compensation of Atmospheric Refraction

Effect of atmospheric refraction is included in the measured data.



Star Right

Ascension, α

Declination,

δ Start End


(Refraction is

included.)

Star 1 α1 δ1 t1a t1b ζ1b’ - ζ1a’

Star 2 α2 δ2 t2a t2b ζ2b’ - ζ2a’

Altitude of the star at observed instant is necessary to calculate the atmospheric refraction.

Relationship between equatorial coordinates and altazimuth coordinates is (see section

5.1.1),

−=

'

'

'

sin0cos

010

cos0sin

e

e

e

h

h

h

N

M

L

N

M

L

φφ

φφ … Equation (5.5.1-15)

Where,

φ is observer’s latitude.

−−

=

h

Ah

Ah

N

M

L

h

h

h

sin

)sin(cos

)cos(cos

… Equation (5.5.1-16)

A is azimuth measured westward from the South and h is “airless” altitude.

δφδφ sincos)cos(cossin)cos(cos −−=−= HAhLh

)sin(cos)sin(cos HAhM h −=−= δ

δφδφ sinsin)cos(coscossin +−== HhN h

h

h

L

MA =− )tan( … Equation (5.5.1-17)

From equation (15.2) in ref. [2] (page 101), atmospheric refraction R is expressed as follows.

R is added to “airless” altitude h to obtain apparent altitude. Note that equation (5.5.1-18) is

valid for the altitude larger than 15 degree.

48/72

−

×−

−

×= hhR

2tan

/18036000824.0

2tan

/1803600276.58 3 π

ππ

π (in radian)

… Equation (5.5.1-18)

Using the following relationship,

h

h

N

N

hh

hh

h22 1

sinsin1

sincos

2tan

−=

−==

−π

… Equation (5.5.1-19)

Equation (5.5.1-18) is expressed as follows.

322 1

/18036000824.01

/1803600276.58

−

×−

−

×=

h

h

h

h

N

N

N

NR

ππ

… Equation (5.5.1-20)

Apparent altitude h0 is expressed as follows.

Rhh +=0 … Equation (5.5.1-21)

Refracted position of the star in altazimuth coordinates

'

'

'

h

h

h

N

M

L

is,

−+

=

+−−−−

≈

+−−−−−−

=

+−+−+

=

h

Ah

Ah

R

N

M

L

hRN

AhRM

AhRL

RhRh

ARhARh

ARhARh

Rh

ARh

ARh

N

M

L

h

h

h

h

h

h

h

h

h

cos

sinsin

cossin

cos

)sin(sin

)cos(sin

sincoscossin

)sin(sinsin)sin(coscos

)cos(sinsin)cos(coscos

)sin(

)sin()cos(

)cos()cos(

'

'

'

… Equation (5.5.1-22)

Then, refracted position of the star in equatorial coordinates

''

''

''

e

e

e

N

M

L

is,

49/72

+

+−+

=

−

−+

−=

−+

−=

−=

hAh

Ah

hAh

R

N

M

L

h

Ah

Ah

R

N

M

L

h

Ah

Ah

R

N

M

L

N

M

L

N

M

L

e

e

e

h

h

h

h

h

h

h

h

h

e

e

e

cossincossincos

sinsin

coscoscossinsin

'

'

'

cos

sinsin

cossin

sin0cos

010

cos0sin

sin0cos

010

cos0sin

cos

sinsin

cossin

sin0cos

010

cos0sin

'

'

'

sin0cos

010

cos0sin

''

''

''

φφ

φφ

φφ

φφ

φφ

φφ

φφ

φφ

φφ

φφ

… Equation (5.5.1-23)

Refracted position of the star in telescope coordinates

'

'

'

N

M

L

is derived from equations

(5.5.1-3) and (5.5.1-23).

+

+−+

−

−=

hAh

Ah

hAh

R

N

M

L

N

M

L

e

e

e

cossincossincos

sinsin

coscoscossinsin

'

'

'

coscossinsinsin

sincoscossincos

0sincos

'

'

'

φφ

φφ


θθ

… Equation (5.5.1-24)

Putting observed data of star 1 at t1b in table 5.5.1-3 to equation (5.5.1-23),

( )( )

( )bbbbb

bbbb

bbbbbbb

hRAhR

AhRH

hRAhRH

111111

11111

11111111

cossincossincossincos

sinsin)sin(coscossin

coscoscossinsin)cos(cossinsin'sin

φφδγδθγ

φφδθγς

++++−+

+−−=

Considering γ and R1b are very small,

50/72

11

1111111

11111

111111

coscos

cossincossincos)sin(cos)cos(sinsin

cossincossincos

sin)sin(coscos)cos(cossin'sin

δδ

φφθγθγδ

φφ

δδθγδθγς

++−+−+=

++

+−+−≅

bbbbbb

bbbbb

bbb

hAhRHH

hRAhR

HH

Considering the following relationship (assuming that ∆ is very small),

11111 cossinsincoscossin)sin( δδδδδ ∆+≅∆+∆=∆+

1

11111111 cos

cossincossincos)sin(cos)cos(sin'

δφφ

θγθγδς bbbbbbb

hAhRHH

++−+−+=

… Equation (5.5.1-25)

Same equation is derived for data of star 1 at t1a in table 5.5.1-3.

1

11111111 cos

cossincossincos)sin(cos)cos(sin'

δφφ

θγθγδς aaaaaaa

hAhRHH

++−+−+=

… Equation (5.5.1-26)

Retracting equation (5.5.1-25) from (5.5.1-26),

CBAab ++=− θγθγςς cossin'' 11 … Equation (5.5.1-27)

Where,

)cos()cos( 11 ab HHA −−−=

)sin()sin( 11 ab HHB −−−=

1

1111

1

1111

1

1111

1

1111

cos

cossintancos

cos

cossintancos

cos

cossincossincos

cos

cossincossincos

δφφ

δφφ

δφφ

δφφ

aaaha

bbbhb

aaaa

bbbb

hhLR

hhLR

hAhR

hAhRC

+−

+=

+−

+=

Same equation is derived for data of star 2.

FEDab ++=− θγθγςς cossin'' 22 … Equation (5.5.1-28)

Where,

)cos()cos( 22 ab HHD −−−=

)sin()sin( 22 ab HHE −−−=

51/72

2

2222

2

2222

2

2222

2

2222

coscossintancos

coscossintancos

coscossincossincos

coscossincossincos

δφφ

δφφ

δφφ

δφφ

aaaha

bbbhb

aaaa

bbbb

hhLR

hhLR

hAhR

hAhRF

+−

+=

+−

+=

Putting θγ sin=u , θγ cos=v in equations (5.5.1-23) and (5.5.1-24),

0)''( 11 =−−++ abCBvAu ςς … Equation (5.5.1-29)

0)''( 22 =−−++ abFEvDu ςς … Equation (5.5.1-30)

Equations (5.5.1-29), (5.5.1-30), (5.5.1-17) and (5.5.1-20) are the equations for declination

drift method with atmospheric refraction.

52/72

5.5.2 Example Calculations

5.5.2.1 Two Star Declination Drift Method with Atmospheric Refraction Neglected

(1) Observed Data

Observed data is shown in table 5.5.2-1.



Star Right

Ascension,

α

Declination,

δ Start End


α Boo α1 =

14h15m49s

δ1 =

19o10’29”

t1a =

2001 May 24,

21h00m00s

t1b =

2001 May 24,

21h50m00s

ζ1b - ζ1a = -34.52”

α Boo α2 =

14h15m49s

δ2 =

19o10’29”

t2a =

2001 May 24,

21h50m00s

t2b =

2001 May 24,

22h23m00s

ζ2b - ζ2a = -65.88”

Observation Location: Latitude, φ = 52o09’20”.32N, Longitude, L = 0o0’38”.36E

All the data is converted to radian.

Table 5.5.2-2 Observed Data (in radian)


Star Right

Ascension,

α

Declination,

δ Start End


α Boo α1 =

3.73420466

δ1 =

0.33466204

t1a =

2001 May 24,

5.49778714

t1b =

2001 May 24,

5.71595330

ζ1b - ζ1a =

-0.00016736

α Boo α2 =

3.73420466

δ2 =

0.33466204

t2a =

2001 May 24,

5.71595330

t2b =

2001 May 24,

5.85994296

ζ2b - ζ2a =

-0.00031940

Observation Location: Latitude, φ = 0.91028772, Longitude, L = -0.00018597

53/72

(2) Sidereal Time

Julian Day number JD corresponding to 2001 May 24, 0h UT is 2452053.5 (chapter 7 in ref.

[2]).

The sidereal time at Greenwich at 2001 May 24, 0h UT is (chapter 11 of ref. [2]),

50139219712.036525

0.24515455.245205336525

0.2451545=

−=

−=

JDT

smh

TTTUThat

95.38061621780288.4deg662304.241deg662304.60138710000

000387933.0770053608.3600046061837.10003

20

====

−++=θ

(3) Hour Angle

From chapter 12 of ref. [2], hour angles are calculated as follows.

αθ −−= LH 0

99662377.5

73420466.3)00018597.0(49778714.550027379093.121780288.41

=−−−×+=aH

21538725.6

73420466.3)00018597.0(71595330.550027379093.121780288.41

=−−−×+=bH

21538725.6

73420466.3)00018597.0(71595330.550027379093.121780288.42

=−−−×+=aH

35977114.6

73420466.3)00018597.0(85994296.550027379093.121780288.42

=−−−×+=bH

(4) Basic Equations


))99662377.5sin()21538725.6(sin(

))99662377.5cos()21538725.6(cos(00016736.0

−−−+−−−=−

v

u

vu 21490953.0038481147.000016736.0 −=− … Equation (5.5.2-1)


))21538725.6sin()35977114.6(sin(

))21538725.6cos()35977114.6(cos(00031940.0

−−−+−−−=−

v

u

54/72

vu 14425712.070006338536.000031940.0 −−=− … Equation (5.5.2-2)

Solving equations (5.5.2-1) and (5.5.2-2),

u = 0.007916 radian = 0.454o = 1633”

v = 0.002179 radian = 0.125o = 449”

55/72

5.5.2.2 Challis’ Method with Atmospheric Refraction Neglected

(1) Observed Data

Observed data is shown in table 5.5.2-3.



Star Right

Ascension,

α

Declination,

δ Start End


ta =

2001 May 24,

21h00m00s

tb =

2001 May 24,

21h50m00s

ζb - ζa = -34.52”

α Boo α1 =

14h15m49s

δ1 =

19o10’29” ta =

2001 May 24,

21h00m00s

tc =

2001 May 24,

22h50m00s

ζc - ζa = -100.40”





Star Right

Ascension,

α

Declination,

δ Start End


ta =

2001 May 24,

5.49778714

tb =

2001 May 24,

5.71595330

ζb - ζa =

-0.00016736

α Boo α1 =

3.73420466

δ1 =

0.33466204 ta =

2001 May 24,

5.49778714

tc =

2001 May 24,

5.85994296

ζc - ζa =

-0.00048675


(2) Sidereal Time


56/72

[2]).


50139219712.036525

0.24515455.245205336525

0.2451545=

−=

−=

JDT

smh

TTTUThat

95.38061621780288.4deg662304.241deg662304.60138710000

000387933.0770053608.3600046061837.10003

20

====

−++=θ

(3) Hour Angle


αθ −−= LH 0

99662377.5

73420466.3)00018597.0(49778714.550027379093.121780288.4

=−−−×+=aH

21538725.6

73420466.3)00018597.0(71595330.550027379093.121780288.4

=−−−×+=bH

35977114.6

73420466.3)00018597.0(85994296.550027379093.121780288.4

=−−−×+=cH

(4) Basic Equations


))99662377.5sin()21538725.6(sin(

))99662377.5cos()21538725.6(cos(00016736.0

−−−+−−−=−

v

u

vu 21490953.0038481147.000016736.0 −=− … Equation (5.5.2-3)


))99662377.5sin()35977114.6(sin(

))99662377.5cos()35977114.6(cos(00031940.0

−−−+−−−=−

v

u

vu 3591666.0037847293.000048675.0 −=− … Equation (5.5.2-4)

Solving equations (5.5.2-3) and (5.5.2-4),

u = 0.008051 radian = 0.447o = 1661”

v = 0.002204 radian = 0.125o = 455”

57/72

5.5.2.3 Two Star Drift Method with Atmospheric Refraction Compensated

(1) Observed Data

Same data as section 5.5.2.1 is used.



Star Right

Ascension,

α

Declination,

δ Start End


α Boo α1 =

14h15m49s

δ1 =

19o10’29”

t1a =

2001 May 24,

21h00m00s

t1b =

2001 May 24,

21h50m00s

ζ1b - ζ1a = -34.52”

α Boo α2 =

14h15m49s

δ2 =

19o10’29”

t2a =

2001 May 24,

21h50m00s

t2b =

2001 May 24,

22h23m00s

ζ2b - ζ2a = -65.88”





Star Right

Ascension,

α

Declination,

δ Start End


α Boo α1 =

3.73420466

δ1 =

0.33466204

t1a =

2001 May 24,

5.49778714

t1b =

2001 May 24,

5.71595330

ζ1b - ζ1a =

-0.00016736

α Boo α2 =

3.73420466

δ2 =

0.33466204

t2a =

2001 May 24,

5.71595330

t2b =

2001 May 24,

5.85994296

ζ2b - ζ2a =

-0.00031940


(2) Sidereal Time

58/72


[2]).


50139219712.036525

0.24515455.245205336525

0.2451545=

−=

−=

JDT

smh

TTTUThat

95.38061621780288.4deg662304.241deg662304.60138710000

000387933.0770053608.3600046061837.10003

20

====

−++=θ

(3) Hour Angle


αθ −−= LH 0

99662377.5

73420466.3)00018597.0(49778714.550027379093.121780288.41

=−−−×+=aH

21538725.6

73420466.3)00018597.0(71595330.550027379093.121780288.41

=−−−×+=bH

21538725.6

73420466.3)00018597.0(71595330.550027379093.121780288.42

=−−−×+=aH

35977114.6

73420466.3)00018597.0(85994296.550027379093.121780288.42

=−−−×+=bH

(4) Basic Equations

Atmospheric refraction is calculated using equations (5.5.1-17) and (5.5.1-20).

51394425.0

)33466204.0sin()91028772.0cos(

)99662377.5cos(*)33466204.0cos()91028772.0sin(1

=−

−=ahL

54264618.0

)33466204.0sin()91028772.0cos(

)21538725.6cos(*)33466204.0cos()91028772.0sin(1

=−

−=bhL

54264618.0

)33466204.0sin()91028772.0cos(

)21538725.6cos(*)33466204.0cos()91028772.0sin(2

=−

−=ahL

59/72

54217341.0

)33466204.0sin()91028772.0cos(

)35977114.6cos(*)33466204.0cos()91028772.0sin(2

=−

−=bhL

81522146.0

3366204.0sin91028773.0sin)99662377.5cos(33466204.0cos91028772.0cos1

=+−=ahN

83752057.0


=+−=bhN

83752057.0


=+−=ahN

83715326.0


=+−=bhN

95311148.081522146.0sinsin 11

11 === −−

aha Nh

99272961.083752057.0sinsin 11

11 === −−

bhb Nh

99272961.083752057.0sinsin 12

12 === −−

aha Nh

99205772.083715326.0sinsin 12

12 === −−

bhb Nh

"37.4100020057.0

81522146.081522146.01

/18036000824.0

81522146.081522146.01

/1803600276.58

322

1

==

−×

−−

×=

radian

R a ππ

"00.3800018421.0

83752057.083752057.01

/18036000824.0

83752057.083752057.01

/1803600276.58

322

1

==

−×

−−

×=

radian

R b ππ

"00.3800018421.0

83752057.083752057.01

/18036000824.0

83752057.083752057.01

/1803600276.58

322

2

==

−×

−−

×=

radian

R a ππ

60/72

"05.3800018448.0

83715326.083715326.01

/18036000824.0

83715326.083715326.01

/1803600276.58

322

2

==

−×

−−

×=

radian

R b ππ


038481147.0)99662377.5cos()21538725.6cos()cos()cos( 11 =−−−=−−−= ab HHA

21490953.0)99662377.5sin()21538725.6sin()sin()sin( 11 −=−−−=−−−= ab HHB

00000769.000019137.000018368.0

33466204.0cos95311148.0cos91028772.0sin

95311148.0tan91028772.0cos51394425.0

00020057.0

33466204.0cos99272961.0cos91028772.0sin

99272961.0tan91028772.0cos54264618.0

00018421.0

coscossintancos

coscossintancos

1

1111

1

1111

−=−=

×+××

×−

×+××

×=

+−

+=

δφφ

δφφ aaah

abbbh

b

hhLR

hhLRC

70006338536.0

)21538725.6cos()35977114.6cos()cos()cos( 22

−=−−−=−−−= ab HHD

14425712.0

)21538725.6sin()35977114.6sin()sin()sin( 22

−=−−−=−−−= ab HHE

61/72

00000012.000018368.000018380.0

33466204.0cos99272961.0cos91028772.0sin

99272961.0tan91028772.0cos54264618.0

00018421.0

33466204.0cos99205772.0cos91028772.0sin

99205772.0tan91028772.0cos54217341.0

00018448.0

coscossintancos

coscossintancos

2

2222

2

2222

=−=

×+××

×−

×+××

×=

+−

+=

δφφ

δφφ aaah

abbbh

b

hhLR

hhLRF

From equations (5.5.1-29) and (5.5.1-30),

000016736.000000769.021490953.0038481147.0 =+−− vu

000031940.000000012.014425712.070006338536.0 =++−− vu

Solving the equations,

u = 0.008015 radian = 0.459o = 1653”

v = 0.002178 radian = 0.125o = 449”

Comparing these values with the result in section 5.5.2.1 (refraction neglected), effect of

atmospheric refraction is small in this example.

62/72

5.6 Dome Slit Synchronization

Computer control of telescope has become popular in amateur astronomy, and now the

advancement expands further. Dome slit control is an example of such advancement. In this

section, equations to perform dome slit control for a telescope on German equatorial mount

are derived. The equations are actually used by John Oliver of University of Florida to

develop his dome control software “DomeSync”.

The center of the telescope tube on German equatorial mount is offset from the center of a

dome as shown in figure 5.6-1. Because of this, the azimuth of the dome slit is not

coincident with the azimuth of the object which the telescope is aimed. The objective is to

develop equations of the azimuth of the dome slit. This problem is a very good example of

matrix method application.

Figure 5.6-1 shows definition of coordinate systems used in this section. Point O is the

center of the dome. Point P is the intersection of the polar axis and the declination axis of

the German equatorial mount. Point Q is the intersection of the telescope tube centerline

and the declination axis. Dimensions of the dome and the mount are also defined in the

figure.

Ydome (East)

Zdome(Zenith)

Object

Ze’ (North Pole)Xe’ (Meridian)

φ : Latitude

P

R

r

Dome

O

Ye’ (East)

Xdome (South)

Xdome0

Zdome0

Q

Azimuth

Figure 5.6-1 Definition of Coordinate Systems – Dome Slit Position

63/72

5.6.1 Object in First Quadrant

Equation when the object is in the first quadrant (180o < H < 270o) and telescope is in east

side of mount is derived in this section. Point S in figure 5.6.1-1 is on the dome surface.

Ydome(East)

Xdome (South)

Zdome(Zenith)

Object


φ : Latitude

H Ye’ (East)

P

QR

r

δ

S

Dome

O

Telescope is in east side of mount.

180o < H < 270o

Figure 5.6.1-1 Object in First Quadrant

From equation (5.1-2), unit vector QS

QS is,

−−

=

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

QS

QS

e

e

e

Vector PàQ in Xe’-Ye’-Ze’ coordinate is,

64/72

−−

=

=

0

)cos(

)sin(

'

'

'

ππ

H

H

r

Z

Y

X

PQ

e

e

e

From equation (5.1-4), vector OàS in Xdome-Ydome-Zdome coordinate is,

+++

=

−+−

−−+−

−−

−+−

−+−

−+−

−+

=

+

−−

+

−−

−

−−

−

−

=

=

FkC

EkB

DkA

HkHrZ

HkHr

HkHrX

Z

X

H

H

kHr

Hr

Z

Y

X

OS

dome

dome

dome

dome

dome

dome

dome

δπ

φδπ

φππ

φ

δπ

δπ

φδπ

φππ

φ

δδδ

ππ

πφ

πφ

πφ

πφ

sin2

cos)cos(cos2

sin)sin(2

sin

)sin(cos)cos(

sin2

sin)cos(cos2

cos)sin(2

cos

0

sin

)sin(cos

)cos(cos

0

)cos(

)sin(

2cos0

2sin

0102

sin02

cos

0

0

0

0


Where,

k: Constant

δπ

φδπ

φ

δ

δπ

φδπ

φ

ππ

φ

π

ππ

φ

sin2

cos)cos(cos2

sin

)sin(cos

sin2

sin)cos(cos2

cos

)sin(2

sin

)cos(

)sin(2

cos

0

0

−+−

−−=

−=

−+−

−=

−

−−=

−=

−

−+=

HF

HE

HD

HrZC

HrB

HrXA

dome

dome


65/72

5.6.2 Object in Second Quadrant

Object is in the second (270o < H < 360o) and telescope is in west side of mount.

Ydome(East)

Xdome (South)

Zdome(Zenith)

Object


φ : Latitude

H Ye’ (East)

P

Q

R

r

δS

Dome

O

Telescope is in west side of mount.

270o < H < 360o

Figure 5.6.2-1 Object in Second Quadrant


QS is,

−−

=

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

QS

QS

e

e

e


66/72

−−

−

=

=

02

3sin

23

cos

'

'

'π

π

H

H

r

Z

Y

X

PQ

e

e

e


+++

=

−+−

−−+

−

−−

−+

−−

−+−

−+

−

−+

=

+

−−

+

−−

−

−

−−

−

−

=

=

FkC

EkB

DkA

HkHrZ

HkHr

HkHrX

Z

X

H

H

kHr

Hr

Z

Y

X

OS

dome

dome

dome

dome

dome

dome

dome

δπ

φδπ

φππ

φ

δπ

δπ

φδπ

φππ

φ

δδδ

π

π

πφ

πφ

πφ

πφ

sin2

cos)cos(cos2

sin2

3cos

2sin

)sin(cos2

3sin

sin2

sin)cos(cos2

cos2

3cos

2cos

0

sin

)sin(cos

)cos(cos

02

3sin

23

cos

2cos0

2sin

0102

sin02

cos

0

0

0

0


Where,

k: Constant

67/72

δπ

φδπ

φ

δ

δπ

φδπ

φ

ππφ

π

ππφ

sin2

cos)cos(cos2

sin

)sin(cos

sin2

sin)cos(cos2

cos

23

cos2

sin

23

sin

23

cos2

cos

0

0

−+−

−−=

−=

−+−

−=

−

−−=

−−=

−

−+=

HF

HE

HD

HrZC

HrB

HrXA

dome

dome


5.6.3 Object in Third Quadrant

Object is in the third quadrant (0o < H < 90o) and telescope is in east side of mount.

Ydome(East)

Xdome (South)

Zdome(Zenith)

Object


φ : Latitude

H

Ye’ (East)

PQ

R

r

δ

SDome

O

Telescope is in east side of mount.

0o < H < 90o

Figure 5.6.3-1 Object in Third Quadrant

68/72


QS is,

−−

=

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

QS

QS

e

e

e


=

=

0

cos

sin

'

'

'

H

H

r

Z

Y

X

PQ

e

e

e


+++

=

−+−

−−+

−−

−+

−+−

−+

−+

=

+

−−

+

−

−−

−

−

=

=

FkC

EkB

DkA

HkHrZ

HkHr

HkHrX

Z

X

H

H

kHr

Hr

Z

Y

X

OS

dome

dome

dome

dome

dome

dome

dome

δπ

φδπ

φπ

φ

δ

δπ

φδπ

φπ

φ

δδδ

πφ

πφ

πφ

πφ

sin2

cos)cos(cos2

sinsin2

sin

)sin(coscos

sin2

sin)cos(cos2

cossin2

cos

0

sin

)sin(cos

)cos(cos

0

cos

sin

2cos0

2sin

0102

sin02

cos

0

0

0

0


Where,

k: Constant

69/72

δπ

φδπ

φ

δ

δπ

φδπ

φ

πφ

πφ

sin2

cos)cos(cos2

sin

)sin(cos

sin2

sin)cos(cos2

cos

sin2

sin

cos

sin2

cos

0

0

−+−

−−=

−=

−+−

−=

−−=

=

−+=

HF

HE

HD

HrZC

HrB

HrXA

dome

dome


5.6.4 Object in Fourth Quadrant

Object is in the fourth quadrant (90o < H < 180o) and telescope is in west side of mount.

Ydome(East)

Xdome (South)

Zdome(Zenith)

Object


φ : Latitude

H

Ye’ (East)

P

Q

R

r

δ

SDome

O

Telescope is in west side of mount.

90o < H < 180o

Figure 5.6.4-1 Object in Fourth Quadrant

70/72


QS is,

−−

=

=

δδδ

sin

)sin(cos

)cos(cos

'

'

'

H

H

N

M

L

QS

QS

e

e

e


−−

−

=

=

02

sin

2cos

'

'

'π

π

H

H

r

Z

Y

X

PQ

e

e

e


+++

=

−+−

−−+

−

−−

−+

−−

−+−

−+

−

−+

=

+

−−

+

−−

−

−

−−

−

−

=

=

FkC

EkB

DkA

HkHrZ

HkHr

HkHrX

Z

X

H

H

kHr

Hr

Z

Y

X

OS

dome

dome

dome

dome

dome

dome

dome

δπ

φδπ

φππ

φ

δπ

δπ

φδπ

φππ

φ

δδδ

π

π

πφ

πφ

πφ

πφ

sin2

cos)cos(cos2

sin2

cos2

sin

)sin(cos2

sin

sin2

sin)cos(cos2

cos2

cos2

cos

0

sin

)sin(cos

)cos(cos

02

sin

2cos

2cos0

2sin

0102

sin02

cos

0

0

0

0

71/72


Where,

k: Constant

δπ

φδπ

φ

δ

δπ

φδπ

φ

ππφ

π

ππφ

sin2

cos)cos(cos2

sin

)sin(cos

sin2

sin)cos(cos2

cos

2cos

2sin

2sin

2cos

2cos

0

0

−+−

−−=

−=

−+−

−=

−

−−=

−−=

−

−+=

HF

HE

HD

HrZC

HrB

HrXA

dome

dome


5.6.5 Intersection

Figure 5.6.5-1 shows vector expression of points on a straight line.

O

P

Q

S

Object

Points P, Q and S are on a straight line.

PQkOPPSOPOS

PQkPS

+=+=

=

Where k is a scalar.

PS

PQ

OP

OS

Figure 5.6.5-1 Vector Expression of Points on a Straight Line

72/72

Since point S is on the dome surface, length of vector OàS is R.

( ) ( ) ( ) 2222 RFkCEkBDkA =+++++

Then,

( ) ( ) ( ) 02 22222222 =−−−−+++++ CBARkCFBEADkFED

Solving this equation for k,

( ) ( ) ( )( )222

22222222

FED

CBARFEDCFBEADCFBEADk

++−−−++++++++−

=


+++

=

=

FkC

EkB

DkA

Z

Y

X

OS

dome

dome

dome


From equations (5.1-5) and (5.1-6), azimuth and elevation of dome slit are,

dome

domedome X

YA =− )tan( … Equation (5.6.5-3)

When Xdome >= 0, (-Adome) is in the first quadrant or the fourth quadrant.

When Xdome < 0, (-Adome) is in the second quadrant or the third quadrant.

Azimuth, Adome is measured from south to westward.

RZ

h domedome =sin … Equation (5.6.5-4)

-π/2 (-90o) <= hdome <= +π/2 (+90o)

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