ISI Theodolite Calculations

ISI Theodolite Calculations Ken Tatebe July 11, 2006 After changing baselines the positions of the telescopes must be measured in order to allow the computer to calculate the position angle, spatial frequency, and proper delay for the various baselines. The locations of the telescopes are measured using a theodolite, which gives the positions of the telescopes in spherical coordinates with the origin at the theodolite. When the theodolite is pointed at each telescope encoders record the orientation, i.e. the θ k and φ k of the k th telescope s position. Additionally, a laser range finder records the distance to the telescope. The theodolite is leveled so that a measurement of zero altitude is along the horizon and, by convention, the north star is used as the zero point for the azimuthal angle. In order to input this information into the computer these numbers must be converted into a northsouth, east-west, up-down cartesian-coordinate system. This is done by first rotating the coordinate system of the theodolite so that the north star is not at zero azimuth, but at its proper location in azimuth. The altitude should need no modification since the theodolite is leveled. Following this the spherical coordinates of the theodolite must be converted into cartesian coordinates. A few notes on convention: The theodolite measures zero altitude when pointed at the horizon and increasing with higher altitudes. The azimuthal angle is set internally to zero when pointed at the north star and increases clockwise as seen from above. These conventions are different from those of celestial coordinates using right ascension and declination. They are also different than the convention in polar coordinates where positive rotations are counter-clockwise. 1 Calculations The problem is decomposed into calculations of the z-component and the x-y plane separately. By doing this only a one-dimensional problem in z and a two-dimensional problem in x & y need to be solved. Further, by using complex notation to describe the x-y plane the computation is made very simple. 1.1 The Z-Component Each telescope is some distance from the theodolite and lies along some slope relative to the theodolite s position. This angle is recorded in the theodolite as the altitude and will be represented here by the variable φ k with k serving as an index that runs over (1,2,3) representing each telescope. Simple trigonometry gives the z-coordinate of each telescope, relative to the theodolite, as z k = L k sin(φ k ) (1) where L k is the point-to-point distance from the theodolite to the telescope. 1.2 The X-Y Plane To compute values in the x-y plane we must first find R k, the radial distance to the k th telescope in the x-y plane. This is given by R k = L k cos(φ k ) (2) Now we are able to do the remaining calculations in the x-y plane in terms of complex numbers, thus greatly simplifying the procedure. The position of a given telescope A k can be equivalently represented as A k = R k e iθ k = y k ix k (3) 1

where R k and θ k specify a location in polar coordinates and y k and x k specify the same location in cartesian coordinates. The variables x k and y k are switched, and the sign of x k reveresed, because the theodolite s coordinates are rotated and have an opposite sense of rotation from polar coordinates. The coordinates of the theodolite are refered to as a horizontal coordinate system as opposed to the polar coordinates of the complex plane. In horizontal coordinates the azimuthal angle θ k is measured from north rather than east. This means we must include a rotation of 90 (counter-clockwise, hence a positive rotation in the Polar Coordinates of the complex plane). This can be done by multiplying by e iπ/2. Also, since positive angles in horizontal coordinates are measured in a clockwise direction, as seen from above, we must reverse the sign of the angle θ k. In all, we may now express a position in the x-y plane in terms of a radius, R k and an angle, θ k, measured clockwise from north, with the following expression A k = R k e i( π 2 θ k) = x k + iy k (4) The real part of A k is the x-coordinate and the imaginary part is in the y-coordinate. One last complication is that the north star is not exactly north. Rather, its x-y coordinates are offset slightly from true north. To account for this we add the azimuth of the north star to our measured position of the telescope. 1 Thus, a location in the horizontal x-y plane in coordinates aligned to the cardinal directions is A k = R k e i[ π 2 (θ k+θ )] (5) and the x and y positions of each telescope are given by 1.3 Baseline Properties x k = Re[A k ] y k = Im[A k ] (6) Of course these are just the vectors from the theodolite to each telescope. We also want to know about the vectors pointing from one telescope to another. The vector pointing from the j th telescope to the k th telescope in the x-y plane, A jk, is found by subtracting the two vectors from the theodolite to each telescope. A jk = A j A k (7) = R j e i[ π 2 (θj+θ )] R k e i[ π 2 (θ k+θ )] (8) = e i( π 2 θ ) [R j e iθj R k e iθ k ] (9) (10) Thus, the horizontal distance between telescopes j and k is A jk. We can then combine this with the z-component derived above to determine the straight line distance between any two telescopes: L jk = L j L k (11) = A jk 2 + (z j z k ) 2 (12) = (x j x j ) 2 + (y j y k ) 2 + (z j z k ) 2 (13) 1 See Section 3.1 to determine the azimuth of a star given its right ascension and declination. 2

In addition, we may also care to know where the baseline points. That is, the φ jk and θ jk values of L jk. The altitude, φ jk, of the vector L jk is simply: ( ) zk z j φ jk = arctan (14) A jk and the azimuth, θ jk of L jk is simply the azimuth of A jk which is given by: [ ] Ajk θ jk = i ln A jk which is equivalent to taking the imaginary part of the log of A jk. 2 Summary For easy reference, the values of interest are given below in terms of the known quantities: θ : The calculated azimuth of the north star, clockwise from north, at the time of measurement. θ k : The change in azimuth, as measured by the theodolite, from the north star to the k th telescope. φ k : The measured altitude of the k th telescope. L k : The distance from theodolite to the k th telescope as measured by the laser range finder. They are: The three cartesian components describing the location of the k th telescope: (15) x k [ = Re (L k cos φ k )e i[ π 2 (θ k+θ ) (16) y k [ = Im (L k cos φ k )e i[ π 2 (θ k+θ ) (17) z k = L k sin φ k (18) The horizontal distance from the j th to the k th telescope: A jk = (x j x k ) 2 + (y j y k ) 2 (19) The straight-line (i.e. point-to-point) distance from the j th to the k th telescope: L jk = (x j x k ) 2 + (y j y k ) 2 + (z j z k ) 2 (20) The altitude of the vector pointing from the j th to the k th telescope: ( ) zk z j φ jk = arctan A jk The azimuth of the vector pointing from the j th to the k th telescope: [ ] Ajk θ jk = i ln A jk (21) (22) 3

3 Measurements Made Without Leveling The Theodolite By measuring the position of two stars, rather then just polaris as suggested above, leveling the theodolite is no longer necessary. Two stars can provide unambiguous orientation about the three axes of rotation. In practice, we will see that it is significantly easier to fix the orientation of the theodolite coordinates by measuring the position of three stars. To do this, we must first convert between right ascension and declination to altitude and azimuth (i.e. horizontal coordinates). 3.1 Converting Between Equitorial and Horizontal Coordinates The following calculations are generally complicated and difficult to derive. I will not attempt to do so here. What follows is mainly taken from Practical Astronomy With Your Calculator by Peter Duffett- Smith. It is an excellent book that has very brief, but understandable formulas and algorithms to convert between a variety of times and coordinate systems. The copy I used is borrowed from the Serendip group here at SSL. To convert between right ascension and declination to altitude and azimuth we must first determine the local sidereal time (no mean feat itself), and then use this in some complicated formulas to convert between coordinate systems. Please note: I do not explicitly mention when to convert units. Conventionally, right ascension is in hours, minutes, and seconds (with 24 hours being a full rotation) while declination is given in degrees, arcminutes, and arcseconds (standard segidecimeal notation). Be aware than an arcminute for degrees is not equal to a minute in hour angle! It is easiest to do the calculation in decimal degrees and decimal hours so as not to confuse the minutes and seconds. Don t forget to convert between hours and degrees though. 3.2 Converting from UT and Date to LST 3.2.1 Julian Date from Calendar Date The steps for the conversion from UT and the date to LST are outlined in Duffett-Smith s book. The first step is to find the Julian date corresponding to 0 h on the calendar date in question. To do this first define the variables y, m, and d as the current year, month, and day. In general, d can be a whole number plus some fraction indicating the time of day. Here, since we are computing for 0 h we will take only the integer portion of d. If m = 1 or m = 2 then y = y 1 m = m + 12 (23) Or, if m > 2 then y = y m = m (24) Also, if the date is later than or equal to 1582 October 15 (the beginning of the Gregorian calendar and the correction to the date of Easter) then ( ) ( ) y A A = Int B = 2 A + Int (25) 100 4 Otherwise B=0. The expression Int(x) indicates taking the integer part of x. 2 If y is negative then calculate 2 Thus, Int(3.98) = 3, and Int( 5.9) = 5. 4

C = Int((365.25 y ) 0.75) (26) Otherwise, C = Int(365.25 y ) (27) Finally, D = Int(30.6001 (m +1)), and the Julian date, JD, is given by: JD = B+C+D+d+1720994.5. Just as an aside: the difference between UT and UTC is kept smaller than 1 sec and has to do with keeping UTC coordinated with solar motions. GMST differs from GST in that it averages out corrections due to nutation. This is a small effect and should be no larger than 1.2 sec over the 18.6 yr nutation period. 3.2.2 From UT to GST Now that we have the Julian date, JD we can compute GST given UT. That is, we can convert from Universal Time, which is based on solar time, to Greenwich Sideral Time, which is based on the fixed stars. To convert from UT to GST we first calculate T = JD 2451545.0 36525.0 (28) Next, find T 0 = 6.697374558 + 2400.051336T + 0.000025862T 2 (29) Then reduce T 0 to a 0 24 range by taking T 0 = T 0 mod 24 and adding 24 if the result is negative to bring T 0 into the range between 0 and 24. Now, compute GST = T 0 + 1.002737909UT (30) where UT is the universal time in decimal hours. Again, take GST = GST mod 24 and add 24 if required to bring GST into the range of 0 24. This is GST in decimal hours. 3.2.3 From GST to LST Now, to compute the Local Sidereal Time (LST), we add the longitude, in hours, to the GST. To find the longitude in hours, take the longitude of your location and divide it by 15. If the longitude is W then subtract this from the GST, if it is E then add it to the GST. That is, LST = GST + θ 15 (31) where θ is the longitude of where the measurements are made. Again, take the result mod 24 to bring the answer into the range 0 24. This is the LST in decimal hours. We are now ready to convert from RA and Dec to Alt and Az. 5

3.3 The Formulae For the Conversion Using the declination, δ and right ascension, α (i.e. equitorial coordinates) at the time it s position was measured we may calculate its altitude, φ and azimuth, θ of a star., i.e. the direction to the star in Horizontal Coordinates. To convert from (δ, α) (φ, θ ) we use the following formulae sin(φ ) = sin(φ ) sin(δ) + cos(φ ) cos(δ) cos(h) (32) cos(θ ) = sin(δ) sin(φ ) sin(φ ) cos(φ ) cos(φ ) (33) where φ is the latitude of the theodolite s position on the Earth and H is the hour angle of the star. 3 The hour angle, H, is computed by H = LST α (34) Finally, one needs to remove the ambiguity inherent in inverse trigonometric functions as to which quadrant the angle lies. If the sin(h) < 0 then do nothing. If sin(h) > 0 then θ = 360 θ. 3.3.1 Compensating for Precession Also, if one wants to be thorough, one can account for the precession of the Earth over time, though the correction is fairly negligible for these purposes. All right ascension and declinations are assumed to be valid for the 2000 equinox. In this case we may precess the coordinates forward in time N years, up to about 50 years, using the following formulae: α 1 = α 0 + N(3.0742 + 1.33589 sin(α 0 ) tan(δ 0 )) (35) δ 1 = δ 0 + N(20.0383 cos(α 0 )) (36) where α 0 and δ 0 are the right ascension and declination for the 2000 equinox, and α 1 and δ 1 are the corrected values. 3.4 Solving for Theodolite to Horizontal Coordinates Transformation We will label the unit vector pointing to each star, as measured by using a theodolite oriented in an arbitrary direction, as û 1, û 2 and û 3. Of course, the theodolite will tell us the direction to the star in terms of some θ t and φ t where the subscript t indicates un-oriented theodolite measurements. These will need to be converted into cartesian coordinates. This is done with the following formulas x = r t sin(θ t ) cos(φ t ) (37) y = r t cos(θ t ) cos(φ t ) (38) z = r t sin(φ t ) (39) (40) Of course, r t equals 1 since we are dealing with unit vectors. These formulas are slightly different from the typical conversions from spherical coordinates since here a zero azimuth angle is due north and increases towards the east. Altitude is the angle made with the horizon. 3 The coordinates of the ISI at Mount Wilson are: φ =34.2249 θ = 118.0564 or Lat:34 13 29.5 N Long:118 3 22.0 W 6

We want to rotate the measured vectors, û 1, û 2 and û 3 into coordinates aligned with the cardinal directions: ê 1, ê 2 and ê 3. Fortunately all of these quantities are known. The û i vectors can be computed using the equations above along with the theodolite measurements. The ê i vectors are also computed using the equations above except that θ t and φ t are replaced by θ and φ which are described in Section 3.1. All that is needed to convert between one coordinate system and the other is a rotation, expressible as a 3 3 matrix, R. Thus ê i = Rû i (41) Since there are three such matrix equations, as i runs from 1 to 3 for each of the three stars, and the vectors are three-dimensional, we have a total of 9 equations and 9 unknown is the matrix R. The system is solvable so long as the vectors pointing to each star are not co-planar. Here we see that we have fully specified the matrix R in terms of linear equations. As mentioned above, only two stars are actually needed. Using two stars would give two 3-dimensional matrix equations. This combined with the fact that the matrix must be orthogonal, and have a determinant of unity would fully specify the matrix. The calculations, however, are very complex and the resulting formulas do not fit into one or even a few lines of text. If we define the matrix R as then we get all 9 equations: R = A B C D E F G H I (42) X i = Ax i + By i + Cz i (43) Y i = Dx i + Ey i + Fz i (44) Z i = Gx i + Hy i + Iz i (45) (46) where i runs from 1 to 3 and ê i = X i Y i Z i and û i = x i y i z i (47) is the x-component of ê i. From this we can construct the linear equation X = M R top (48) where R top is a vector containing the top row of the matrix R and X contains the x-components of all three ê i vectors. Expanding this out we get X 1 X 2 = x 1 y 1 z 1 x 2 y 2 z 2 A B (49) X 3 x 3 y 3 z 3 C All of the elements in M are known, as are the components of X. Thus 4, we simply invert the matrix M to find R t : 4 If the reader is curious to find the components of R t using only two stars they would need to solve the system of three equations consisting of the top two rows of M t and the equation A 2 + B 2 + C 2 = 1. The last equation can be derived from demanding that R be orthogonal, i.e. R T R = I. 7

M 1 X = Rtop (50) Similar equations can be constructed for the the other 6 elements, the middle R mid and bottom R btm rows of the matrix R: Y = M R mid (51) Z = M R btm (52) (53) and following the same procedure as outlined for the top row of R the entire matrix may be solved for. As a double check, the matrix R should be tested to confirm that it is orthogonal and has a unit determinant. The matrix R is over-specified for a rotation (which only has an axis and a rotation angle, i.e. 4 parameters instead of 9). This means that we have not strictly demanded that the determinant be unity or that the matrix R be orthogonal. As a result there may be small deviations in the matrix R from an exact rotation. 3.5 Transformation of Telescope Positions to Horizontal Coordinates Having fully constructed the matrix R we may now apply it to the vectors pointing to each telescope as measured by the theodolite, t k to find the vectors pointing to the telescopes in a coordinate system aligned with the cardinal directions, T k. T k = R t k (54) From here simple vector addition can be used to find the vectors that point from one telescope to another as outlined in the Summary above (Section 2). 4 The IDL Program All of this is done in the program theodolite.pro found in my home directory: /home/tatebe/. Since, as mentioned above, R is not required to be orthogonal, and hence does not exactly preserve the norm of vectors transformed by it, small errors will be introduced by the transform. These errors come from errors in the measured star positions and in roundoff errors in the computer s calculations. For this reason, as a double check, the determinant of the matrix R and the matrix R T R are output to confirm that they are unity and the identity matrix respectively. Typical errors using this method are on the order of 0.15%. All of the parameters, the star positions measured by the theodolite, the right ascension and declination of each star, the times of observation, etc. must all be input in the code of the program. The program is commented to indicate where the quantities should be input. 4.1 Suggested Stars to Use The Royal Stars are 4 good stars to use, in addition to Polaris would be Aldebaran (α Tau), Regulus (α Leo), Antares (α Sco), and Fomalhaut (α PsA). Of these Fomalhaut is probably the hardest to see, is it is about 30 south of the ecliptic. Since this star is generally overhead in the dead of winter, and few observations of any kind are made due to weather, this should not present much problem. In recent measurements at the ISI, the stars Polaris, Arcturus, and Dubhe were used. 8

5 References Practical Astronomy With Your Calculator, Duffett-Smith, Peter. Third Ed. Cambridge University Press 1990. ISBN: 0 521 35699 7 9