<p>I am reading/studying the instrument description of Airy's Water Filled Telescope (Airy, G. B., &quot;History and Description of the Water Telescope of the Royal Observatory, Greenwich&quot;, Greenwich Observations in Astronomy, Magnetism and Meteorology made at the Royal Observatory, Series 2, vol. 33, pp.CXVII-PI). Needless to say this was a significant experiment in the history of Physics and the subsequent development of the Theory of Special Relativity.</p>
<p>I am unable to fully understand the calibration procedure Airy used. Any clarification which will help me understand the procedure used is greatly appreciated.  Here is the problem:</p>
<p>To calibrate the instrument, Airy measures the time required for the star <span class="math-container">$\gamma$</span>-Draconis to traverse 26 wires (25 gaps, total width one inch) placed in the focal plane of the telescope. The time interval measured was <span class="math-container">$9^m 39.97^s$</span>. Airy take half of the time measured (to find the half angle) and converts it to the corresponding polar angle <span class="math-container">$1^o 12^m 29.775^s$</span>. So far, so good.</p>
<p>In the next step, Airy in his own words, (page cxxx)</p>
<blockquote>
<p>The mean of the declinations of <span class="math-container">$\gamma$</span>-Draconis on July 23 and July 24 is <span class="math-container">$51^o 30^m 31.64^s$</span>. Hence the half-arc of great circle, joining the mean of the first five wires with the mean of the last five wires, expressed in seconds of arc, is equal to <span class="math-container">$$\frac{sin(1^o 12^m 29.775^s)*cos(51^o 30^m 31.64^s)}{sin(1^s)}$$</span></p>
</blockquote>
<p>Why is Airy projecting the measured angle to the great circle (equator)? Didn't the transit time measurement and the subsequent polar angle calculation gave him the focal length of the telescope which may be used to figure out the angles prospective stars are subtending? (Of course, for the measurement of aberration angle, Airy would have rotated the telescope by <span class="math-container">$90^o$</span>, an action which will not change the focal length of the telescope).</p>

CC BY-SA 4.0
2022-04-05T13:15:39.820Z
A guess: rotating the path $\gamma-Drac$ follows to extract the longitudinal component (if the wires were aligned along longitude lines) ?
Carl Witthoft
CC BY-SA 4.0
2022-04-05T23:35:43.377Z
I guess, the wires are aligned east-west so that the star goes from one wire to the next.  I understand any star make a rotation of $2 \pi$ radians in a sidereal day.  If Airy had an imaginary telescope which could have imaged the whole sidereal day traverse of $\gamma$-Draconis, it would have subtended $2 \pi$ radians. The telescope would have faithfully recorded this image. Of course, one can do this with a star close to the poles.
JKrsl
<coordinate><positional-astronomy><observatory><astrometry>
Airy's Water Filled Telescope: help required to fully understand the calibration procedure followed
JKrsl
# Airy's Water Filled Telescope: help required to fully understand the calibration procedure followed

(Asked by: username_0 on 2022-04-05)

I am reading/studying the instrument description of Airy's Water Filled Telescope (Airy, G. B., "History and Description of the Water Telescope of the Royal Observatory, Greenwich", Greenwich Observations in Astronomy, Magnetism and Meteorology made at the Royal Observatory, Series 2, vol. 33, pp.CXVII-PI). Needless to say this was a significant experiment in the history of Physics and the subsequent development of the Theory of Special Relativity.

I am unable to fully understand the calibration procedure Airy used. Any clarification which will help me understand the procedure used is greatly appreciated. Here is the problem:

To calibrate the instrument, Airy measures the time required for the star $\gamma$-Draconis to traverse 26 wires (25 gaps, total width one inch) placed in the focal plane of the telescope. The time interval measured was $9^m 39.97^s$. Airy take half of the time measured (to find the half angle) and converts it to the corresponding polar angle $1^o 12^m 29.775^s$. So far, so good.

In the next step, Airy in his own words, (page cxxx)

> The mean of the declinations of $\gamma$-Draconis on July 23 and July 24 is $51^o 30^m 31.64^s$. Hence the half-arc of great circle, joining the mean of the first five wires with the mean of the last five wires, expressed in seconds of arc, is equal to $$\frac{sin(1^o 12^m 29.775^s)\*cos(51^o 30^m 31.64^s)}{sin(1^s)}$$

Why is Airy projecting the measured angle to the great circle (equator)? Didn't the transit time measurement and the subsequent polar angle calculation gave him the focal length of the telescope which may be used to figure out the angles prospective stars are subtending? (Of course, for the measurement of aberration angle, Airy would have rotated the telescope by $90^o$, an action which will not change the focal length of the telescope).

**Tags:** coordinate, positional-astronomy, observatory, astrometry
> **Comments:**
> **username_1**: A guess: rotating the path $\gamma-Drac$ follows to extract the longitudinal component (if the wires were aligned along longitude lines) ?  
> **username_0**: I guess, the wires are aligned east-west so that the star goes from one wire to the next.  I understand any star make a rotation of $2 \pi$ radians in a sidereal day.  If Airy had an imaginary telescope which could have imaged the whole sidereal day traverse of $\gamma$-Draconis, it would have subtended $2 \pi$ radians. The telescope would have faithfully recorded this image. Of course, one can do this with a star close to the poles.  

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## 0 answers

