DIY History | Transcribe | Scholarship at Iowa | Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no.26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903 | Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no. 26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903, Page 51

Transcribe
Translate

Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no.26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903

Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no. 26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903, Page 51

obtained by a perfect observation with a perfect and perfectly adjusted instrument."

     The Effect of the Level Error
     Upon the Time of Transit.

     If the east end of the transit instrument's axis is too high the instrumental meridian will be to the west of the true meridian and the correction for level error will be negative. Let b[superscript]s[/superscript] be the level error. Let S be the position of a star on the true meridian and S' the position it will have on the instrumental meridian. We wish to find the number of seconds it will take the star to go from S to S'. This time is measured by the hour angle P or < Z P S'. In D S'PA we have ((sin P)/(sin A)) = ((sin S'A)/(sin PS'))
or ((sin P[superscript]s[/superscript])/(sin b[superscript]s[/superscript])) = ((sin h)/(sin co - δ)) , S'A being taken approximately equal to h. Now P[superscript]s[/superscript] and b[superscript]s[/superscript] being very small we may take ((sin P[superscript]s[/superscript])/(sin b[superscript]s[/superscript])) = ((P[superscript]s[/superscript])/(b[superscript]s[/superscript])) .
Then we have 
((P[superscript]s[/superscript])/(b[superscript]s[/superscript])) = ((sin (90° - φ + δ))/(sin (90° - δ))) = ((cos (φ - δ))/(cos δ)) ;

Saving...

obtained by a perfect observation with a perfect and perfectly adjusted instrument." The Effect of the Level Error Upon the Time of Transit. If the east end of the transit instrument's axis is too high the instrumental meridian will be to the west of the true meridian and the correction for level error will be negative. Let b[superscript]s[/superscript] be the level error. Let S be the position of a star on the true meridian and S' the position it will have on the instrumental meridian. We wish to find the number of seconds it will take the star to go from S to S'. This time is measured by the hour angle P or < Z P S'. In D S'PA we have ((sin P)/(sin A)) = ((sin S'A)/(sin PS')) or ((sin P[superscript]s[/superscript])/(sin b[superscript]s[/superscript])) = ((sin h)/(sin co - δ)) , S'A being taken approximately equal to h. Now P[superscript]s[/superscript] and b[superscript]s[/superscript] being very small we may take ((sin P[superscript]s[/superscript])/(sin b[superscript]s[/superscript])) = ((P[superscript]s[/superscript])/(b[superscript]s[/superscript])) . Then we have ((P[superscript]s[/superscript])/(b[superscript]s[/superscript])) = ((sin (90° - φ + δ))/(sin (90° - δ))) = ((cos (φ - δ))/(cos δ)) ;