DIY History | Transcribe | Scholarship at Iowa | Theory of least squares applied to the problems arising in our observatory by Arthur George Smith, 1895 | Theory of Least Squares Applied to the Problems Arising in our Observatory by Arthur George Smith, 1895, Page 15

[[page#]]11[[/page#]]
In measuring 18.71 / 100 of a revolution of the azimuth screw the probable error is 0.063 of a graduation, then, as from the nature of the quantity under consideration, it may be treated as a linear function, the probable error of one revolution would be expressed by
.063 : [[?E?]] :: [[square root of:]] .1871 : 1
[[?E?]] = .063 / [[square root of:]] .1871 =
[[?+- 0.145?]] graduation
Therefore the value of one revolution of the azimuth screw = 5.345 +- 0.007 intervals of the transit reticule. One wire interval of the transit reticule
= 5[[superscript]]s[[/superscript]].422+-[[superscript]]s[[/superscript]] .005 [see page 29]
Then, [[? 5,345 ?]] intervals would
= 28[[superscript]]s[[/superscript]],979+-0[[superscript]]s[[/superscript]].02 = 7' - 14".685 +- 0".3 arc
This last probable error is found from formula
R[[superscript]][[?s?]][[/superscript]] = (((dX / dx) squared)(r squared)) + (((dX/d(x[[subscript]]1[[/subscript]])) squared)(r squared)) + in probable error for X in function expressed by X = [[integral sign]] (x x[[subscript]]1[[/subscript]] x[[subscript]]2[[/subscript]] . . . ) [Churnvent, Art. 22.]

[[page#]]11[[/page#]]
In measuring 18.71 / 100 of a revolution of the azimuth screw the probable error is 0.063 of a graduation, then, as from the nature of the quantity under consideration, it may be treated as a linear function, the probable error of one revolution would be expressed by
.063 : [[?E?]] :: [[square root of:]] .1871 : 1
[[?E?]] = .063 / [[square root of:]] .1871 =
[[?+- 0.145?]] graduation
Therefore the value of one revolution of the azimuth screw = 5.345 +- 0.007 intervals of the transit reticule. One wire interval of the transit reticule
= 5[[superscript]]s[[/superscript]].422+-[[superscript]]s[[/superscript]] .005 [see page 29]
Then, [[? 5,345 ?]] intervals would
= 28[[superscript]]s[[/superscript]],979+-0[[superscript]]s[[/superscript]].02 = 7' - 14".685 +- 0".3 arc
This last probable error is found from formula
R[[superscript]][[?s?]][[/superscript]] = (((dX / dx) squared)(r squared)) + (((dX/d(x[[subscript]]1[[/subscript]])) squared)(r squared)) + in probable error for X in function expressed by X = [[integral sign]] (x x[[subscript]]1[[/subscript]] x[[subscript]]2[[/subscript]] . . . ) [Churnvent, Art. 22.]