in magnitude and direction, and make it represent the resultant of All to north and to east, thus including part of the ship's force, namely (A-1)H to north and A to east, along with the earth's horizontal force in one circular dygogram, the residue of the horizontal component of the ship's force has also a circular dygogram. The construction thus obtained is fully described and illustrated by a diagram under the heading "Dygogram No. II., above. The proof of this is very simple. The following is the analytical problem of which it is the solution :-In the general equations (2) suppose Z to be constant, and put X'-(p, z)Z-P=X", Y'-(1, 2)Y-Q=Y". ... (12) We have } (13) X" =[1+(p, x)]X+(p, y)Y, Now imagine two dygogram curves (ellipses or circles) to be constructed as the locus of points (x, y,), (x', y') given by the equations X2+Y2=H2, x=X+(aX+ẞY); y=Y+(yX+dY); x'=X" — X−(aX+ßY); y'=Y" — Y−(yX+dY); (14) and let it be required to find a, ß, y, d so that these two curves may be circles; we have four equations for these four unknown quantities. Then, as x'+x=X", y'+y=Y", the resultant of the radius vectors of the two concentric circles thus obtained is the resultant of the constituent (X", Y") of the force on the compass; and by (12) we have only to shift the centre of one of them to the point whose coordinates are (p, z) Z+P, (q, z) Z+Q, to find two circles such that the resultant of corresponding radius vectors through the centre of one of them shall be the whole horizontal component of the force on the compass. Thus we have Smith's beautiful and most useful Dygogram of two Circles.-W. T., January 1874. |