324 



PETER GUTHRIE TAIT 



image. But, when the elevation of the piece is greater than 45, the vertex of the 

 path lies in the tipper half of the ellipse, where the tangent leans back over the 

 gun ; and a small increase of elevation shortens the range. Two contiguous paths, 

 therefore, intersect one another again above the horizon. And, in the optical 

 problem, this corresponds to an inverted image. In symbols, if the eye be taken as 

 origin and the axis of x horizontal, there will be a direct image for a ray at whose 

 vertex dyjdx and x (in the curve of vertices) have the same sign, an inverted image 

 when the signs are different. 



Hence, whatever be the law of refractive index of the air, provided only it be 

 the same at the same distance from the earth's surface (i.e. the surfaces of equal 

 density parallel planes, and therefore the rays each symmetrical about a vertical 

 axis), all we have to do, in order to find the various possible images of an object 

 at the same level as the eye, is to draw the curve of vertices for all rays passing 

 through the eye, in tlte vertical plane containing the eye and the object, and find its 

 intersections with the vertical line midway between the eye and the object. As soon as 

 this simple idea occurred to me, I saw that it was the very kernel of the matter, 

 and that all the rest would be mere detail of calculation from particular hypotheses. 

 Each of the intersections in question is the vertex of a ray by which the object can 

 be seen, and the corresponding image will be erect or inverted, according as the 

 curve of vertices leans from or towards the eye at the intersection. Thus, in Fig. 2, 



Fig. 2. 



let E be the eye, and the dotted line the curve of vertices for all rays in the plane 

 of the paper, and passing through E. Let A be an object at the level of the eye, 

 A 1 A*A* the vertical line midway between E and A. Then A 1 , A*, A* are the 

 vertices of the various rays by which A can be seen. If we make the same con- 

 struction for a point B, near to A, we find that whereas the contiguous rays through 

 A 1 , B 1 and through A*, B 1 do not intersect, those through A', B 3 do intersect. At 

 A 1 and A* the curve of vertices leans from the eye, and we have erect images ; at 

 A* it leans back towards the eye, and we have an inverted image. And thus, if 

 this curve be continuous, the images will be alternately erect and inverted. The 

 sketch above is essentially the same as one given by Vince, only that he does not 



