390 
ON THE TRANSFORMATION OF UNICURSAL SURFACES. 
[523 
Hence comparing the two expressions in question we have 
2 (n — 4) b = ^ (n — 1) (n — 2) (n — 3) — 2b + q + Ы, 
that is, 
0 = ^ (n — 1) (n — 2) (n — 3) — 2 (n — 3) b + q + 4t, 
or, as I prefer to write it, 
0 = £ (n — 1) (n — 2) (n — 3) — (n — 3) b + \q 4- 2t; 
which agrees with a more general formula in my “ Memoir on the theory of Reciprocal 
Surfaces,” Phil. Trans, vol. CLix. (1869), [411], see p. 227, [Coll. Math. Papers, vol. vi. 
p. 356]. I consider any two residues, a &-thic residue and a Z-thic residue; to each 
intersection of these there corresponds an intersection of their projections: or the number 
of intersections of the two residues must be equal to that of the two projections. Now 
the projections being (as above) 
order (k — n + 4) n' — 3 passing (k — n + 4) r — 1 times through each point a r , 
„ (Z-%+4)%'-3 „ (J-w + 4)r-l 
the number of the intersections in question is 
= [(& — n + 4) v' — 3] [(Z — n + 4) n — 3] — £ [(& — n + 4) ?— 1] [(l — n + 4) r — 1] a,. + &>, 
where for a reason which will be afterwards explained I have added the term со: this is 
= (k — n + 4) (l - n + 4) (n' 2 — Sr 2 a r ) + (k + l — 2n + 8) (— 3n' + %rOr) + 9 — (2a r — со), 
viz. it is 
= (k — n + 4) (l — n + 4) n + (k + l — 2n + 8) (n — 6 — 20) + 9 — (Sa,. — со), 
viz. substituting for © its value, = — ^ (n — 2) (n — 3) + b, and reducing, the number is 
= kin — 2 (к + l) b — n z + 8?г 2 — 16n + 9 + 4 (n - 4) b — (2a r — со). 
But the surfaces n, k, l, having in common the curve b which is a nodal curve on n, 
besides intersect in 
kin — b(n + 2k+2l — 4t) + 2q + qt 
points (Salmon’s Geometry of three Dimensions, 2nd Ed. p. 283, except that in the 
formula as there given the singularity t is not taken account of); that is, the number 
of intersections of the two residues is 
= kin — 2 (k + l)b - (n — 4) b + 2q + 9t, 
which is equal to the number of intersections of the two projections ( J ): or comparing 
the numbers in question we have 
that is, 
— n 3 + 8?i 2 — 16n + 9+ 4 (n — 4) b — (Sa r — со) — — (n — 4) b + 2q + 9t, 
2q + 9i = 5 (n — 4) b — n 3 + 8n 2 — 16% + 9 — (2a r — g>). 
1 I remark that n + X being positive or not less than ji-3, two (n + X) thic residues meet in 
n (X + 4) (X + 6) — 12X - 39 — 4 (X + 4) - (2a r - w) points : in particular, two (n — 3)-thic residues meet in 
3n-3 - 49 - (2a r -w) points; and two (n - 2)-thic residues meet in 8n -15 - 89 - (2a r -w) points.