RamBaut—Rotation-Period of “ Garnet”? Spot on Jupiter. 395 
and it is clear that the amount of displacement CC’ is half the 
defect AD. 
To calculate the amount of this correction, we remark that in 
fig. 4 (in which the circle 4D’B represents the equator of Jupiter, 
and JS and JE the directions in which the Sun and Earth 
respectively lie—cf. fig. 2), the arc BD’ measures the proportion 
of the illuminated hemisphere visible at the time, and since the 
Fig. 4. 
apparent disc is an orthogonal projection of the visible portion 
of the illuminated hemisphere of the planet, we see that JC” is 
the amount of the displacement of the centre due to phase, and 
this is equal to 34D. But AD=4AB (1-cos AJD’) = 4AB 
x (1-cos SJE). Also, the angle CJC’ being very small, we may 
take its arc CC’ as equal to its chord, i.e. CC’ = JC” =i AB 
x (1 -—cos SJH). Now the time of moving from C’ to C is the 
same fraction of the whole period that the are CC’ is of the 
whole circumference of the equator, or 
P 
the correction for phase = ;AB (1 — cos SJE) x 
7tAB 
Also, SJE is the angle we have already denoted by a, so that 
we have, finally, 
3 P 
the correction for phase =sin’ 3a x a (C.) 
It is obvious, too, that before opposition this correction will 
be positive, and after opposition, negative. 
