on the Planet Mars. — 259 
an arch EN of 60° 11’; in the triangle NEO, right-angled 
at E, there is alfo given the angle ENO, according to the fame 
author, 1° 51’, which is the inclination of the orbit of Mars 
to the ecliptic. Hence we find the angle EON 89° 5’, and 
fide ON 60° 12’. Again, when Mars is in the node of its 
orbit N, we have, by calculation from our principles, the angle 
PNE = 63° 7’, to which, adding the angle ENO=1° 51’, we 
have PNO = 64° 58’; from which two angles PON and PNO 
with the diftance ON, we obtain the inclination of the axis of 
Mars, and place of its node with refpect to that planet’s own 
orbit ; the inclination being 61° 18’, and the place of the node 
of the axis 58° 31’ preceding the interfection of the ecliptic 
with the orbit of Mars, or in our 19° 28’ of Pifces. 
. Being thus acquainted with what the inhabitants of Mars 
will call the obliquity of their ecliptic, and the fituation of 
their equinoctial and folftitial points, we are furnithed with the 
means of calculating the feafons on Mars; and may account, 
in a manner which I think highly probable, for the remarkable 
appearances about its polar regions. 
But firft it may not be improper to give an inftance how te 
refolve any query concerning the martial feafons.. Thus, let it 
be required to compute the declination of the Sun on Mars, 
June 25, 1781, at midnight of our time. If vy ¥ m9, &c. fig. 31. 
(tab. X.) reprefent the ecliptic of Mars, and + a+ vg the eclip- 
tic of our planet, Aa, 4B, the mutual interfection of the mar- 
tial and terreftrial ecliptics, then there is given the heliocentric 
longitude of Mars, vrm=9s. 10° 30’; then taking away fix 
figns, and = 4, or va=1s. 17° 58’, there remains 6m = 
1s. 22° 32’, From this arch, with the given inclination, 1° 51’, 
of the orbits to each other, we have cofine of inclination to 
radius, as tangent of dm to tangent of BM=1s, 22° 33’. And 
M m 2 taking 
