Cor. In a given lat. the error in time is least when the sine of the azimuth is greatest, that is, when the azimuth is 90°, or when the body is in the prime vertical. Hence for finding the apparent time (art. 301) the observation should be made on or near the prime vertical.

Prop. XIV. To find when the part of the equation of time, which arises from the obliquity of the ecliptic to the equator, is a maximum.

If the sun moved equably in the ecliptic, the difference between its longitude (AL) (fig. 59) and the right ascension (AR) would be the equation of time. The longitude and right ascension are equal at the equinox and also at the solstice: and somewhere between, the difference is a maximum. It is evident the maximum will be when the difference ceases to increase, that is, when the increase of AL the increase of AR.

Draw the circle of declination Ir very near LR and meeting LR produced in the pole P.

Draw also lv a parallel of the equator. Then

Hence

Ll: lv: rad: sin ALR.

lv: Rr sin Pv rad. The sine Pv=cos LR nearly.

:

Therefore L: Rr:: cos2 LR : cos AX rad.*

Hence when the equation is a maximum,

cos

LR= cos AX rad

or cos 2 decl cos ob. ecl.

a See note, page 189.

Otherwise thus:-The difference of AR and AL is a max.

therefore, d. AL· d. AR=0.

Now tan AR = cos A. tan AL,

Prop. XV. To deduce the sun's change in declination near the solstice.

Let Oob. ecl.

S= sun's distance from the solstice in seconds,

change in declination.

By circular parts

sin ob. ecl X sin long = sin decl,

or, sin Ox cos S sin (O-a) sin O. cos acos O. sina, (261).

=

=

cos S1-S2 sin 21"; sin x = a. sin l", and because x is very small compared with S, cos a = 1.

Hence by substitutions

or x=

If D

S2 sin 2 1". sin Oxsin 1". cos 0,

S2 sin 1". tan 0,00000 1052 S2.

sun's distance from solstice in degrees, a" 13, 63 D2,

a will thus be had sufficiently exact for several days before and after the solstice, (see art. 3/76)

Prop. XVI. To deduce the change of altitude of the sun or a star, when near the meridian, in a given time.

Let p SP the polar distance of the object (fig. 60), c=ZP the co-latitude, and m+x=ZS the zenith distance, m being the zenith distance when on the meridian, and the change required.

(1).. By trig, cos (m + x) = cosp. cos c + sinp. sin c. cos P, (2)..cos (m + x) = cos m―sin x. sin m, because

a being supposed very small cos x = 1 nearly.

(3)..cos m=cos (p—c) = cosp. cos c + sinp. sin c.

Hence equating the second members of (1), (2), and substituting for cos m as in (3),

Where D and L = the declination and latitude respectively, and P= the time from the meridian reduced into space.

When L and D are nearly equal, and of the same kind, this expression can only be used at a very small distance of time from the meridian.

This problem is of much importance, when meridional altitudes are taken by the repeating circle.

Prop. XVII. To investigate nearly the quantity and law of atmospherical refraction.

Let LI (fig. 61) be a ray of light falling on the atmosphere at I, and refracted in the curvilineal course IS. The object appears to a spectator at S in the direction ST, a tangent to the curve, and VST is the apparent zenith distance.

The space in the figure between the concentric circles represents all the atmosphere, which has any effect on the ray of light, so that the light may be considered as passing out of a vacuum into this space.

If the surface of the earth were a plane, the different strata of air might be considered as parallel thereto, and by the principles

of optics, the refraction would be the same as would take place were the ray of light to pass from a vacuum into air of the same density as that at the surface. It is therefore evident that if we take into account the spherical form of the earth and atmosphere, the error resulting from the supposition of an uniform atmosphere will, necessarily, be very small compared with the change occasioned by considering the atmosphere spherical, provided that change be small.

:

Let m: 1 sin of incidence: sin of refraction, when a ray of light passes from a vacuum into air of the density of that at the surface of the earth. Suppose all the air contracted into an uniform atmosphere, then SI is a right line. Let HIL=¿, SIC=r. VSI=z, SC=a, the height of the uniform atmosphere =1, or CI=a+l.

Let ir+R, then R is the quantity of refraction, sin (r+R) = sini.

or because R is small, sin r + cos r. sin R= sin i,

or sinr+R. sin l′′. cos r = sini,

substituting in this equation for sinr and sini as above, also for

Taking 2=80°, l=5, and a 4000 miles, the second term (arising from the spherical figure of the atmosphere) = 10" nearly. If a were infinite, that is if the surface of the earth were a plane, this second term would vanish. Hence we may safely conclude, that as far as 80° zenith distance, the error arising from supposing the atmosphere of uniform density must be much less than 10", and that consequently the above expression gives the refraction as far as 80° from the zenith with sufficient accuracy. If we neglect the second term, the refraction will vary as the tangent of the zenith distance.

The exact experiments of MM. Biot and Arago have determined the value of m—1—, 0002946 when the barometer is at 29,93 in (Metre) and Far. Therm, at 32°. From their experiments and the law of expansion of air it may be inferred that

height of the barometer, and t that of Farenheit's thermometer. When t=50° and b = 29,60 inches, this expression gives

m-1

sin 1"

-= 57",82, a result independent on astronomical observations. The French tables of refraction, by Delambre, founded on astronomical observations, give

By upwards of 500 observations made by myself with the eight feet astronomical circle,