r r r - MatAppferrario/e/notesgeo2/images.pdf · r r r r r r r r r 45 r r r r r r r r r THE THEORY OF CONIC SECTIONS more represented-though, for the sake of sim- plicity, in horizontal

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r rrr r rr r rr 44 r rrr r rr r rr

r rrr r rr r r 45 r rrr r rr r rTHE THEORY OF CONIC SECTIONS

more represented-though, for the sake of sim-

plicity, in horizontal position. A line perpen- dicular to its plane at the center O must meet the point of the gnomon G. Hence CG is the extended gnomon. From figure 2 it follows that OCG = CGL = 6.

It is now easy to express the angle / = HGC

as function of a = COH. If r = OH is the radius of the parallel circle we have on the one hand (fig. 3)

CH = 2rsina CH = 2r sin-, 2

r Because CG = HG = we have on the

cos 6 '

other hand 2r ./

CH = s5sln- cos 6 2

FIG. 2.

of the ecliptic. Figure 2 represents the celestial

sphere with the shadow-casting point G of the

gnomon as center. ELW is the equator, RESWR' the horizon, RHCR' the daily orbit

of the sun, culminating at C. Hence CGL = 6 the declination, and CG the direction of the

gnomon. Assume the sun in H. The angle a,

measuring the distance from noon is given by a = COH, where O is the center of the parallel circle RHCR'. The length s of the shadow is

then given by s = tan ,, where / = HGC is the

angle between the ray HG and the direction of

the gnomon GC. All we have to do is to find / as function of a.

In figure 3 the parallel circle RHCR' is once

. C . a Thus sin = cos 6sin . Because s = tan /

2 2

we compute also tan . By a simple computa-

tion one finds iv

(1) tan =

2 1 - cos2 a sin2 -

This answers our question.

CONSEQUENCES

(A) Because cos 5 = cos (- 6) we obtain the

same shadow length for declinations symmetric to the equator, especially for 6 = e and 5 = - E. Our sundial shows equal shadow lengths for both

solstices.

(B) Because the geographical latitude has no

influence on 6, our sundial gives the same shadow

lengths for all localities on the earth.

(C) Because shadows are only cast on the

receiving plane when /3 < 90? we find for

161 = e the following limit for a. We substitute

in formula (1) the value tan = 1 and find 2

(2)

FIG. 3.

1 cos ao = 1 - o

cos" 2

Because 1/cos2 e > 1 we see that cos ao < 0 and therefore ao > 90?.

The angle c = ROC measures the half length of daylight for a solar declination 6. If c > ao the sundial does not operate for the interval

VOL. 92, NO. 3, 1948] 137

Crnos sn -

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r r r - MatAppferrario/e/notesgeo2/images.pdf · r r r r r r r r r 45 r r r r r r r r r THE THEORY OF CONIC SECTIONS more represented-though, for the sake of sim- plicity, in horizontal