Revision of Chapter 5 of Sundials by René R.J. Rohr, New York 1996declining inclined dials

part D Declining and Inclined Dials by Mathematicsusing a new figure

by: Armyan Fennerwick, the Netherlands

In 1965 appeared René R.J. Rohr's standard volume on Gnomonics, in French: Les Cadrans Solaires; in 1970, in English: Sundials; in 1982, in German: Die Sonnenuhr; and in 1989, in Italian: Meridiane. Since it was completely sold out, Dover Publications luckily issued a reprint of the English volume in 1996.

In the mid-eighties, I learned to know the German version and was subsequently intrigued by the chapter on spherical trigonometry. Closer study (as encouraged by Mr. Rohr: "[...] it may lead to further research." page 77) showed that the derived formulae resulted in incorrect data. I couldn't locate the error(s). The derivations I recalculated showed no errors. A late friend, a mathematician, gave me two important hints:

1. keep all the angles in the figure smaller than 90°
2. choose an algebraic direction and strictly stick to it

In light of the first tip I tried to adapt Mr. Rohr's figure but I had to make a new one.

In light of the second I chose a coordinate system with two axes. Spherical trigonometry deals with a plane, be it a curved one, so two axes suffice. (This implies that the method can only be applied to pole style sundials which are not influenced by the declination of the sun.)

For the axes I chose the horizon and the meridian; for the origin I chose the northern point of intersection. This resulted in a positive latitude, smaller than 90°. On the horizon I took the direction of the sun to be positive from N to E, resulting in a positive value for time. The inclination of the dial plane is the angle with the horizon plane, the declination is the angle in the horizon plane between the meridian plane and the plane of the angle of inclination.

I stuck to the Mr. Rohr's letters and other symbols as much as possible, both in the drawing and in the text.

The formulae derived in this way yield correct results. Mr. Rohr's original text follows; where necessary it has been changed with regard to the new figure.

Some minor corrections. The chapter in the book contains a few printing errors:

• page 77, righthand column, line 5: "angle in O" should read "angle in R";
• page 78, line 1: "cosine formula" should read "cotan formula";
• page 78, line 7 from bottom; "S'M" should read "SM";
• page 79, line 4 from bottom: "tan d sin / φ" should read "tan d / sin φ"; and in figure 75 "angle HA" should be: "180° HA".

The Calculation of the Angles between the Hour-lines and the Noon Line(recalculation with a new figure of: SUNDIALS by René R.J. Rohr, New York 1996, page 77 a.f.)

Let HMPZ be the celestial sphere (figure 75), C being its centre and also the centre of a declining and inclined dial; the great circle of diameter M'M is the plane of the dial. HH' is the plane of the horizon, CP the line of the pole and also the style of the dial; and CZ the vertical at the locality. If CN is the normal to the plane of the table MO, then the angle ZN equals the inclination i of the dial. If the vertical plane H'ZN goes through ZN, the angle NZM' is the declination of the dial. If the shadow of the style falls on a point O of the dial, PO is the hour-circle of the sun and the angle MPO is the hour-angle HA of the latter. Let MC be the noon line of the dial.

The shadow CO makes an angle OCM with the noon line. This angle OCM = ZZ is determined by the arc MO, the elevation of which is the goal of all our calculations. The great circle H'ZN divides it in two parts v and w, giving Z = v + w. The intersection R of the plane of the dial with the vertical plane H'ZN is a right angle since the latter contains the vertical CN to the plane MO of the table. An examination of the figure shows that v is the angle between the line of noon and the line CR of the plane MO which is its line of greatest slope, and w is the angle between the hour-line corresponding to the hour-angle HA of the sun and the same line of greatest slope.

Let A be the intersection of the vertical H'ZN with the hour-circle PO. In the right-angled triangle MRZ, we have:

tan v = tan d.sin RZ

Since RZ = 90° – i, this becomes
tan v = tan d.cos i
The formula confirms the obvious fact that v is independent of time. In the triangle ARO, which has a right angle in R, we have:
tan w = sin AR.tan μ,
if we call μ the angle PAZ. We note that, in the triangle PZA
(1)
cos μ = cos d.cos HA + sin d.sin HA.sin φ,
where φ stands for the latitude PCH, the arc PZ then being equal to 90° – φ. Also
ZA + AR = 90° – i
AR = 90° – i – ZA
and therefore
tan w = tan μ.cos (i + ZA)
We can thus write
(2)
tan w =
cos i.cos ZA.sin μ – sin i.sin ZA.sin μ
cos μ
But in triangle APZ, we have
sin μ
cos φ
=
cos HA
sin AZ
i.e.
(3)
sin AZ.sin μ = cos φ.cos HA
Multiplying this equation by cot AZ, we obtain
(4)
cos AZ.sin μ = cos φ.cos HA.cot AZ
If we replace cos AZ.sin μ and sin AZ.sin μ in (2) by their values in (3) and (4), we have
(5)
tan w =
cos i.cos φ.cos HA.cot AZ – sin i.cos φ.cos HA
cos μ
To obtain cot AZ, we apply the cotan formula to the triangle PZA:
cos d.sin φ = –cot ZA.cos φ + sin d.cot HA
which yields
(6)
cot ZA = cos d.tan φ –
sin d.cot HA
cos φ
Substituting this value in (5), after some simplifications we obtain
tan w =
cos i.sin φ.cos d – cos i.cot HA.sin d – sin i.cos φ
cos μ/sin HA
We again use (1), which gives
cos μ/sin HA = cos d.cot HA + sin d.sin φ.
so that
tan w =
cos i.cos d.sin φ – sin i.cos φ – cos i.sin d.cot HA
cos d.cot HA + sin d.sin φ
We now have explicit values for tan v and tan w, from which we can infer v and w, as well as
Z = v + w.
Despite appearances, this calculation is simple since v is calculated once and for all, while for w there are only two factors containing the variable term.

Calculation of the Angle between the Noon Line and the Line of Greatest Slope.

We recall the first formula obtained above
tan v = tan d.cos i
which gives the angle v between the line of greatest slope CR and the line of noon where d and i are arbitrary. This formula is of fundamental importance for the calculation of the elements of a declining and inclined dial because it determines the position of the noon line which serves as a reference co-ordinate for the sketching of the other hour-lines.

Calculation of the Angle between the Style and the Substyle

In order to avoid confusion, we recall that the axis PC is not perpendicular to the plane MOC in our drawing; the plane MOC should not be confused with the plane of the equator which is not shown.
The plane perpendicular to the plane of the dial MOC, containing PC and CN, the normal to that plane, cuts it along CS the substyle.
CS makes an angle with the noon line which is measured by the arc SM and which we call β. The style PC makes an angle α with the substyle, which has the arc PS for measure NS, by definition, is a right angle.
PN is the complement of SP = α. In the triangle PNZ we have
cos PN = cos(90° – φ).cos i + sin(90° – φ).sin i.cos(180° – d)
in other words
(7)
sin α = sin φ.cos i – cos φ.sin i.cos d

Calculation of the Angle between the the Substyle and the Noon-line

Let us look again at the triangle ZPN. We note that
cot [angle]ZPN =
cos φ.cot i + sin φ.cos d
sin d
This may be written as
tan [angle]ZPN = tan [angle]SPM =
sin d
cos φ.cot i + sin φ.cos d
If, as before, we call β the angle SM, in the right angled triangle SPM we have
tan β
sin α
= tan ∠[angle]SPM
or
(7a)
tan β
sin α
=
sin d
cos φ.cot i + sin φ.cos d
For sin α we take the value found in (7) and therefore,
tan β = sin i.cos d.
1 – cos d
cos φ. cot i + cos d

Calculation of the Hour-angle of the Sun at the Instant when its Shadow falls on the Substyle

The shadow falls on the substyle at the instant when the hour-angle of the sun equals MPS. We call this angle γ. We then have
tan γ =
tan β
sin α
or, after some simplifications,
tan γ =
sin i.sin d
cos i.cos φ + sin φ.cos d.sin i
or
|
sin d
cot i.cos φ + sin φ.cos d

Arrangement of the Hour-lines with Respect to the Substyle

If we reckon time from the substyle (which would be highly impractical), the angle of the desired hour-line would be OPS = λ. The shadow would make an angle OCS = δ whith the substyle, and in the triangle OSP (which has a right angle in S).
tan λ = sin α.tan δ
Since α is independent of time, the formula expresses the fact that the hour-lines are arranged symmetrically with respect to the substyle.

Particularization to some Special Cases

The corresponding formulae pertaining to the particular cases are deduced by assigning to d or i their appropriate values. These are shown in the following table:

SPECIFIC VALUES OF i, d, α, β AND γ FOR SOME PARTICULAR TYPES OF DIALS(Move the mouse pointer over the symbols to see their meaning)

 i d Z α β γ Non-Declining Dials Horizontal 0° 0° tan Z = sin φ.tan HA φ 0° 0° Meridional vertical 90° 0° tan Z = cos φ.tan HA 90° – φ 0° 0° Septentrional vertical 90° 0° tan Z = cos φ.tan HA φ 0° 0° Equatorial 90° – φ 0° Z = HA 90° 0° 0° Polar φ 0° Z = 0° 0° 0° 0° Meridional inclined Arbitrary 0° tan Z = sin(φ – i).tan HA φ – i 0° 0° Declining Dials Vertical declining 90° Arbitrary (d) tan Z =  cos φ cos d.cot HA + sin d.sin φ sin α = cos d.cos φ tan β = sin d.cot φ tan γ =  tan d sin φ Oriental (E) 90° 90° 0° 0° 0° –90° Occidental (W) 90° –90° 0° 0° 0° 90° φ is the latitude and v, the angle between the noon line and the line of the greatest slope,is zero for all the dials mentioned exept for the occidental and the oriental dials