垂直、偏向垂直 和地平日晷的小時線 Hour Lines for Vertical, Declining Vertical and Horizontal Sundials

假設有個赤道日晷面(F1)，其南面的小時線分布如Fig. 1，每小時15度，其晷針垂直於紙面。如果將此圖用平行於晷針的光投影至一平面( F2)，其形狀將視兩平面相互之間的角度而定。
Fig. 1 shows the hour lines on the face (F1) of an equatorial dundial. Those lines are separated equally by 15 degrees. The gnomon, which is not shown here, should be perpendicular to the screen. If some parallel light is cast through this diagram onto a plane (F2), the shape of the new diagram formed by the shadows those lines varies according to the angle between F1 and F2.
若F2垂直，F1單純向南傾斜一角度M，則F2上之投影結果必然左右寬度不變，上下長度變長。以各自晷心為原點，在F1上某一點P1(x1,y1)投至F2上成為P2(x2,y2)點；其中x2 = x1, y2 = y1/cos M。就是Fig. 1 中所有的點(線)相對於6-18線的距離都同時向下移動，新的距離是原來距離的1/cos M倍，以形成Fig. 2。
If F2 is kept vertical and F1 is simply declined due south with M degree, then the resultant new diagram must be longer in height but without any change in width. Taking each dial center as its own origin, point P1 (x1,y1) on F1 generates point P2(x2,y2) on F2 with x2 = x1, y2 = y1/cos M. That means all points in Fig. 1 move downwards, referred to 6-18 line, to their new distances, which equal to the old distances multiplied by 1/cos M, to yield the shadow diagram, such as Fig. 2.
Fig. 2 是個M = 40 的例子。因赤道日晷傾斜40度，所以此圖即是北緯40度適用的垂直日晷小時線圖。晷針必與投影光線平行，故與水平面成50 (= 90 - M)度。
Fig. 2 is a result of M = 40 degrees. It is also the hour line diagram for a vertical sundial face on latitude 40N. Its gnomon ought be in the direction of the casting light, so it has an angle of 50 ( = 90 - M) degrees to F2 in this example.

如果將垂直面F2逆時鐘旋轉D度，D在此稱為偏向角，形成偏向垂直面F3，此時F1上的線條會在F3面上產生新圖樣，如Fig. 3。其小時線的角度分配決定於緯度和偏向角。
If the vertical plane F2 is rotated D degrees counterclockwise, (this rotation is called declination), plane F3 results. Those shadow lines on F2 are now on F3 and may have the hour line pattern as shown in Fig. 3. The pattern depends on latitude and declination.


Fig. 4說明了P2和P3的關係，其中藍色線是投影光線，距水平面角度 φ。P1和P2的關係已如前述。 P1 和赤道日晷原點(晷心)的關係是:x1 = sin ha, y1 = cos ha, ha = 時角 = (12 - hour) × 15。故可得垂直偏向日晷的時角h12公式
x3 = x1 ÷ cos D
y3 = y2 - x2 × tan D × tan φ
= y1 ÷ cos φ - x1 × tan D × tan φ
即
tan h12 = x3 ÷ y3
= cot φ ÷ {cot ha × cos D ÷ sin φ - sin D}


The mathematical relations between P2 and P3 are depicted if Fig. 4, where the blue line represents the cast light with an angle of φ from horizon. Those between P2 and P1 are given above. With x1 = sin ha, y1 = cos ha, and ha = hour angle for equatorial dial = (12 - hour) × 15, we have the formula for the hour angle h12 for declining vertical dial as
x3 = x1 ÷ cos D
y3 = y2 - x2 × tan D × tan φ
= y1 ÷ cos φ - x1 × tan D × tan φ
tan h12 = x3 ÷ y3 = cot φ ÷ {cot ha × cos D ÷ sin φ - sin D}

如果將垂直面F2向下轉成地平，形成平面F4，P1在此新平面F4上投至P4(x4,y4)。則 x4 = x1, y4 = y1 / sin φ，φ為緯度。將Fig. 1赤道日晷小時線各端點的(x1,y1)換成(x4,y4)，再由這些新的端點連線至晷心，即得地平日晷之小時線。
If the vertical plane is reclined so to give a horizontal plane F4, the shadow of P1 on F1 casts on F4 at P4(x4,y4) and x4 = x1, y4 = y1 / sin φ. That is
x4 = sin ha
y4 = cos ha / sin φ
Connect P4 points obtained with proper ha to the dial center with straight lines. These are the hour lines for horizontal dial at latitude φ.