Cylindrical Slide Rules for Aviators and Navigators

Fig. 1 shows Rule I for computing the zenith distance for the “Marcq Saint-Hilaire” method, and consists of seven parts:

1.	The outer cylinder.	1-2,	the outer cylinder slides back and forth on the inner cylinder with a snug fit.
2.	The inner cylinder.
3.	The rotating collar.	3-4,	the rotating collar and the ferrule are fitted with about a turn and a half of a screw thread of the same pitch as that of the spiral scales on the two cylinders,
4.	The ferrule.
5.	The index arm.	5-6,	the index arm is attached to the rotating collar so that its left edge will be parallel to the elements of the cylinders so that it may be placed in a plane passing through the common axis of the two cylinders, in order that when the index arm is set on a reading on one cylinder the corresponding reading may be obtained from the other cylinder. The sliding marker, when properly set, shows the exact point on the scale on the outer cylinder from which the scale is read.
6.	The sliding marker.
7.	The handle.

The scales used on Rule I are shown in Fig. 2. The length of these scales, when developed into straight lines, must be proportional to the outer diameters of the outer and inner cylinders in order that when laid off spirally around the cylinders at the same pitch they may preserve the proper alignment.

Description of Scales A, B, C, D and E

A is a scale of log haversines with the corresponding hours and minutes.

B is a scale of natural haversines corresponding to the log haversines of equal angles on

scales A and C and may be used with either of the scales, A or C, to find the    angles in        hours

and minutes or degrees and minutes, corresponding to any given natural haversine, or to find the natural haversine corresponding to any given angle.

C is the same as scale A with the corresponding degrees and minutes.

D is a scale of log cosines with the corresponding degrees and minutes.

E is a scale of log secants with the corresponding degrees and minutes.

Scale A may be dispensed with if the hour angle be converted into degrees and minutes, thereby saving some space on the outer cylinder which would allow the scales to be longer₍

667

U. S. Naval Institute Proceedings

[Aug.

though with scale A the rule will be more convenient. This is a matter for the aviators to decide.

To Compute the Zenith Distance

Let L = latitude, d= declination, t = hour angle, ?= the usual auxiliary angle, and z=the computed zenith distance.

Forumula (1)

Referring to Fig. 2:

1. Set the index arm to the latitude on the inner cylinder by revolving the collar until the left hand edge of the index arm coincides exactly with the latitude on scale E.
2. Next by turning the outer cylinder, holding the handle in the right hand, set the left hand edge of the index arm to the hour angle on scale A and move the sliding marker to this point.
3. Set the index arm to the declination on scale D by rotating the collar and the marker will indicate on scale A the point opposite which on scale B will be found hav ? (natural haversine ?).
4. From scale B take the hav (L+d); add this to hav ? for hav z and opposite this number on scale B will be found z on scale C.

Example.—Latitude 40°N., hour angle 2 hrs. 40 min., declination 30°N., find the zenith distance.

Take the distance between 30° on scale D and 40° on scale E in a pair of dividers, transfer this distance to scale A by placing the right leg at 2 hrs. 40 min. on scale A and the left leg will indicate the nat hav ? = 0.0774 on scale B. As M. Z. D. = L —d=10° is not found on the portion of scale C given, take it from the tables = 0.0076, and nat hav z = 0.0850, from scale C we find z = 33°58'. The value computed by logarithms is 33°56'30". As the slide rule from which scales A, B, C, D and E were taken was only 30 inches long, it is readily seen that with the cylindrical rule the computation may be made with great precision.

Description of Scales F, G, and E

F is a scale of log sines with the corresponding hours and minutes.

G is a scale of log sines with the corresponding degrees and minutes.

H is a scale of log secants with the corresponding degrees and minutes.

As in the case of scale A, F may be dispensed with if it is desired to save space on the outer cylinder.

Formula (2)

in which Z= the azimuth of the observed body and h = its true altitude.

To Compute the Azimuth

Given the hour angle, declination and altitude.

1. Set the index arm to the declination, d, on scale H on the inner cylinder.
2. Turn the outer cylinder and set the hour angle, t, to the left hand edge of the index arm, scale F, outer cylinder, and move the marker to this point.
3. Revolve the collar and set the index arm to the true altitude, h, on scale H.
4. The marker will designate the point on scale F opposite which on scale G will be found the azimuth.
5. Name the azimuth as directed below.

Example.—In latitude 26°S., hour angle 3 hrs. 00 min, West, declination 15°N., true altitude 30°, find the azimuth.

Take the distance on scale H between 15° and 30° on a pair of dividers; transfer this to scale G by placing the left leg at 45° and the right leg will show the azimuth, 52°, on the same scale. By precept 4, below, this is marked N. 52° W.2'

To Identify an Unknown Star

Given the latitude, L, the true altitude, h, and the azimuth, Z, of an unknown star, to find its declination, d, and hour angle, t, and from the local sidereal time the right ascension of the star.

To Find the Star’s Declination

Let b be an auxiliary angle and r the polar distance.

1. Set the index arm to L on scale E.
2. Turn outer cylinder and set index arm to Z on scale C, move marker to this point.
3. Rotate collar and set index arm to h on scale D, and the marker shows the point on scale C opposite which on scale B read nat hav b. Find the nat hav (L~h) from scales C and B.
4. Nat hav b+nat hav (L~h) = nat hav r; find r from scales B and C.
5. When r< 90°, d has the same name as the latitude, and d =90° — r.
6. When r>90°, d has the contrary name to the latitude and d=r — 90°.

To Find the Star’s Hour Angle

1. Set the index arm to h on scale H.
2. Turn outer cylinder and set index arm to Z on scale G and move the marker to this point.
3. Rotate collar and set index arm to d on scale H, then the marker will show the point on scale G opposite which on scale F find the hour angle, t.

Apply the hour angle to the local sidereal time and find the right ascension of the star; thus we have the right ascension and the declination to determine the star observed.

Naming the Azimuth

1. Mark the azimuth E. or W. according to whether the observed body is east or west of the meridian of the observer.
2. When L = 0° give the azimuth the same name as the declination.
3. When the latitude and the declination are of the same name and the latitude is less than the declination, the azimuth has the same name.
4. When the latitude and the declination are of contrary names, the azimuth has the name of the declination.
5. When the latitude and the declination have the same name, and the declination is less than the latitude, the body will cross the prime vertical. When it is between the P. V. and the elevated pole, that is, when its hour angle is greater than the hour angle on the P.V., the azimuth will have the same name as the latitude. When the body is between the P. V. and the equator, that is, when its hour angle is less than t_v, the hour angle on the P. V., the azimuth will have the contrary name to the latitude.

The hour angle on the P. V. may be found with Plate I as shown below.

The reasons for the above rules will be seen from an examination of Fig. 3, which shows the celestial sphere projected on the plane of the horizon. M, M', M" and M'", are the positions of four bodies.

By naming the azimuths in this way we avoid the use of angles greater than 90°.

Example 1.—Z, = 0°; t = 4^h 30^m (west); d = 28° N.; h = 20°.

Find Z.                            Ans. N. 60° 12' W.

Example 2.—L = 19° N.; t = 4^h 18^m (east); d= 29° N.; h = 31°.

Find Z.                                                      Ans. N. 67° 4’ E.

Example 3.—L = 25° S.; t = 3^h 20^m (east); d = 28° N.; h = 19°.

Find Z.                                             Ans. N. 45° 38' E.

Example4.—L = 48° N.; t = 5^h 55^m (west); d = 28° N.; h = 21°.

Find Z.                                                        Ans. N. 71° W.

Here t_v = 4^h 6^m and as t>t_v, Z has the same name as L.

Example 5.—L = 45° S.; t = 3^h 10^m (east); d= 26° S.; h = 48°.

Find Z.                                                        Ans. N. 83° E.

Here t_v = 4^h 3“ and, as t<t_v, Z has the contrary name.

TABLE I—LOG SECANTS AND COSECANTS

		Enter at top for Secants							Minutes for Log Secants							Diffs. for minutes—Add
Sec.	O' .	10'	20'	30'	40'	50'		Sec.	0'	10'	20'	30'	40'	50'		1'	2'	3'	4'	5'	6'	7'	8'	9'
0	0.0000	0000	0000	0000	0000	0000	89°	1°	0.0001	0001	0001	0002	0002	0002	88°
2	0.0003	0003	0004	0004	0005	0006	87	3	0.0006	0007	0007	0008	0009	0010	86							i	i	i
4	0.0011	0012	0012	0013	0014	0016	85	5	0.0017	0018	7001	0020	0021	0023	84						i	l	1	1
6	0.0024	0025	0027	0028	0030	0031	83	7	0.0033	0034	0036	0037	0039	0041	82				i	i	1	l	1	1
8	0.0042	0044	0046	0048	0050	0052	81	9	0.0054	0056	0058	0060	0062	0064	80			1	1	1	1	l	2	2
10	0.0066	0069	0071	0073	0076	0078	79	11	0.0080	0083	0085	0088	0091	0093	78			1	l	1	1	l	2	2
12	0.0096	0099	0101	0104	0107	0110	77	13	0.0113	0116	0119	0122	0125	0128	76		i	1	l	1	2	2	2	3
14	0.0131	0134	0137	0141	0144	0147	75	15	0.0151	0154	0157	0161	0164	0168	74		i	1	l	2	2	2	3	3
16	0.0172	0175	0179	0183	0186	0190	73	17	0.0194	0198	0202	0206	0210	0214	72		i	1	l	2	2	2	3	3
18	0.0218	0222	0226	0230	0235	0239	71	19	0.0243	0248	0252	0256	0261	0266	70		i	1	2	2	2	3	3	3
20	0.0270	0275	0279	0284	0289	0294	69	21	0.0298	0303	0308	0313	0318	0323	68		i	1	2	2	3	3	4	4
22	0.0328	0333	0339	0344	0349	0354	67	23	0.0360	0365	0371	0376	0381	0387	66	i	i	2	2	3	3	4	4	5
24	0.0393	0398	0404	0410	0416	0421	65	25	0.0427	0433	0439	0445	0451	0457	64	l	i	2	2	3	3	4	5	5
26	0.0463	0470	0476	0482	0488	0495	63	27	0.0501	0508	0514	0521	0527	0534	62	l	i	2	3	3	4	4	5	6
28	0.0541	0547	0554	0561	0568	0575	61	29	0.0582	0589	0596	0603	0610	0617	60	l	i	2	3	3	4	5	6	6
30	0.0625	0632	0639	0647	0654	0662	59	31	0.0669	0677	0685	0692	0700	0708	58	l	2	2	3	4	5	5	6	7
32	0.0716	0724	0732	0740	0748	0756	57	33	0.0764	0772	0781	0789	0797	0806	56	i	2	2	3	4	5	6	6	7
34	0.0814	0823	0831	0840	0849	0857	55	35	0.0866	0875	0884	0893	0902	0911	54	l	2	3	3	4	5	6	7	8
36	0.0920	0930	0939	0948	0958	0967	53	37	0.0976	0986	0996	1005	1015	1025	52	i	2	3	4	5	6	7	8	9
38	0.1035	1045	1054	1065	1075	1085	51	39	0.1095	1105	1116	1126	1136	1147	50	l	2	3	4	5	6	7	8	9
40	0.1157	1168	1179	1189	1200	1211	49	41	0.1222	1233	1244	1255	1267	1278	48	l	2	3	4	5	7	8	9	10
42	0.1289	1301	1312	1324	1335	1347	47	43	0.1359	1370	1382	1394	1406	1418	46	l	2	4	5	6	7	8	9	11
44	0.1431	1443	1455	1468	1480	1493	45	45	0.1505	1518	1531	1543	1556	1569	44	i	2	4	5	6	7	9	10	11
46	0.1582	1595	1609	1622	1635	1649	43	47	0.1662	1676	1689	1703	1717	1731	42	l	3	4	5	7	8	9	11	12
48	0.1745	1759	1773	1787	1802	1816	41	49	0.1831	1845	1860	1875	1889	1904	40	2	3	5	6	7	9	10	12	13
50	0.1919	1934	1950	1965	1980	1996	39	51	0.2011	2027	2043	2058	2074	2090	38	2	3	5	6	8	9	11	12	14
52	0 2107	2123	2139	2155	2172	2189	37	53	0.2205	2222	2239	2256	2273	2290	36	2	3	5	7	8	10	12	13	15
54	0.2308	2325	2343	2360	2378	2396	35	55	0.2414	2432	2450	2469	2487	2506	34	2	4	• 5	7	9	11	13	14	16
56		2543	2562	2581	2600	2619	33	57	0.2639	2658	2678	2698	2718	2738	32	2	4	6	8	10	12	14	16	18
58	0.2758	2778	2799	2819	2840	2861	31	59	0.2882	2903	2924	2945	2967	2988	30	2	4	6	8	10	12	15	17	19
60		3032	3054	3077	3099	3122	29	61	0.3144	3167	3190	3213	3237	3260	28	2	5	7	9	11	14	16	18	20
62	0 3284	3308	3332	3356	3380	3405	27	63	0.3429	3454	3479	3505	3530	3556	26	2	5	7	9	12	15	17	20	22
64		3608	3634	3660	3687	3714	25	65	0.3741	3768	3795	3823	3851	3879	24	3	5	8	11	13	16	19	22	24
66		3935	3964	3993	4022	4052	23	67	0.4081	4111	4141	4172	4202	4233	22	3	6	9	12	15	18	21	24	27
68	o!4264	4296	4327	4359	4392	4424	21	69	0.4457	4490	4523	4557	4591	4625	20	3	7	10	13	17	20	24	26	30
70		4694	4730	4765	4801	4837	19	71	0.4874	4910	4948	4985	5023	5062	18	4	7	11	14	18	22	25	29	33
72		5139	5179	5219	5259	5300	17	73	0.5341	5382	5424	5467	5510	5553	16	4	8	12	16	21	25	29	33	37
74		5641	5686	5731	5777	5823	15	75	0.5870	5918	5965	6014	6063	6113	14	5	9	14	19	24	28	33	38	42
76	0.6163	6214	6266	6318	6371	6425	13	77	0.6479	6534	6590	6647	6704	6762	12	5	11	16	22	27	32	38	43	49
	60'	50'	40'	30'	20'	10'	Cosec		60'	50'	40'	30'	20'	10'	Cosec	r	2'	3'	4'	5'	6'	7'	8'	9'.
Minutes for Log Cosecants.								Enter at bottom for Cosecants								Diffs. for minutes—Subtract

 

To Find the Hour Angle on the Prime Vertical

Enter Plate I with the latitude on the top or bottom margin; the intersection of the vertical line through this point with the curve corresponding to the declination of the body fixes a horizontal line which determines the hour angle of the body on the left hand margin.

The “Time Sight or Chronometer” Problem

Fig. 4 shows Rule II designed to find the hour angle of the observed body, given the latitude, declination and true altitude of the body.

Fig. 5 shows the arrangement of the scales.

Description of Scales I, K, L and M I is a scale of log sines with the corresponding degrees and minutes.

K is a scale of log haversines with the corresponding hours and minutes.

L is a scale of log secants with the corresponding degrees and minutes.

M is a decimal scale of natural numbers on which (log sec E+log sec d) may be found.

Formula (3)

in which s =½(L+h+r) and r=90°±d, the other quantities being the same as before.

To Compute the Hour Angle

Set the left edge of the index arm to s= 1/2 (L+h+r) on scale L, inner cylinder; turn outer cylinder and set index arm to (s — h) scale I, outer cylinder; move sliding marker to this point, then rotate collar until index arm, left edge, coincides with (log sec r+log sec d) from Table I, on scale M, inner cylinder, and the sliding marker will show the point on scale I opposite which on scale K will be found the hour angle.

The azimuth is found as with Rule I.

To Find the Great Circle Distance—Rule I

Given the latitudes, L₁ and L₂, of A and B, the points of departure and destination and the difference of longitude, l, between their meridians, find the great circle distance, D, between A and B.

Formula (4)  in which a is an auxiliary angle.

1. Set the index arm to L₁ on scale E.
2. Turn the outer cylinder and set the index arm to the difference of longitude, l, on scale A; set marker.
3. Rotate collar and set index arm to L₂ on scale D; the marker will indicate on scale A the point opposite which on scale B will be found hav a.
4. From scale B take hav (L₁ L₂); add this to hav a for hav D and opposite this number on scale B will be found D on scale C.

To Find the Great Circle Course

Formula (5)  , in which C₁ is the initial great circle course.

1. Set the index arm to L₂ on scale H.
2. Turn the outer cylinder and set the index arm to the difference of longitude,  l, on scale F or G, and set marker.
3. Revolve collar and set index arm to (90°~D) on scale H and the marker will show the great circle course on scale G.

Rule I may be also used to find the hour angle, t, and declination, d, of an unknown star, given the latitude of the observer, L, the true altitude of the star, h, and its azimuth, Z, reckoned from the elevated pole.

The “altitude azimuth” may be solved with Rule II.—Given the latitude of the observer, the true altitude and declination of the body, to find its azimuth.

Set the left edge of the index arm to s=1/2(L+z±d) on scale L, inner cylinder; turn outer cylinder and set index arm to (s ? d) on scale /, outer cylinder; move sliding marker to this point, rotate collar until left edge of index arm coincides with (log sec L+log sec h), from Table I, on scale M, inner cylinder, then the marker will show the point on scale I opposite which on scale K will be found the azimuth in hours and minutes; convert to degrees.

1. The upper signs in the above expressions are used when L and d are of the same name and the lower when of contrary names.
2. The azimuth is given the same name as the latitude and is marked East or West like the hour angle.
3. When this method is used it is not necessary to have recourse to Plate I, the azimuth being always reckoned from the elevated pole.

Rule II may be used also to find the initial and final great circle courses when the great circle distance between two places is known.

Given the latitudes, L₁ and L₂, of the points of departure and destination with the G. C. D., D, between them, find the initial and final G. C. courses, C₁, and C₂.

4. To find the initial course. Set the index arm to s=1/2(L₁+D±L₂) on scale L, turn the outer cylinder and set index arm to (s ? L₂) on scale I, set marker at this point, rotate collar and set index arm to (log L+ log cosec D) from Table I, on scale M and the marker will show the point on scale I opposite which on scale K will be found C₁ in hours and minutes; convert to degrees.
5. To find the final course. Proceed as above using s= l/2(L₂+D ± L₁) on scale L and (s ? L₁) on scale I with (log sec L+log cosec D) on scale M.
6. In the above expressions use the upper signs when L₁ and L₂ have the same name and the lower signs when they have contrary names.
7. Give C₁ and C₂ the names of L₁ and L₂ respectively; C₂ is thus reckoned as an initial course; reverse its direction for the final course.
8. The G. C. D. may be found from time to time by deducting the distance made good from the port of departure from the original distance.

Scales F, G and H may be omitted from Rule II, Fig. 4, and with the four scales L,M, I and K the following problems may be solved:

1. The Time Sight.
2. The Altitude-Azimuth.
3. Finding the great circle course.

With (a) and (b) a Sumner Line may be obtained by the tangent method.

With Rule I and a minimum of six scales the following problems may be solved:

4. Computing the zenith distance.
5. Time-Altitude Azimuth.
6. Finding the great circle distance.
7. Finding the great circle course.
8. Indentifying an unknown star by its altitude and azimuth.

With (d) and (e) a line of position may be obtained by the Marcq Saint-Hilaire method.

It appears that there is not so much difficulty in providing the means for the aviator to solve the astronomical triangle quickly as there is in measuring the altitude of a heavenly body accurately from an airship; to obtain the correct altitude is the real problem.