Astronomia Britannica exhibiting the doctrine of the sphere, and theory of the planets decimally by trigonometry, and by tables : fitted for the meridian of London ... / by John Newton ...

CHAP. 18. Of the Semidiameters of the Sun, Moon, and shadow of the Earth.

THe angle of the Suns apparent Semidiameter, in his nearest di∣stance to the Earth, Bullialdus hath by observation found to be 16′ 45″, or in Decimall numbers 27917. And by an Eclipse of the Moon, December 1638, he found her Semidiameter to be 16′ 54″ or 28167, and the Semidiameter of the Earth; shadow 44′ 9″, or 7 583, at which time (being the time of incidence) her distance from the Earth by his computation was 97908 parts of the Semiaxis of the Elipsis 100. 000. By this and another observation in the same Eclipse, he shew∣eth how to finde her apparent semidiameter, in all the other intervalls. The inferiour limbe of the Moon and the first Starre in the foot of the for∣mer Twin, (whose place then according to Tycho was Gemini 28. 25′ 17″, or Gemini 28. 42138 with South Latitude, 0 d. 58′ or 0. 96667.) being in the same Azimuth, was 8′ or 13333 higher then the Star and the Alti∣tude of the heart of Hydra then taken by him at Paris was 30 deg. 37′, or 30 d. 61667. From whence the hour was found, 30 h. 40′, or 13 h. 66667 and the houre being given the altitude of the Starre is also given, deg. 56. 42′ 15″, or deg. 56. 70416. The apparent altitude of the center of the Moone was deg. 57 7′ 9″, or deg. 57 11916, but by her latitude and place it should have beene deg. 57 40′ 4″, or deg. 57 66778 and there∣fore her parallax of altitude 32′ 55″, or 54861. The situation of the Moone and Azimuth in which her interiour limbe and the Stars were, being given, her aparent Longitude was almost in Gemini, deg. 28 38′ 30″, or Gemini deg. 28 64167, her parallax of longitude 18 min. or 30000 and therefore the center of the Moon in her true motion in Gemini 28 d. 57 min. fere. or in Gemini 28 d. 95000, her parallax of Latitude is 19 min. or 31667. to which 21′ or 35007, the difference of the observed latitude of the Moon and Stars, being added the true difference is 50 min. or 83333 min. and thence the Moons Latitude 8 min. or 13333 S.

Now then to finde the distance of the Moon from the Earth, in this E∣clipse, the Earths semidiameter being one degree, Let FEC represent the true Horizon, BDE, the vertical at Paris E the center of the earth, D the City of Paris: the Moons true altitude, AEF, deg. 5766778, the observed altitude ADG, deg. 5711916. The parallacticall angle DAE, deg. 0. 34861. Therefore in the Triangle ADE we have given all the angles, and the finde DE one Semidiameter of the Earth, to finde AE, for which the anolagy is.

[illustration]

As the sine of DAE 0 d.	54861 co. ar.	2. 0188745
To the side DE 1		0. 0000000
So is the sine of ADB 32 d.	88084	9. 7347147
To the side AE	56. 70	1. 7535892

This foundation being laid, he proceedeth to the rest: and to shew how we may possibly fall into some absurditie, he supposeth the Moons distance from the Earth in this Ecclipse to be but 55 semidiameters, or the side BC in the following figure, the apparent angle of the semidiameter of the Earths shadow CHI, 0. 73583 AEF represents the Sun, his semidiame∣ter AE, the angle of his apparent Semidiameter when he is Perig. AGE 16. 45, or in decimalls 27916 BHG represents the Earth. BG the Semi∣diameter thereof, hence to finde HI in the triangle HIC the proportion is.

As the sine of HIC	89. 26417 co. ar.	0. 0000358
To the side HC	54	1. 7323937
So is the Radius HCI	90	10. 0000000
To the Hypothenusal HI	54. 004	1. 7324295

2. In the triangle HBI we have given the sides BH. 1. and HI 54. 004 with the angle BHI 179. 26417, hence to finde the Angle BHI, the Analogie is.

As the greater side H I	54. 004 co. ar.	8. 2675705
Is to the less H B	1.	0. 0000000
So is Radius		10. 0000000
To the tangent of	1. 06082	8. 2675705
Adde	45.
As Radius		10. 0000000
To the Cotang. of	46. 06082	9. 9839145
So Tang. halfe summe	36791	7. 8075980
To tang. halfe differ.	35454	7. 7915125
Angle B I H	01337. And the angle C B I	72246

3. In the triangle C B I, we have given the angles and the side B C 55 to find C I. Therefore say,

As Radius

To the tangent of C B I	72246	8. 1007064
So is B C	55	1. 7403627
To C I	0. 6935 = to B K	1. 8410691

and therefore K G 3065, and the angle K I G 0. 31857.

For, •••• I K	55 co. ar.	8. 2596373
To Radius		10. 0000000
So is K G	0. 3065	1. 4864305
To the tangent of K I G	0. 31857	7. 7450678

and the angle B D G is equal thereunto, but so the angle of the Suns ap∣par••n•• Semidiameter A G E 27916 by observation, is lesse then the angle A D E, which is absurd, and therefore some part assumed is false. The Semidiameters of the Sun and Moon must not be changed, constant experience agreeing with these observations.

In this Eclipse therefore Bullialdus doth take for the distance of the Moone from the earth, B C 57. 85 Semidiameters of the earth, and the Semidiameter of the earths shadow, C B I 75111. Hence to find C I, the analogie is.

As Radius

To the side B C	57. 85	1. 7623034
So the tang. of B C	0. 75111	8. 1176019
To C. I.	0. 75841	1. 8799053

Let B K be equal to C. I. So is K G 24159. The••

As I K	57. 85 co. ar.	8. 2376066
To Radius		10. 0000000
So is K G	24159	1. 3830789
To the tang. of K I G.	23928 or B D G	7. 6207755

equal to E D A, and the Suns apparent Semidiameter being given A G E, 27916, the angle G A B, or the difference between the angles A G E, and E D A shall be given also, viz. 03988. the Suns Horizontal parallax when he is Perigaeon. And the Moones Perigaeon distance from the earth, in Syzigiis, 56. 50 Semidiameters of the earth.

For, as	97908 co. ar.	5. 0091819
To	57. 85	1. 7623034
So is	95638	4. 9806304
To B C	56. 50	1. 7521157

Hence to find the Moones Horizontal parallax when she is perigaeon, the analogie is, in the preceding Diagram.

As E G or B C	56. 50 co, ar.	8. 2478843
Is to Radius: So is D E	1.	0. 0000000
To the sine of E G D	1. 01399	8. 2478843

The Horizontal parallax of the Sun when he is perigaeon or the angle

B A G was found to be

. 03988

The Moones Horizontal parallax is

1. 01399

Their aggregate

2. 05387

Semidiameter of the Sun subtract

. 27916

There rests the angle C B I

. 77471

or the apparent semidiameter of the earths shadow in loco transitus Lun••, Perig.

In the triangle therefore B C I, we having the angles and the side B C given, C I shall be also given.

For, As the sine of	90 deg.
Is the side B C	56. 50	1. 7521157
So tang. of C B I	77472	8. 1310339
To the side C I	7640 = B K	1. 8831496
And therefore K G	••360

And in the triangle A G B having the angles and B G given the side A B is also given, for

As the tang. of B A G	03988 co. ar.	3. 1585620
Is to B G 1. So is Radius		10. 0000000
To A B	1440. 66	3. 1585520

which is the distance of the Earth from the Sun, when he is Perihelion. And because the Suns eccentricity is 1784, his Apogaean distance is 101784, hence to find his distance, in Semidiameters of the earth, say,

As his Perigaean distance	98216 co. ar.	5. 0078178
Is to his distance	1440. 66	3. 1585620
So is his Apog. distance	101784	5. 0076794
To his Apog. dist.	1493. 03	3. 1740692
Then as B S	1493. 03. or E G	6. 8259308
To Radius, so is E D.	1	10. 0000000
To the sine of E G D	0. 3855	6. 8259308

The Suns Horizontal parallax when he is Apogaeon.

As Radius, to A B:	1440. 66	3. 1585620
So is tang. of A B E	27916	7. 6877120
To A E	70189	0. 8462740
Then as B S	1493. 03 co. ar.	6. 8259308
To Radius, so is S T	70189▪ ☉ Semid.	0. 8462740
To the tang. of S B T	26936	7. 6722048

The apparent Semidiameter of the Sun when he is Apogaeon.

The Sun being Perigaean, we have given B G 1. K B. 75841. KG. 24159 and B C. 56. 50, the distance of the Moon from the Earth when she is Pe∣rigaean; from whence the longitude of the earths shadow may thus be found.

As K G	2360 co. ar.	10. 6270880
To K I	56. 50	1. 7521157
So is C I	7640	1. 8831496
To C D	182. 93	2. 2623533

Add B C 56. 50 then is B D 239. 43. the longitude of the earths sha∣dow.

Let B S be the Apogaean distance of the Sun,

1493. 03

The angle of the Suns apparent Semidiameter S G T

26936

The Perigaean Semidiameter or the angle A G E

27916

Their difference is the angle Z G E

00980

Let TG be produced to N, then shall the angle I G N be equal to the an∣gle Z G E, but the Sun being Perigaean, the angle B D G was found to be o. degrees 239••8. whose complement is the angle B G D 89. 76072 therefore when the Sun is Apogaean, it shall be 89▪77052, therefore B X G 0. 22948, equal to K N G.

Hence to find K G the analogie is.

As Radius	90	10. 0000000
To K I	56. 50	1. 7521157
So tang. of K I G	0. 22948	7. 6006035
To K G	22632	1. 3547192
And K B	77368. Then to find C B N. Say.
As B C	56. 50 co. ar.	8. 247••843
To Radius	90	10. 0000000
So C I or rather C N.	77368	1. 8885613
To the tang. of C B N	7844••	8. 1364456

The Sun being Apogaean; and the angle C B I, the Sun being Paerigaean, was before found to be 77471, and therefore the difference of the earths shadow between the Suns Apogaean and Perigaean is, 00971. Then,

As K G	22632 co. ar.	10. 6452808
To K I	56. 50	1. 7521157
So is C N	77368	1. 8885613
To C X	19••. 18	2. 2859578
Add B C	56. 50. Then is B X	249. 68.

The semidiameter of the Moon, when she is Perigaean, is greater then the semidiameter of the Sun, being Apogaean, and therefore Bullialdus doth make it 17. or 28333, and because the eccentricity of the Moon is given 4362, her Apogaean distance in Syzygiis 104362, the Moon being Perigaean her distance from the earth is, 95638, and in semidiameters of the earth 56. 50 and therefore her Apogaean distance in semidiameters of the earth, by the analogy following,

As	95638 co. ar.	5. 0193696
To	56. 50	1. 7521157
So is	104362	5. 0185423
To	61. 66	1. 7900276
As her Apogaean dist.	61. 66 Co. ar.	8. 2099724
To the Moons Perig. semid.	28333	1. 4522925
So is the Moon Perig. dist.	56. 50	1. 7521157
To the Apog. semid.	25964	1. 4143806

We have the semidiameter of the Cone C I 76400, and her Perigaean distance 56. 50, and D C 182. 93, but when the Moon is Apogaean, D C will be no more then 177. 77. found by abating K I or K N 61. 66. from B D 239. 43. Hence to find C I or C N in the same parts say.

As D C	182. 93 co. ar.	7. 7376467
To C I	7640	1. 8831496
So is C D	177. 77	2. 2498584
To C I	7424	1. 8706547
Then as B C	61. 66 co. ar.	8. 2099724
Is to Radius	90	10. 0000000
So is C I	7424	1. 8706547
To the tang. of C B I	. 68981	8. 0806271

Here then we have determined
	Apogaeon	26936
The Suns Semidiameter
	Perigaeon	27916
	Apogaeon	1493. 03
His distance from the earth
	Perigaeon	1440. 66
	Apogaeon ☉	249. 68
The Axis of the earths shadow
	Perigaeon ☉	239. 43

The Semidiameter of the shadow, when the Sun is Apogaeon. In loco ••ransitus Lunae, Apog 78442 Perig. 77471.

	Apogaeon	25964
The Semidiater of the Moone in Syzygiis
	Perigaeon	28333
	Apog.	61. 66
The distance of the Moon from the earth in Syzygiis
	Perig.	56. 50
	Perig.	23928.
Semiangle of the Cone
	Apog.	22948.