Harriot, Thomas - Mss. 6789

3317r

[Commentary:

This folio contains Harriot's proof of the angle preserving properties of stereographic projection. For the diagram see Add MS 6789, f. 18
For further details, including a transcript and translation, see Jon Pepper, 'Harriot's calculation of the meridional parts as logarithimic tangents', Archive for History of Exact Sciences, 4(1968), 359–413 (pages 366–367, and ]

Quod rumbus in planispærio nostro facit angulum cum meridiano
æqualem angulo facto a rumbo cum meridiano in
[Translation: That a rhumb in our planisphere makes an angle with the meridian equal to the angle made by the rhumb with the meridian on the sphere.

Sit

A B E

semicirculus pro
meridiano quolibet in globo
terrestri.
Cuius centrum

e

. poli

A

E

.
In illo meridiano sumatur quoduis
punctum

B

.
Sit,

e X

, communis sectio plani
æquatoris et plani meridiani,
quæ meridianus est in nostro
planisphærio correlatiuus meridiano
in globo terrestri.
In semicirculo pro meridiano
sumatur quodvis punctum

B

.
Agatur recta

A B

producta quæ
secabit,

e X

, in puncto

b

,
et punctum

b

in planisphærio erit
correllatium puncto

B

in superficie
sphæræ.
Agatur etiam recta,

e B

, et

G H

tangens peripheriam in puncto,

B

.
Dico primo quod anguli

H B b

H b B

sunt æquales.

H B b + A B e =

recto.

e b A + e A B =

recto.
Ergo:

H B b + A B e = e b A + e A B

.
Sed:

A B e = e A B

.
ergo:

H B b = H b B

[Translation: Let

A B E

be a semicircle on any meridian of the terrestrial sphere, with centre

e

, and poles

A

and

E

. Let

e X

be the intersection of the equator and the meridional plane, which meridian in our planisphere corresponds to the meridian in the terrestrial sphere.
In the semicircle take any point

B

on the meridian. Construct the extended line

A B

which will cut

e X

in the point

b

, and the point

b

in the planisphere will correspond to the point

B

on the surface of the sphere. Also construct the lines

e B

and

G H

touching the circumference in the point

B

.
I say first that the angles

H B b

amd

H b B

are equal.

H B b + A B e =

a right angle.

e b A + e A B =

a right angle.
Therefore

H B b + A B e = e b A + e A B

.
But

A B e = e A B

.
Therefore

H B b = H b B

Iam intelligitur

B

punctum esse centrum
circuli nautici qui vulgo compassus
dicitur; sive circuli horizontis visibilis.
cuius plano

e B

, est ad angulos rectos.

G H

est communis sectio plani meridiani
et illius circuli nautici, et est linea dicta
meridiana in plano horizontis circuli nautici.
Agatur quælibet linea

B D

in plano horizontis faciens quemvis angulum

D B G

cum meridiana

B G

et in illa linea sumatur quodvis punctum

D

et Agatur

D C

perpendicularis ad

G H

.
Conectantur puncta

A C

: et

A D

. et constituatur
pyramis cuius basis

B C D

, et vertex,

A

.
Quoniam

C D

, perpendicularis est in plano
erecto ad planum meridionis circuli productum
erit etiam perpendicularis ad

C A

, et planum
trianguli

D C A

erit erectum ad planum
meridiani.
Linea

A C

, secat,

e X

, in puncto

c

(non opus est ducere lineas

D F

, et

C F

. ideo
[Translation: Now it is to be understood that the point

B

is the centre of a nautical circle, which is commonly called the compass; or circle of the visible horizon, which is at right angles to the plane of

e B

G H

is common to the plane of the meridian and that nautical circle and is said to be the meridional line in the horizontal plane.
Construct any line

B D

in the horizontal plane making any angle

D B G

with the meridian

B G

and in that lane take any point

D

, and construct

D C

perpendicular to

G H

. Join the points

A C

and

A D

to make a pyramid whose base is

B C D

and vertex

A

. Since

C D

is a perpendicular in a plane at right angles to the plane of the circle of the meridion, it is also perpendicular to

C A

, and the plane of triangle

D C A

will be at right angles to the plane of the meridian.
The line

A C

cuts

e X

in the point

c

.
(there is no need to draw the lines

D F

and

C F

, therefore they are omitted).

Et in plano

A C D

a puncto

c

in linea

A C

erigatur,

c d

, ad
angulos rectos, quæ secabit

A D

in puncto

d

. Et erit communis
sectio æquatoris et trianguli

A C D

.
Connectantur puncta

d

b

.
Dico quod angulus

d b c

in plano
æquatoris est qualis angulo

D B C

in plano horizontis
sive circuli nauticis.
producatur

A B

et sit

C F

perpen-
dicularis ad illam.
fiat

F P = F B

et agantur

F C

F D

P C

, et

P D

.
Quoniam

D C

est perpendicularis
ad planum meridiani, facit
rectos angulos cum

C A

C B

C F

,
et

C P

.
Cum etiam

C B

C P

, sunt æquales
triangula rectangula

B C D

P C D

, sunt æqualia et
æquiangula, et angulus

C B D

æqualis est

C P D

.
Sed angulus

c b d

in æquatore est æqualis

C P D

, ob parallelismum
triangulorm

c b d

C P D

nam

c d

est parallela

C D

c b

C P

, quia:

c b A = C P A = C B P = H B b = H b B

ergo tertium latus

d b

est parallelum

D P

ergo plana triang-
ulorum

c b d

et CPD

sunt parallela. Et
similia, et angulus

c b d = C P D

.
Ergo

c b d = C B D

.
quod demonstrandum
fuit.
Aliter

d b

est parall.

D P, q u i a

[Translation: And in the plane

A C D

, there is constructed

c d

from point

c

at right angles to the line

A C

, which will cut

A D

in the point

d

. And it will be common to the equatorial plane and the triangle

A C D

. Join the points

d

and

b

.
I say that the angle

d b c

in the equatorial plane is equal to angle

D B C

in the horizontal plane, or nautical circle.
Let

A B

be extended, and let

C F

be perpendicular to it. Make

F P = F B

and construct

F C

F D

P C

, and

P D

.
Since

D C

is perpendicular to the plane of the meridian, it makes right angles with

C A

C B

C F

, and

C P

.
Since also

C B

and

C P

are equal, the right-angled triangles

B C D

and

P C D

are equal and equiangluar, and the angle

C B D

is equal to

C P D

. But the angle

c b d

in the equatorial plane is equal to

C P D

, because the triangles

c b d

C P D

are parallel, for

c d

is parallel to

C D

, and so are

c b

and

C P

since

c b A = C P A = C B P = H B b = H b B

Therefore the third side

d b

is parallel to

D P

, and so the planes of the triangles

c b d

and

C P D

are parallel. And similarly also angles

c b d = C P D

.
Therefore

c b d = C B D

, as was to be proved.
Another way,

d b

is parallel to

D P, s i n c e :

10 (5v)	11 (6r)
12 (6v)	13 (7r)
14 (7v)	15 (8r)
16 (8v)	17 (9r)
18 (9v)	19 (10r)

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