The Construction of the Wonderful Canon of Logarithms/Notes on Trigonometrical Propositions

by John Napier

SOME NOTES
BY THE LEARNED
HENRY BRIGGS
ON THE FOREGOING PROPOSITIONS.

[a]Iven an arc, to find the logarithm of its versed sine.

To the end of this proposition ⁂ I should like to add the following:—

Conversely, given the logarithue of a versed sine, to find its arc.

Add the known logarithm of the required versed sine to the logarithm of 30°, viz, 693147, and half the sum will be the logarithm of half the arc sought for.

Thus let 35791 be the given logarithm of an unknown versed sine, whose arc is also unknown.

To this logarithm add 693147, and the sum will be 728938, half of which, 364469, is the logarithm of 43° 59′ 33″. The arc of the given logarithm is therefore 87° 59′ 6″, and its versed sine is 9648389.

Again, let a negative logarithm, say —54321, be the known logarithm of the required versed sine. To this logarithm add, as before, 693147, and the sum, that is the number remaining since the sines are contrary, will be 638826, half of which, 319413, is the logarithm of 46° 36′ 0″. The arc of the given logarithm is therefore 93° 12′ 0″, the versed sine of which is 10558216, and since this is greater than radius it has a negative logarithm, namely —54321.

Demonstra- tion
c	b	$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$	versed sine of arc	$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$	c	d
a	b				a	d

x	c	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$	cont.	x	a	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$	cont.	½x c, sine of 30° 0′	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$	cont.
c	g		pro-	a	e		pro-	c g, sine of ½ arc c d		port.
c	h		port.	a	f		port.	c b, double of line c h		port.

Of the spherical triangle A B D ]

In finding the base we may pursue another method, namely:—

Add the logarithm of the versed sine of the given angle to the logarithms of the given sides, and the sum will be the logarithm of the difference between the versed sine of the difference of the sides and the versed sine of the base required. This difference being consequently known, add to it the versed sine of the difference of the sides, and the sum will be the versed sine of the base required.

For example, let the sides be 34° and 47’, their logarithms 581261 and 312858, and the logarithm of the versed sine of the given angle −409615. The sum of these three logarithms is 484504, which is the logarithm of the difference between the versed sine of the base and the versed sine of the difference of the sides.

Now the line corresponding to this logarithm, whether a versed sine or a common sine, is 6160057, and consequently this is the difference between the versed sine of the base and the versed sine of the difference of the sides. If to this you add the versed sine of the difference of the sides, that is 256300, the sum will be the versed sine of the base required, namely 6416357, and this subtracted from radius leaves the sine of the complement of the base, namely 3583643, which is the sine of 21°. Consequently the base required is 69°.

Conversely, given three sides, to find any angle.

If from the logarithm of the difference between the versed sine of the base and the versed sine of the difference of the sides you subtract the logarithms of the sides, the remainder will be the logarithm of the versed sine of the angle sought for.

As in the previous example, let the logarithms of the sides be 581261 and 312858. Subtract their sum, 894119, from the logarithm 484504, and the remainder will be the negative logarithm −409615, which gives the versed sine of the required angle 120° 24’ 49’.

Of five parts of a spherical triangle ]

This proposition appears to be identical with the one which ts inserted at the end, and distinguished like the former by (*). The latter proposition I consider much the superior. There are, however, three operations in it, the first two of which I throw into one, as they ave better combined. Thus:—

Let there be given

the base 69°,
the angles at the base	$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$	42°	29′	59″
		31°	6′	5″
‍		73°	36′	4″	sum.
		36°	48′	2″	half sum.
		73°	36′	4″	complement of ½ sum.
		11°	23′	54″	difference.
		5°	41′	57″	half difference.
		84°	18′	3″	complement of ½ diff.

						Logarithms.
Propor- tion 1.	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine half difference	5°	41′	57″	23095560
		Sine half sum	36°	48′	2″	5124410
		Sine difference	11°	23′	54″	16213641
		Sum of sines				−1757509
Propor- tion 2.	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine of sum	73°	36′	4″	415312
		Sum of sines				−1757509
		Tangent half base	34°	30′	0″	3750122
		Tangent ½ sum of sides	40°	30′	0″	1577301
Propor- tion 3.	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine ½ sum of angles	36°	48′	2″	5124410
		Sine ½ diff. of angles	5°	41′	57″	23095560
		Tangent ½ base	34°	30′	0″	3750122
		Tangent ½ diff. of sides	6°	30′	0″	21721272

40°	30′
6°	30′	‍
47°	0′	$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$	sides
34°	0′

These are the operations described by the Author. But
I replace the first two by another, retaining
the third.

						Logarithms.
Propor- tion	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine compl. ½ sum of angles	53°	11′	58″	2222368
		Sine compl. ½ diff. of angles	84°	18′	3″	49553
		Tangent ½ base	34°	30′	0″	3750122
		Tangent ½ sum of sides	40°	30′	0″	1577307

Another Example.

Let there be given

the angle 47°,
the sides containing it	$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$	50°	35′	11″
		31°	6′	5″
‍		90°	41′	16″	sum.
		45°	20′	38″	half sum.
		44°	39′	22″	compl. of half sum.
		28°	29′	6″	difference.
		14°	14′	33″	half difference.
		28°	29′	6″	compl. of half diff.

						Logarithms.
Propor- tion 1.	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine compl. ½ sum of sides	44°	39′	22″	3526118
		Sine compl. ½ diff. of sides	75°	45′	27″	312192
		Tan. compl. ½ vert. angle	66°	30′	0″	−8328403
		Tan. ½ sum of angs. at base	72°	30′	0″	−11452329
Propor- tion 2.	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	Sine ½ sum of sides	45°	20′	38″	3406418
		Sine ½ diff. of sides	14°	14′	33″	14023154
		Tan. compl. ½ vert. angle	66°	30′	0″	−8328403
		Tan. ½ diff. of angs. at base	38°	30′	0″	2288333

72°	30′
38°	30′	‍
111°	0′	$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$	angles at the base.
34°	0′

And these relations are all uniformly maintained, whether there be given two angles with the interjacent side or two sides with the contained angle. In each oe the important point is what occupies the third place in the proportion, In the former it is the tangent of half the base, in the latter the tangent of the complement of half the vertical angle. In these examples, if the tangent or the sum of the sines be greater than radius, the logarithm is negative and has a dash preceding, for example −8328403.

Another way of the same ]

[d]Then divide the sum of the first and second found by the square of radius, and you will have )

To make the sense clearer, I should prefer to write this as follows:—

Then divide both the first and second found by the square of radius, add the quotients, and you will have the tangent, &c.

This proposition ts absolutely true, as well as the one
preceding; but while the former may most conveniently be
solved by logarithms, the latter will not admit of the use
of logarithms throughout, as the quotients must be added
and subtracted to find the tangents; for the utility of
Logarithms ts seen in proportionals, and there-
fore in multiplication and division,
and not in addition or
subtraction.

THE END.