In the late manner of Fortification the ordinary Care is:
- That the Angle of the Bulwark may be either a right Angle or near unto it.
- That this Angle may be defended from the Flank and Cortin on either Side.
- That the Lines of defense may not exceed the reach of a Musket, which is said to be twelfe score Yards, and those make 720 foot.
- That the depth of the Flanks aand the bredth of the Rampard be sufficient to resist battery; and that may be about 100 foot at the ground.
Upon these considerations depend the rest of Lines and Angles: whereif I will set down some Propositions, beginning with that which may resolve the works of others.
A regular Fort is that, which is made by equal Sides and Angles, each Bulwark like unto other.
Suppose, that by observation or otherwise, we have found that in a square Fort, the Side was 700 foot, the Gorge 140, the Flank 100, and the Face 335; In a Pentagonal, Hexagonal, Heptagonal, as in this Table.
|
Quadr. |
Pentag. |
hexag. |
Heptag. |
Octagon. |
The Side |
AB |
700 |
800 |
900 |
950 |
1000 |
The Gorge |
AD |
140 |
180 |
190 |
200 |
230 |
The Flank |
DE |
100 |
120 |
140 |
150 |
158 |
The Face |
EF |
335 |
352 |
370 |
360 |
420 |
And that it were required to find the rest of the Lines, and the quantity of the Angles belonging to each Fort, beginning with the Quadrate.
First, we may protract this Fort, by making a Square whose Side AB shall be 700 foot by the Scale: then taking but 140 for the Gorge, and set the off from A unto D, and from A unto H: at D and H raise two Flanks perpendicular to the Sides of the Fort, and there prick down 100 from D unto E, and from H unto G. That done, take 335 out of the same Scale, and setting one foot of the Compasses in the point E, make an occult Ark of a Circle. Again, setting one foot of the Compasses in the point G, make another occult Ark, crossing the former in the Point F; So the Lines EF, FG shall represent the Face of the Bulwark.
In like manner for the Bulwark at B, we may set off the Gorge from B unto N, &c. So have we divers Triangles, which may be resolved by the first three Propositions of right-lined Triangles: and the manner of it shall be to set down, as that Precept may be easily distinguished from the Example, and applied to any other number by this Canon and Table of Logarithms, but by the old Canon of Sines and Tangents, and by the Lines of Sines and Tangents both upon the Sector and the Cross-staff.
- In the Rectangle ADE, having the Sides AD, AE, we may find the Angles at A and E, and the third Side AE, by the former part of the third Proportion of Right-lined Triangles.
As the Gorge |
AD |
140 |
2.1461280 |
|
|
To the Flank |
DE |
100 |
2.0000000 |
So the Radius |
|
90 00 00 sec. |
10.0000000 |
|
|
To the Tangent of |
DAE |
35 32 ¼ |
9.8538720 |
Take the Angle DAE out of 90 gr. the Complement will give the Angle DEA; and then, having two Sides and three Angles, we may well find the third Side AE by the first Proposition of the right-lined Triangles.
As the Sine of |
DAE |
35 32 ¼ |
9.7643542 |
|
|
To the Side |
DE |
100 |
2.0000000 |
So the Sine of |
|
90 00 00 sec. |
10.0000000 |
|
|
To the Side |
AE |
172047 |
2.2356458 |
- Because the Fort is supposed to be sure, the Angle HAD, must be 90 gr. and the half Angle CAD 45 gr. if we add this Angle CAD unto the Angle DAE and take the Sum out of 180 gr. the Remainder 99,27¾ shall be the Angle EAF. Then in the Triangle AEF, having the Angle at A, and the two Sides FE, AE, we may find the other Angles a E and F, by the third Proposition of right-lined Triangles.
To the Face |
EF |
335 |
2.5250448 |
To the Sine of |
EAF |
99 27 ¼ |
9.9940502 |
|
|
|
7.4690054 |
|
|
So the Line |
AE |
172047 |
2.2356458 |
To the Sine of |
AFE |
30 26 1/5 |
9.7046513 |
Add this Angle AFE to the Angle EAF, and take the Sum out of 180 gr. the Reaminder 50 gr. 6 m.3 sec. shall be the Angle AEF. And then we have two Sides and three Angles, to find the Head-line AF.
As the Sine of |
EAF |
99 27 ¼ |
9.9940502 |
To the Face |
EF |
335 |
2.5250448 |
|
|
|
7.4690054 |
|
|
So the Sine of |
AEF |
50 6 1/20 |
9.8848958 |
To the Head-line |
AF |
26055 |
2.2356458 |
- If we produce the Face FE until it meet the Cortin in O, we shall have the Triangle AEO: wherein, knowing the Side AF, and the three Angles (for knowing two Angles, the third is always known by the Complement unto 180 gr.) we may find the other two Sides FO, AO.
As the Sine of |
AOF |
14 gr. 33 m. 48 sec. |
9.4004548 |
To the Head-line |
AF |
26055 |
2.4158904 |
|
|
|
6.9845644 |
|
|
So the Sine of |
FAO |
45 00 00 |
9.8494850 |
To the Line |
FO |
7369 |
9.8649206 |
|
|
And the Sine of |
AFO |
30 26 12 |
9.7046513 |
To the Line |
AO |
524212 |
2.7200869 |
Take the Gorge NB 140, out of the Side AB 700, there remains 560 for the Line AN. Take this Line AO out of AN, and there remains 35622 for ON that part of the Cutain from it hence the Face of the Bulwark may be defended.
- In the Triangle AFN, having two Sides AF, AN, and the Angle between them FAN, we may find the other two Angles at F and N, by the tater part of the third Proposition of right-lined Triangles.
As the Sum of the Sides AF, AN |
82035 |
|
2.9141050 |
Is to the Difference of those Sides |
29949 |
|
2.4763245 |
|
|
|
4377805 |
|
|
So the Tang. of the half sum of opp. Ang. at FON |
22 30 |
|
9.6176153 |
To the Tang. of half the Diff. between those Ang. |
8 36 1/3 |
|
9.1798348 |
This half Diff. added to the half sum gives the greater Ang. AFN |
31 6 1/3 |
and subtracted the lesser ANF
| 13 53 4/5 |
|
|
As the Sine of |
ANF |
|
9.3805157 |
To the Head-line |
AF |
26055 |
2.4158904 |
|
|
|
6.9646253 |
|
|
So the Sine of |
FAN |
45 00 00 |
9.8494850 |
To the Line of Defense |
FN |
767114 |
2.4158904 |
- In the Triangle ABC we have the Side AB, and the three Angles, to find the Side CA or CB from the Center to the Angles of the Fort.
As the Sine of |
|
90 00 00 sec. |
10.0000000 |
|
|
To the Side |
AB |
700 |
2.8450980 |
To the Sine of |
ABC |
45 00 00 |
9.8494850 |
To the Line |
AC |
494121 |
2.6945830 |
This Line AC added to the Head-line AF, gives the whole CF, from the Center of the Fort to the uttermost point of the Bulwark to be 755525.
- In the Triangle CFL (the Side FL being parallel to AB the Side of the Fort) we have the three Angles and the Side CF; by which we may find FL the Distance between the points of the two next Bulwarks.
As the Sine of |
CLF |
45 00 00 |
9.8494850 |
|
|
To the Line |
CF |
755525 |
2.8782498 |
So the Sine of |
FCL |
90 00 00 |
10.0000000 |
|
|
To the Line |
FL |
1068.464 |
3.0287648 |
Thus by resolving of six Triangles we have found
|
gr. |
m. |
sec. |
The Angle at the Gorge |
DAE |
35 |
32 |
15 |
The Angle of the Bulwark |
GFE |
60 |
52 |
24 |
The Angle |
FED |
104 |
33 |
48 |
The Angle |
ANF |
13 |
53 |
48 |
|
Foot. |
The length of the Line |
AE |
172 |
047 |
The Head-line |
AF |
260 |
540 |
The Line on the Cortin |
ON |
35 |
088 |
The Line of Defence |
FN |
767 |
115 |
The Semidiameter |
CA |
494 |
975 |
The Line from the Center to the Bulwark |
CF |
755 |
525 |
The Distance between the the Bulwark |
FL |
1068 |
464 |
The principal Lines and Angles belonging to the Bulwark at A: |
The rest of the Lines are either parallel unto these, or else they may be found in the same manner.
And all these may be understood by the same in the rest of the Bulwark belonging to this Fort.
Again, what is said of a Square Fort, the same may be applied to all regular Forts.
And so, resolving the works of other men, it may appear how near they have come to the former grounds.
But that we may not altogether insist upon Examples, I will set down some profitable Suppositions, and from them proceed to find the rest of the Linesn and Angles belonging to any Regular Fort.
- The Angle at the center ACB, between the Lines CA, CB drawn from the center to each Bulwark, is found by dividinf 360 gr. by the number of the Sides. So in a Square Fort, this Angle will be 90 gr. In a Pentagonal Fort, where there are five Sides, it will be 72 gr. &c:
- Take this Angle at the center, out of 180 gr. there remains the Angle of the Fort HAD.
- The Angle ADE between the Flank and the Cortin, may be always 90 gr.
- The uttermost Angle of the Bulwark EFG, must be less than the Angle of the Fort, yet not less than 60 gr. nor doth it ness to be much more than 90 gr. If we allow it to be 2/3 of the Angle of the Fort, it may be defended from the Flank and Cortin on either side.
- The Angle at the Grge DAE, which forms the Flank DE, may be allowed between 35 and 40 gr. For in small regular Forts it may be 40 gr. But where the Angle of the Fort is great, it may be less.
These five Anales being first setled, the most of the other Angles will depend upon them, as in the Table following.
Or howsoever there may be other Angles found to be more convenient, yet these are sufficient to explain the use of Triangles.
In a Regular Fort. |
|
Quadr. |
Pentag. |
Hexag. |
Heptag. |
Octagon. |
Cortin |
|
Gr. |
M. |
Gr. |
M. |
Gr. |
M. |
Gr. |
M. |
Gr. |
M. |
Gr. |
M. |
|
Angle at the Center |
ACB |
90 |
0 |
72 |
0 |
60 |
0 |
51 |
25 |
45 |
0 |
0 |
0 |
Angle of the Fort |
HAD |
90 |
0 |
108 |
0 |
120 |
0 |
128 |
34 |
135 |
0 |
180 |
0 |
Angle of the Flank |
ADE |
90 |
0 |
90 |
0 |
90 |
0 |
90 |
0 |
90 |
0 |
90 |
0 |
Angle of the Bulwark |
GFE |
60 |
0 |
72 |
0 |
90 |
0 |
84 |
42 |
90 |
0 |
90 |
0 |
Angle of the Gorge |
DAE |
40 |
0 |
39 |
0 |
38 |
0 |
37 |
0 |
36 |
0 |
35 |
0 |
|
The half of HDA is |
CAD |
45 |
0 |
54 |
0 |
60 |
0 |
64 |
17 |
67 |
30 |
90 |
0 |
The half of GFE is |
AFE |
30 |
0 |
36 |
0 |
40 |
0 |
42 |
51 |
45 |
0 |
45 |
0 |
The Complement of CAD is |
DAF |
135 |
0 |
126 |
0 |
120 |
0 |
115 |
43 |
112 |
30 |
90 |
0 |
AFE out of CAD leaves |
AOF |
15 |
0 |
18 |
0 |
20 |
0 |
21 |
25 |
32 |
30 |
45 |
0 |
The Complement of AOF is |
OED |
75 |
0 |
72 |
0 |
70 |
0 |
68 |
35 |
67 |
0 |
45 |
0 |
The Complement of OED is |
DEF |
105 |
0 |
108 |
0 |
110 |
0 |
111 |
26 |
112 |
30 |
135 |
0 |
|
The Complement of DAE is |
AED |
50 |
0 |
51 |
0 |
52 |
0 |
53 |
0 |
54 |
0 |
55 |
0 |
AED out of DEF leaves |
AEF |
55 |
0 |
57 |
0 |
58 |
0 |
58 |
26 |
58 |
20 |
80 |
0 |
AEF out of AFE leaves |
FAE |
95 |
0 |
87 |
0 |
82 |
0 |
78 |
43 |
76 |
30 |
55 |
0 |
|