The General Use of the Canon and Tables of Logarithms.

CHAP. IV.
Conteining some Use of right-lined Triangles in the Practice of Fortification
.

In the late manner of Fortification the ordinary Care is:

  1. That the Angle of the Bulwark may be either a right Angle or near unto it.
  2. That this Angle may be defended from the Flank and Cortin on either Side.
  3. 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.
  4. 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.

PROP. I.
Having the Side of a Regular Fort, with the length of the Gorge, the Flank and the Face of the Bulwark, to find the rest of the Lines and Angles.

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.

Fort

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.

  1. 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
  2. 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
  3. 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.
  4. 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
  5. 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.
  6. 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.

  1. 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:
  2. Take this Angle at the center, out of 180 gr. there remains the Angle of the Fort HAD.
  3. The Angle ADE between the Flank and the Cortin, may be always 90 gr.
  4. 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.
  5. 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

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