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bases of any other points, as a, b, c, &c., as they may happen to be hereafter required. Also the measure of the angle È CA, being the exterior angle of a triangle, checks the measurement of the angles at A and D, taken together; the same of D C B as to the angles at E and B. Should it so happen, however, that the intersection C falls in a narrow valley, then there will be no opportunity from such a point to fix such stations as a, b, c, &c., as above mentioned; but if the opposite angles at C prove to be equal, and if ACD equal to CAD, plus CDA, and ACE equal to A EC, plus E A C, and the same of the others, we may consider C as safe; so much so that if circumstances, such as obstructions, for instance, had hitherto prevented such a line as CB from being set out, so that if it were purely imaginary the position of A C, relative to D E, combined with the measurements of the angles as above mentioned, would enable us to sight out C B, and if correctly done it will fall in at the point B, and no other.
Amongst the many methods for obtaining the length of inaccessible distances without instruments, the following will, I think, be found the most useful.
PROBLEM 13. Fig. 34.
On the line A B to obtain the inaccessible length CD across a river.
On the line A B take any point E, say two chains from C, and sight the line DF, making any angle, not too acute, with the line A B; from C and E set up the perpendiculars CG and E F, to intersect D F at G and F; and from G let fall the perpendicular G H; then, because of the similar triangles
HF: HG:: FE: ED.
and D E minus C E equal to D C, the length required.
PROBLEM 14. Fig 35.
Another method of obtaining inaccessible distances.
Let A B be the required inaccessible distance. Produce A B to any convenient point C; set out any two lines, BD and E C, making F D equal to B F, and E F equal to CF; through the point of intersection F, lay out the line A F G to intersect the line ED produced to G; GE will be equal to CA, and CA minus C B equal to B A, which was required.
PROBLEM 15. Fig. 36.
To find the intersection of two lines meeting in an inaccessible point.
Let BA and CA be the two lines meeting at the inaccessible point A. From any two convenient points, as D E on BA, set out any two lines, D F and E G intersecting each other on the line CA, as at I; make IG equal to EI, and I F equal to DI. Produce F G to intersect Ĉ A, as at K; then K L equal to I A, and K equal to D A.
But whenever there is a theodolite or box-sextant at hand, the best way of obtaining inaccessible distances is given in the two following problems.
PROBLEM 16. Fig. 37.
To find an inaccessible distance on a chain line with the box
Let the chaining on the line A B be interrupted by the coming across a block of buildings. From the point C set out an angle of 60°, as B C D, being careful to observe that the line CD will clear the obstruction. Measure on to D, from whence you can perceive you can return to the line A B clear of the obstruction; set out again the angle of 60°, CD E, and make DE equal to CD, when E will be on the line as required, and CD or DE will be equal to the distance required, CE. Over a sheet of water some point on C B would probably be visible from C, whereby to set out the angle of 60°, but with a block of buildings or a wood this would not be the case; we must, therefore, to make ECD equal to 60°, set out A C D equal to 120°, and with the sextant it will be better to do this by setting out the angle of 60° twice, which will give the angle of 120°. It may also happen, that on reaching the point E, all forward points to guide us are invisible; then produce D E to F, and at F set out another angle of 60°, and make F G equal to E F, which will again place us on the base or main chain line A B, and E G will be equal to EF or F G.
PROBLEM 17. Fig. 38.
To find an inaccessible distance, as the width of a river, with the box-sextant.
Let A B be the length required; set out BC at right angles to A B, and make BC any convenient length; measure the
COMPUTATION OF AREAS.
angle BCA, from the natural trigonometrical tables, take out its natural tangent and multiply it by B C, which will give the required length, AB; in this way any number of inaccessible distances on the same line may be obtained as a, b, c, &c. In the latter cases a boat would be used with two persons in it, the boatman to guide it, and an assistant to direct him; the assistant keeping himself in line by means of two points, as B and D on the main line.
COMPUTATION OF AREAS.
In the article on "Gunter's Chain," it has been shown how the product of one side of a figure by another side is treated to obtain acres, roods, perches, and odd links, as also the principles of division of this chain; in the following remarks, therefore, in order to keep the subject within limits, we shall only give one further example, and confine our remarks to the division and treatment of irregular figures mostly, to the practice of ascertaining areas in the office, and at the end of this work to the description and use of an instrument lately introduced by Messrs. Elliott, and called the planimeter, by means of which the labour of this kind of work is very considerably reduced.
The area of any plane figure is the measure of the space contained within its boundaries. The most simple of these is the square or parallelogram composed of two pairs of parallel boundaries and four right angles; and if one side of such figure be multiplied by one of its adjacent sides, the product will be the included area.
The next most simple figure for admeasurement is the triangle, in which one side, being multiplied by half the perpendicular, let fall thereon from the opposite angle, the product will be the area. Half the perpendicular is used because the area of every triangle is equal to half a parallelogram upon the same base, and of equal altitude, or between the same parallels; and "if a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle."
A trapezium is an irregular four-sided figure, of which the sides and angles may all be unequal; and the area of a trapezium is equal to the product of the diagonal drawn from one angle to another, by half the perpendiculars let fall thereon from the other two opposite angles, by which operation the figure is divided into two triangles with one side common to both.
A polygon is a figure bounded by more than four sides. In a
regular polygon the sides and angles are all equal, and in an irregular polygon they may be all unequal. The interior angles of a polygon are together equal to twice as many right angles as the figure has sides, less four. On this is based the theory of the traverse, of which further explanation will be given in another place. The area of any polygon may be ascertained by dividing the figure into trapezia and triangles, as in Fig. 39, where the polygon AB, CD.... GA, is so divided, and the whole area is equal to
9.28000, or 8-2-9.28. (See Page 3.)
The above division of a polygon into trapeziums and triangles is very commonly adopted in most offices; and where the figure is carefully divided, and the lengths carefully scaled or read off, it is very accurate. When the area to be computed consists of several fields, then the whole should be computed first, and then the fields separately. If the computation has been correctly made, the areas of the fields taken together will be equal to the area of the whole.
A little scaling, adding, and dividing may be saved in computing the trapeziums in the following manner :-With a pair of parallel rulers draw Ea Bb, and with a scale twice the size of that of your plan measure ab, with which to multiply the diagonal CF. The same operation may be repeated at Bc and G d.
To reduce a trapezium to a triangle of equal area.
In Figs. 40 and 41, set the edge of the parallels on A C, and through B draw the parallel line BE; join AE. The triangles DAE are equal to the trapeziums A C.
REDUCTION OF FIGURES.
To reduce a pentagon, or five-sided figure, to a triangle of equal area.
In Fig. 42 produce A B both ways; set the parallels on the points D and A, and make EG parallel to DA; join DG; set the parallels on the points D and B, and make CF parallel to DB; join DF. The triangle F D G is equal to the pentagon ABCDE.
To reduce hexagons, or six-sided figures, to triangles of equal area.
In Figs. 43 and 44 produce AB both ways; join two sides containing an angle, as A F and FE, by the line E A, and draw Fe parallel to EA; draw the diagonal Ee. The hexagon will be thus reduced to an equal pentagon, the side Ee being substituted for the sides AF and E F. Now make EF parallel to De, and join Df; proceed in a similar manner on the other side of the figure, and the triangle g Df will be the triangle required, and equal to ABCDEF.
To reduce a figure bounded by any number of sides to a triangle of equal area, Fig. 45.
Let ABCDEFGHIKLA be such a figure; produce A B on both sides; join the two sides K L and LA containing an angle, and make La parallel to KA; join K a, thus for the present substituting Ka for the sides K L and LA. Now draw Kb parallel to Ia, and join Ib; Ib is now substituted for IK and Ka; proceed thus throughout on both sides until the figure is reduced to the triangle g Gh, which will be the triangle of equal area required.
This method of computing is by far the easiest and most accurate; it is not necessary to draw all the dotted lines we have shown to explain the process; it is merely requisite to find the intersections on the line A B produced. After a very little practice with a good pair of parallel rulers any figure is very quickly reduced, and only one scaling and one computation are required. We recommend the system not only from our own experience, but with the knowledge that one of the principal computers in one of the Government offices, who has passed a large portion of twenty years in doing such work, finds it the most accurate and simple, after having tried every method that has been brought out. As to the many "succedaneums" so often proposed, we will not refer to them, as we know that they are not practised even by those who have advocated them.