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up a rod exactly in line with EG, in the direction of H, and make G F equal to EG; FP will be the perpendicular required. If you have a line equal to two chains in length instead of only one, so much the better, only then the triangle EGP will be isosceles instead of equilateral. Also you will have the following proportion :-ED: DG :: EP: PF.

PROBLEM 4. Fig. 19. From a given point to let fall a perpendicular to a given line.

From P as centre with any radius greater than PC, describe the arc D E, and from D and E, with the same radius, describe the arcs intersecting at R; join P R, and PC will be the perpendicular required.

PROBLEM 5. Fig. 20. On the ground, from a given point to let fall a perpendicular

to a given line. Let P be the given point, and A B the given line; from P towards A B lay out any two lines, as PC and PD; set up two rods at C and D), taking care that they are exactly on the line A B; make P E and PF each equal to one chain, and stick a couple of pegs at E and F, taking care that they are exactly on the lines PC and PD respectively ; measure PC, and make PG equal to PC; measure CG, EF, and G D, and take the following proportion :

CG: EF :: GD : FH. From F to H set out this fourth proportional just found, being careful that H is exactly on the line PD. PE being equal to one chain, take a line exactly equal to it, fasten one end to an arrow, or peg, at P, and bring the line round so that it shall exactly intersect E H at I. This can be done by a person standing at E sighting the one holding the end of the line at I until it is exactly on the line EH. Now measure E I, and bisect it at K; a line produced through PK will fall on L, and be perpendicular to A B.

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PROBLEM 6. Fig. 21. On the ground, through a given point to set out a straight

line parallel to another straight line. Let P be the given point, and A B the given straight line. From any convenient point, C on the line A B, set out through

INACCESSIBLE DISTANCES,

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the point P the line C D, making PD one or two chains in length. From D set out the line D E, carefully placing E on the line A B; make D F equal to P D, and D G equal to DC. Now measure C G and G E, and take the same proportion as that given above, though in other letters; that is,

AG : PF :: GE : FH. Set out the length of F H thus found. PH is parallel to A B. If the required parallel is to be of any length, set up a rod as Tin line with PH. If the prolongation of PH is of any considerable length the work may be tested; produce CG to intersect P T at I; measure off I L and I K equal to two or three chains each, and measure LK. Then proceeding to C, make C M and C N each equal to I K and I L, and N M should be equal to L K.

Or, use the following means. Let P, Fig. 22, be the given point through which it is required to set out a parallel to the straight line A B. With rods only set up the line A P; make AC two or three chains in length, and set off by some of the foregoing rules the line C D perpendicular to AP, and produce it towards G. Make P E equal to AC, set up the perpendicular E F, and make E F equal to C D, through F sight P F G, which will be parallel to AB. Measure EH and CG, which should be equal to it. Should there be found any slight inaccuracy, correct it.

Should A B, Fig. 23, be inaccessible, sight the line with ranging rods to any convenient distance, as to C, and with ranging rods set out CP; measure off two or three chains along CP as C D, and from D set out D E perpendicular to CP, and measure D E. Make PF equal to CD, and the perpendicular FG equal to D E. Through P set off GPI, which will be parallel to A B. Produce the perpendicular D E towards H, and take the following proportion,

CD : DE :: PD: D H. Measure off DH accordingly, when H I will be in line with the points I, P, and G.

By similar means we may often obtain the length of inaccessible distances, as in the following problems

PROBLEM 7. Fig. 24. To ascertain the length of the inaccessible distance A B

through a sheet of water. Sight A E on any convenient object, and measure on to C,

where

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observe you can run clear of the obstruction up to B, and leave a mark at C; make CE equal to A C, and leave a mark at E; now measure BC, and make CD, being B C produced, equal to BC. The parallel line D E will be equal to the inaccessible length A B.

Now let there be another obstruction, as in Fig. 25, such that you cannot make C E equal to A C. Measure out C E as far as you can conveniently, and having measured BC, take the following proportion,

AC: CE :: BC: CD. Then make C D equal to this fourth proportion thus found. Now measure D E, and take the following proportion,

EC : AC :: ED : AB; which fourth proportional gives the length required.

Again, let there be such an obstruction at C, Fig. 26, that we cannot proceed beyond. Produce C B to D, and make BD equal to BC; now from C B cut off C F, and from C A cut off CE, so that F E shall not exceed one chain in length. Make D G equal to CF, and on G D set off the triangle DHG equal to the triangle CEF; this may be done with the chain and a tape, making the length of one equal to C E as D H, and the other equal to FE as G ; their intersection will give the point H. Through this point produce DH to I on the main line A B, when I В will be equal to A B. This may be tested by measuring D I, which should be equal to CA.

It may also happen that there is another obstruction on the other side of the main line, as at F, Fig. 27, so that what we require cannot be accomplished by the above means; then pro. ceed in the following manner :- Make BF equal to CD, and in the same manner as before make the triangle F G B equal to the triangle CED; produce F G on to I on A B produced, when

FB: CB: IB: BA,

І В which fourth proportion, therefore, gives the length required.

Or, at the other end of A B produce C A , and make A K equal to AC; at K set off the triangle equal and similar to C DE, and produce K L on to M; M A is equal to A B.

. The above are based on the 4th, 15th, 27th, and 28th propositions of the first book, and 4th, 5th, and 6th of the sixth book of Euclid.

PROBLEM 8. Fig. 28. To bisect a given angle on the ground. Let A B C be the given angle ; make B D and B E equal, and

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GEOMETRY ON THE GROUND.

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measure DE, and make D F equal to half D E, B F will bisect the angle.

PROBLEM 9. Fig. 29. To lay out on the ground, from a given point, an angle equal

to a given inaccessible angle.

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Let P be the given point, and A CB the given angle. With rods sight AC to E, and BC to F, and from any convenient points, as E and F on these lines, sight EP and FP; then, by the second part of Problem 6, set out the parallels Pd and Pe; e Pd will be the angle required.

PROBLEM 10. Fig. 30. To luy out on the ground two lines parallel to two given inaccessible lines containing any angle, and at a given distance from the two given lines.

Let A B and C B be the two given lines; with the ranging rods produce A B to any point D, and C B to any point E. At any convenient point on B E, as F, set up the perpendicular FG, which make equal to the given distance, and leave a mark midway at c; leave marks also at a and b, making a F equal to Fb; also make Gd equal to a For Fb. At any other con

; venient point, I on B, repeat this operation ; K G produced will be parallel to BC, and at the given distance from it, and if the work is correct, a c, b c, and c d will all be equal to each other. Exactly the same operation repeated on A B produced, D will give the parallel as required to A B.

PROBLEM 11. Fig. 31. To lay down on the ground an angle equal to a given angle.

Let A B C be the given angle ; from B to C measure off any convenient distance as B C, and from C set up the perpendicular CD, intersecting BA at D, and measure CD, as also B D. lay down the required angle, set off E F equal to BC, and perpendicular to it F G, which make equal to C D, and lay off the line EG;FEG will be equal to the angle C BA as required, and E G will be equal to B D.

PROBLEM 12. Fig. 32. To bisect on the ground a given inaccessible angle, as, for

instance, to find the capital of a bustion. Let A B C be the given angle which it is required to bisect; produce C B indefinitely as to D, and A B to E; parallel to these, by the second part of Problem 6, set out the parallels FO and GO;a line through O B will bisect the angle as required. To check this, make 0 1 and O K equal, and from I and K set off the perpendiculars I L and KL; their point of intersection at L will be in line with O B.

Amongst the many propositions of Euclid which are of the greatest service in engineering field-work, the four following are of everyday use in surveying :

The 13th, Book I. The angles which one straight line makes with another upon one side of it, are either two right angles, or · are together equal to two right angles.

The 14th, Book I. If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

The 15th, Book I. If two straight lines cut one another, the vertical or opposite angles shall be equal.

The 17th, Book I. Any two angles of a triangle are together less than two right angles.

The 18th, Book I. The greater side of every triangle is opposite to the greater angle.

The 32nd, Book I. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.

Reference to Fig. 33 will illustrate the value of the above in triangular surveying. The line DC falls upon the straight line A B at C; then if the angles A CD and DCB measure equal to two right angles, then A B is a straight line. And if the angle A C D measure equal to the angle ECB, then A B and D E are straight lines. Now where, in triangulating a district, it so happens that such lines intersect on a narrow ridge of high ground, over which it has been difficult to sight them, and to make certain of the exact position of intermediate stations along them, this circumstance enables us, in the first place, to ascertain that C is the true intersection of the bases, and also to fix any number of points on the straight lines down in the valleys below. From such a point we may also fix the direction relative to the

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