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found and multiplied by the divisor, the products will be the latitude and departure of the whole course. It is also plain, that the latitude or departure of two or more courses, having the same bearing, is equal to the sum of the latitudes or departures of the courses taken separately.

Hence, if we have any number greater than 100, as 614, we have only to regard the last figure as a cipher, and recollect that, 610+4=614; and also, that the latitude and departure of 610, are ten times greater, respectively, than the latitude and departure of 61: that is, equal to the latitude and departure of 61 multiplied by 10, or with the decimal point removed one place to the right.

1. To find the latitude and departure for the bearing 291, and the course 614. Latitude for 610 530.90 Departure for 610 300.40 Latitude for 4 3.48 Departure for 4

1.97 Latitude for 614 534.38 Departure for 614 302.37

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In this example, the latitude and departure answering to the bearing 291', and to the distance 61, are first taken from the table, and the decimal point removed one place to the right: this gives the latitude and departure for the distance 610; the latitude and departure answering to the same bearing and the distance 4, are then taken from the table and added.

2. To find the latitude and departure for the bearing 621, and the course 7855 chains. Latitude for 7800 3602.00 Departure for 7800 6919.00 Latitude for 55 25.40 Departure for 55

48.79 Latitude for 7855 3627.40 Departure for 7855 6967.79

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REMARK. When the distances are expressed in whole numbers and decimals, the manner of finding the latitudes and departures is still the same, except in pointing off the places for decimals : but this is not difficult, when it is remembered that the column of distances in the table, may regarded as decimals, by removing the decimal point to the

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3. To find the latitude and departure for the bearing 47, and the course 37.57. Latitude for 37.00 • 24.88 | Departure for 37.00 27.39 Latitude for 57 38 Departure for 57

42 Latitude for 37.57 25.26 Departure for 37.57 27.81

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Of Balancing the work. 138. The use of the traverse table being explained, we can proceed to compute the area of the ground.

The field notes having been completed, rule a new table, as below, with four additional columns, two for latitude, and two for departure.

Then find, from the traverse table, the latitude and de. parture of each course, and enter them in the proper columns opposite the station.

Then add up the column of northings, and also the column of southings : the two sums should be equal to each other. If they are not, subtract the less from the greater, and the remainder is called the error in latitude. This error takes the name of that column which is the least. For example, if the sum of the northings is less than the sum of the southings, the error is called, error in northing : but if the sum of the southings is less than the sum of the northings, the error is called, error in southing. We find the crror for each particular course by the following proportion.

As the sum of the courses
Is to the error of latitude,
So is each particular course

To its correction. The error of each course, thus found, may be entered in a separate column; after which, add it to the latitude of the course, when the error and latitude are of the same name, but subtract it, when they are of different names. This will make the sum of the northings equal to the sum of the southings, and is called balancing the work. The northings and southings, thus corrected, are entered in columns on the right, under the head, balanced. Having done this, balance the eastings and westings in the very same manner. The difFor an example, we will resume the same example that has already been considered.

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4 S 45° W 10.40 7.29 7.41 -0.19 +0.02 Sum of courses, 37.25 12.76 13.44 12.69 12.63

12.76 12.63 Error in Northing, :

0.68

0.06 Error in Westing.

13.11/13.11 12.67 12.67

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As 37.25 : 0.68 :: 10 : 0.18 error in lat, of 1st course. As 37.25 : 0.68 :: 9.25 : 0.17 error in lat. of 2d course. As 37.25 : 0.68 : : 7.60 : 0.14* error in lat. of 3d course. As 37.25 : 0.68 :: 10.40 : 0.19 error in lat. of 4th course.

As 37.25 : 0.06 :: 10 : 0.02* error in dep. of 1st course. As 37.25 : 0.06 :: 9.25 : 0.01 error in dep. of 2d course. As 37.25 : 0.06 :: 7.60 : 0.01 error in dep. of 3d course. As 37.25 : 0.06 :: 10.40 : 0.02 error in dep. of 4th course.

139. REMARK I. In finding the error in latitude or departure, for a particular course, the last figure is sometimes doubtful; in which case it is best to mark it, as in the third proportion for error in latitude, and the first for error in departure; and then, if the figures taken do not balance the work, let each be increased or diminished by 1.

140. REMARK II. It has already been observed (Art. 136), that if the measurements on the field are correctly made, the sums of the northings and southings will be equal to each other, as also those of the eastings and westings. It is the opinion of some surveyors, that when the error in latitude or departure exceeds one link for every five chains of the courses, the field notes ought not to be relied on. This, perhaps, is a higher degree of accuracy than can be attained. The error, however, should always be made considerably less than one link to a chain.

Of the double meridian distances of the courses. 141. After the work has been balanced, the next thing to be done is to calculate the double meridian distance of each course.

For this purpose, a meridian line is assumed, lying either wholly without the land, or passing through any point within it. It is, however, most convenient to take that meridian which passes through the most easterly or westerly station of the survey; and these two stations are readily determined by inspecting the field notes.

Having chosen the meridian, let the station through which it passes, be called the principal station, and the course which begins at this point, the first course. Care, however, must be taken, not to confound this with the course which begins at station 1, and which is the first course that is entered in the field notes.

It has already been remarked (Art. 132), that all departures in the direction east, are considered as plus, and all departures in the direction west, as minus ; then, through whatever station of the survey the assumed meridian be taken, we shall have for the calculation of the double meridian distances, the following

RULE, I. The double meridian distance of the first course is equal to its departure.

II. The double meridian distance of the next course is equal to the double meridian distance of the first course, plus its des parture, plus the departure of the second course.

III. The double meridian distance of the third course is equal to the double meridian distance of the second, plus its departure, plus the departure of the third course.

IV. And, the double meridian distance of any course is equal to the double meridian distance of the preceding course, plus ils departure, plus the departure of the course itself.

REMARK. It should be recollected that plus is here used in its algebraic sense, and that when double the meridian distance of a course and the departure which is to be added to it, are of different names, that is, one east and the other west, they will have contrary algebraic signs ; hence, their algebraic sum will be expressed by their difference, with the

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Demonstration of the Rule. Let the figure ABCD, which we

N have already surveyed with the compass, be resumed. By inspecting the field notes, it will be seen that B, or station 2, is the most westerly B station. Through this point let the assumed meridian NS be supposed 6 to pass. Then, B will be the principal station, and BC the first course, By what has been already said, every

Л departure towards the east is to be S considered as plus, and every departure towards the west, as minus.

Now, since p, k, d and a, are the middle points of the courses BC, CD, DA and AB, we have, by similar triangles,

> qp=2 şx=sC=the first departure.
2 Cr=2 hk=Cy=the second departure.
2 fg=2 GA=Af=the third departure.

2 At=? ab=Ac=the fourth departure, We also have,

2 qp=sC=doub. mer. dis. of BC.
2 p+2 xC+2 Cr=2 kn=doub, mer. dis. of CD.
2 kn + 2 kh 2 gf=2 de=doub. mer. dis. of D.A.

2 de-2 g.A-2 Ai=2 ab=doub. mer, dis. of AB. The depasture of the courses BC, CD, are east, and therefore positive ; while tho departures of the courses DA, AB, are west, and consequently negative.

Since the course of reasoning just pursued is applicable to all figures, we may regard the rule as demonstrated for every case which can occur.

REMARK. The double meridian distance of the last course should be equal to the departure of that course.

A verification of the work is, therefore, obtained by comparing this double meridian distance with the departure of the course.

142. To apply the above rule to the particular example already considered, rule a new table, as below, in which are entered the balanced northings and southings, and the bal

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