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At the end of the first station, or at 2, which is the beginning of the second, with the pin place the centre of the protractor, turning the arc up, because the bearing of the second station is more than 180, viz: 348. Place your protractor parallel as before, and by the edge of the semicircle, with the pin prick at that degree, thro' which and the end of the foregoing station, draw a blank line, and on it set the distance of that station.
In the like manner proceed thro' the whole, only observe to turn the arc of your protractor down, when the degrees are less than 180.
If you lay off the stationary distances by the edge of the protractor, it is necessary to observe, that if your map is to be laid down by a scale of 40 perches to an inch, every division on the protractor's edge will be one two-pole chain; } a division will be 25 links, and of a division will be 12; links.
your map is to be laid down by a scale of 20 perches to an inch two divisions will be one twopole chain ; one division will be 25 links ; } a division 123 links, and of a division will be 61 links.
In the general, if 25 links be multiplied by the number of perches to an inch the map is to be laid down by, and the product be divided by 26 för which is the same thing, if you cut off one and take the half) you will have the value of one division on the protractor's edge, in links and parts.
1. How many links in a division, if a map be laid down by a scale of 8 perches to an inch?
10 links. Answer.
2. How many links in a division, if a map
a be laid down by a scale of 10 perches to an inch?
12.5 or 123 links. Answer.
And so of any other.
To protract a field-book, taken by the angles of the
Note. We here suppose the land surveyed is kept on the right-hand as you survey.
Draw a blank line with a ruler of a length greater than the diameter of the protractor; pitch upon any convenient point therein, to which apply the centre-hole of your protractor with your pin, turning the arc upwards if the angle be less than 180, and downwards if more; and observe to keep the upper edge of the scale, or 180 and 0 degrees
upon the line; then prick off the number of degrees contained in the given angle, and draw a line from the first point through the point at the degrees; upon which lay the stationary distance. Let this line be lengthened forwards and backwards, keeping your first station to the right, and second to the left ; and lay the centre of your protractor over the second station, with your pin, turning the arc upwards, if the angle be less than 180, and downwards, if more ; and keeping the 180 and 0 degrees on the line, prick off the number of degrees contained in the given angle, and thro' that point and the last station draw a line, on which lay the stationary distance : and in like manner proceed through the whole.
In all protractions, if the end of the last station falls exactly in the point you began at, the fieldwork and protraction are truly taken and performed, if not, an error must have been committed in one of them : in such case make a second protraction : if this agrees with the former, and neither meet nor close, the fault is in the field-work, and not in the protraction; and then a re-survey must be taken.
Containing two Methods by which the Areas of
right-lined Figures may be determined.
HE area or content of any plane surface in
perches, is the number of square perches, that surface contains.
Plate VII, fig. 1..
Let ABCD represent a rectangular parallelogram or oblong : let the side AB, or DC, contain 8 equal parts: and the side AD, or BC, three of such parts ; let the line AB be moved in the direction of AD, till it has come to EF; where AE, or BF (the distance of it from its first situation) may be equal to one of the equal parts. Here it is evident, that the generated oblong ABEF, will contain as many squares as the side AB contains equal parts, which are 8; each square having for its side one of the equal parts, into which AB, or AD, is divided. Again; let AB move on till it comes to GH, so as GE, or HF, may be equal to AE, or BF; then it is plain that the oblong AGHB, will contain twice as many
squares as the side AB contains equal parts. After the same manner it will appear, that the oblong ADCB will contain three times as many squares as the side AB contains equal parts; and in general, that every rectangular parallelogram, whether square or oblong, contains as many squares as the product of the number of equal parts in the base, multiplied into the number of the same equal parts in the height, contains units, each square having for its side one of the equal parts.
Hence arises the solution of the following problems.
To find the content of a square piece of ground.
1. Multiply the base in perches, into the perpendicular in perches (or square the base) the product will be the content in perches; and because 160 perches make an acre, it must thence follow, that
Any area, or content in perches, being divided by 160, will give the content in acres; the remaining perches, if more than 40, being divided by 40, will give the roods, and the last remainder, if any, will be perches.
Or thus :
2. Square the side in four-pole chains and links, and the product will be square four-pole chains and links ; divide this by 10, or cut off one more than the decimals, which are five in all, from the right towards the left : the figures