PROBLEM III.

From an angle, in a given triangle, to draw lines to the opposite side, dividing the triangle into any number of parts, which shall be in any assigned proportion to each other.

Divide the base into the same number of parts, and in the same proportion, by Prob. I.; then from the several points of division draw lines to the proposed angle, and they will divide the triangle as required.*

EXAMPLE.

Let the triangle ABC, of 20 acres, be divided into five parts, which shall be in proportion to the numbers 1, 2, 3, 5, 9; the lines of division to be drawn from A to CB, whose length is c 1600 links.

Here 1+2+3+5+9=20, and 1600 20 80; which being multiplied by each of the proportional numbers, we have 80, 160, 240, 400, and 720. Therefore make Ca80, ab160, bc240, cd400, and dB 720; then, by drawing the lines Aa, Ab, Ac, Ad, the triangle is divided as required.

PROBLEM

* DEMONSTRATION. For the several parts are triangles of the same altitude, and which therefore are as their bases, which bases are taken in the assigned proportion.

PROBLEM IV.

From any point, in one side of a given triangle, to draw lines to the other two sides, dividing the triangle into any number of parts, which shall be in any assigned ratio.

K

I

B

H

E F D

From the given point D draw DB to the angle opposite the side AC, in which the point is taken; then divide the same side AC into as many parts AE, EF, FG, GC, and in the same proportion with the required parts of the triangle, as in the last Problem; and from the points of division draw lines EK, FI, GH, parallel to the line BD, and meeting the other sides of the triangle in K, I, H; lastly, draw KD, ID, HD; so shall ADK, KDI, IDHB, HDC, be the parts required."

G

An example of this is performed like that of the last Problem.

LEVELLING.

* DEMONSTRATION. The triangles ADK, KDI, IDB, being of the same height, are as their bases. AK, KI, IB; which, by means of the parallels EK, FI, DB, are as AE, ÈF, FD; in like manner, the triangles CDH, HDB are to each other as CG, GD but the two triangles IDB, BDH, having the same base BD, are to each other as the distances of I and H from BD, or as FD to DC; consequently the parts DAK, DKI, DIBH, DHC, are to each other as AE, EF, FG, GC.

LEVELLING.

EY LEVELLING we find how much one point on the surface of the earth is higher or lower than another; or the difference in their distances from the centre of the earth.

PROBLEM I.

To find the difference between the true and apparent levels at a given distance.

its points is called the line of true level. Hence DB is the height of the apparent above the true level at the distance AD or AB, being the difference of these heights above the centre of the earth. It is therefore evident, that the difference of the true and apparent height is always equal to the difference of the radius of the earth and the secant of the arc of distance.

By a property of the circle, we have 20D+DBX DB AB, or because the diameter of the earth is so great with respect to the line BD at all distances, to which an operation of levelling commonly extends, that 20D may safely be taken for 20D-DB, without any sensible

error; 20D X DB AB', or DB nearly.

AB'

AD'

2OD

20D.

That:

That is, the difference between the true and apparent level is equal to the square of the distance between the places, divided by the diameter of the earth; and consequently it is always proportional to the square of the dis

tance.

Now suppose, for example, we want to know the dif ference between the true and apparent level at the distance of an English mile, or 1760 yards: the square of this number is 3097600, which, being divided by the diameter of the earth 13953280, expressed in the same measure, gives o 222 of a yard, nearly; which being multiplied by 36, the number of inches in a yard, the product is 7′992, or nearly 8 inches, for the said difference.

NOTE 1. As the correction for reducing the apparent to the true level at the distance of 1 mile is very nearly 8 inches; therefore, as the square of 1 mile : 8 inches :: the square of any other distance the correction for curvature.

NOTE 2. A table of corrections for given distances may be easily computed by means of logarithms. For the logarithm of the diameter of the earth 13953280, expressed in yards, is 7.14468, (more figures being unnecessary) which being constantly subtracted from double the logarithms of the distances, the differences will be the logarithms of the corrections, expressed in decimal parts of a yard; which, being multiplied by 36, the number of inches in a yard, gives the corrections in inches and decimals of an inch.

PROBLEM

PROBLEM II.

To find the level of two places; or the ascent or descent from one to the other.

This is best done with a spirit level, having telescopic sights; which may be set horizontal by screws, that raise or lower the ends of the tube; and station staves with sliding vanes.

To take a level from A to I-let one assistant stand at A and another at C, and at a convenient place, as P, between A and C, place the level, and set it horizontal by means of the screws. Let the assistant at A hold the staff upright in his hand at A, while from P you look through the sights toward A, and direct him to slip the vane up or down till the white line at B be level with the sights; then note AB. Direct the other assistant to stand at C, and to hold the staff upright, then turning the instrument round at P, and looking toward C, direct him to raise or lower the vane till, through the sights, you can see the white line at D; then note the height above the ground, CD. And the difference of the heights, AB and CD, is the ascent or descent with respect to the apparent level, if there be no other station.

But, if there be many stations, make a table with two columns, one for the back stations, and the other for the fore stations. Now to proceed, set down the two heights AB, CD; and let the assistant at A go to F with his staff, and remove the instrument to Q and level it; then direct

the