or after noon. The difference between the computed and observed azimuth, is the variation. The variation may also be found from the amplitude of the sun or a star, which is an arc of the horizon, intercepted between the true east or west point, and the centre of the sun or a star, at its rising or setting. The amplitude is found by the following rule : As the cosine of the latitude is to radius, so is the sine of the sun's or star's declination, to the sine of the amplitude. Magnetical amplitude is an arc of the horizon, contained between the sun or a star, at its rising or setting, and the magnetical east or west point of the horizon, as pointed out by the azimuth compass. The difference between the observed and computed amplitude, is the variation of the needle. In the latitude of Dublin, viz. 53, when the sun's declination is 23° 28′, then That is, the sun then rises or sets 41° 26′ from the east or west point to the north or south, according as the declination is north or south. If the magnetic amplitude be observed by the compass to be 14° 11', then the difference is the variation west. The same conclusion may be arrived at by taking two equal altitudes and azimuths of the sun. For the meaning of the astronomical terms here introduced, see the author's Epitome of Astronomy. PROP. 2. Given the area, and map of a survey, to find the scale to which the map was laid down. Construct another map, exactly similar to the given one, to any scale you please; find its area: then say, as this area is the given area of the survey, so is the square of the scale to which this plan was laid down, to the square of the scale required; the square root of which will be the required scale. The area of a survey is 48A. 3R. 10p., and the area of a similar plan, constructed to a scale of 3 chains, or 12 perches to the inch, is 12A. OR. 32P.; required the scale to which the map was constructed. 48A. 3R. 10p. 7810 perches. Hence the map was laid down to a scale of 6 chains, or 24 perches, to one inch. PROP. 3.- To cut off any required number of acres from a given survey, by one line drawn through it. B E D C Let it be required to cut off 56 acres, 3 roods, and 8 perches, from the piece of land ABDC, commencing from AB. Draw FE at right angles to AC, cutting off, as nearly as you could guess, A the required area. Survey the figure ABEF, and find the difference betwixt its contents and 56 acres, 3 roods, 8 perches. Multiply this difference by 2, and divide the product by the length of EF in chains, then GFH measure the quotient from the point F, towards G or H, according as the space ABEF is found greater or less than 56 acres, 3 roods, 8 perches. Join GE, if ABEF was found greater than the required quantity; but if the space ABEF was found to contain less than the required quantity, join, HE. In the former case, ABEG contains 56A. 3R. 8P.; but in the latter case, the figure ABEH contains the required quantity. Having surveyed the figure ABEF, it was found to contain 58A. 3R. 20P., the line EF being 2000 links. 415000+2000=207.5 links. Here, after having measured the space ABEF, it was found to contain 2 acres, O roods, and 12 perches more than the required quantity; therefore the quotient, 207.5 links, is measured from F to G, and the space ABEG is the required space. But if the space ABEF was found to contain only 54 acres, 2 roods, and 36 perches, then the quotient, 207.5 links, should be measured from F to H, and the space ABEH would contain the exact quantity required. PROP. 4.- To divide a farm or estate into two parts, having a given ratio to each other. Survey the whole estate, and divide the acreage into two parts, having the given ratio to each other; then by the last proposition, cut off one of those parts from either extremity of the estate, by a line passing through it. If the last figure be made to represent a farm, containing 132A. 2R. 5 P., it is required to divide it into two parts, which shall be to each other in the ratio of 3 to 4. Here, having ascertained that one of the parts contains 56A. 3R. 8P., we proceed to cut it off from the survey by the line GE, exactly as shown in the last problem. PROP. 5.- To straighten a crooked boundary between two estates, without prejudice to either proprietor. Let the two estates be AP, AN; and the crooked line ABCDE, the boundary; it is required to draw a straight line GE to separate both estates, so that the sum of all the parts cut off by it from one estate may be equal to the sum of all the parts which it cuts off from the other. Measure the line FE, cutting off equal portions of each estate, as nearly as you can guess. Commencing at the point F, take insets and offsets to every corner and bend in the crookedline; the insets being the perpendiculars towards the right, and offsets the perpendiculars towards the left. Then find the sum of the areas of all the crooked spaces to the right of the line FE, and likewise the sum of the areas of all the crooked spaces to the right of it; and if these sums be equal, the right line FE is the required boundary; but if these sums be not equal, deduct one from the other, and divide twice the difference by the length of the line FE in links; lay off the quotient from F to G, or from F to H, according as the case may require; then GE or HE is the straight boundary which equalizes the portions taken from each estate. In laying out the first line FE, it should always, if possible, be at right angles to the line OM. When the line ABCDE is not visible from the point A, it becomes a matter of great difficulty, if not of impossibility, to find the direction of the new boundary line on the ground. The author met such a case in the course of his practice. Let us suppose that the points D, E, F, G, cannot be seen from the point A. In this case, determine the i |