elevation of the base was 6,025 ft. above sea level, what was the elevation of the summit? SOLUTION.-From the first table, Height for 17.92 in. = 14,039.6 ft. Height for 24.15 in. = 5,910.5 ft. Approx. difference in elevation= 8,129.1 ft. As T+1=90°, from the second table, C=-.0058, and z=8,129.1X(1 − .0058) = Elevation, top of mountain Use of Barometer.-The mining engineer ordinarily has but one aneroid and this is not commonly provided with a thermometer. With a single instrument reliable results are difficult to obtain, and depend as much on good and uniform weather conditions as on the skill and carefulness of the engineer. In fact, the results obtained during storms, whether of wind or rain, are not to be relied on at all. The single aneroid, then, should be used only when weather conditions are of the best. In the morning, before starting out, its reading should be noted and recorded. If a thermometer is available, it, too, should be read. As the aneroid is usually employed in exploratory work in which more or less time is required to examine various coal openings, it should be read a few minutes after reaching an opening and again on leaving, the times and temperatures being noted as well. After the examination of one opening is completed, the engineer should hasten as rapidly as possible to the next; that is, he should move rapidly from place to place but should remain a sufficient time at each to estimate the changes in pressure (and consequently in apparent elevation) that are taking place. By taking two observations at each opening at intervals of, say, hr., a correction curve showing the changes in pressure may be worked out by means of which an allowance may be made for these changes while the barometer is being carried from place to place. Thus, a reading at Sta. A at 9.15 A. M., showed an elevation of 810 ft., and a reading at Sta. B at 9.30 A. M. showed one of 860 ft., and a second at B at 10 A. M. indicated 875 ft. This last reading shows a change in apparent elevation of 15 ft. in 30 min., or at the rate of .5 ft. per min. The apparent difference in elevation between A and B (first readings) is 50 ft., but a part of this difference is due to a change in the atmospheric pressure. The two readings at Sta. B show that the elevations are apparently increasing at the rate of .5 ft. per min.; therefore, as it took 15 min. to go from A to B, the elevation of B over A apparently increased 7.5 ft. Hence, the actual difference in elevation between A and B is not 50 ft. but 50-7.5-42.5 ft. Very satisfactory results may be obtained in this way; and if the time between stations is short, corrections for changes in temperature, etc., need not be made, provided the difference in elevation is not great. By taking double readings at each station, a continuous curve can be worked out and applied to correcting the day's observations. If possible, the aneroid should be reread at the various openings on the way home, and the mean of the afternoon and morning readings taken as the true reading. The instrument should be read upon reaching the starting point at night and again when leaving the next morning, When two barometers are available, their readings are compared at starting, one being carried into the field and the other retained at headquarters where it is read by an assistant throughout the day at intervals of 10 or 15 min., a record being kept of the time, temperature, and pressure. The field man, then, need take but one reading at a station, preferably just before leaving, and should likewise note the temperature and time. These double readings, unless the field-man is so far from the base that weather conditions are markedly different, afford a complete check on fluctuations in pressure due to changed atmospheric conditions. Care of the Barometer.-The aneroid should not be removed from its case; should not be subjected to violent jars; nor exposed to or read while affected by artificial heat. It should be read in a horizontal position and on sunny days should be allowed to remain in the shade for, say, 5 min. before being read, that all its parts may have time to assume the temperature of the air at the station. PRACTICAL PROBLEMS IN SURVEYING 1. To Prolong a Straight Line. Let AB, Fig. 1, be a straight line whose position on the ground is fixed by stakes set at A and B, and let it be required 4 B FIG. 1 to prolong the line to C. This can be done in two ways; namely, by foresight only, or by backsight and foresight, the latter method being commonly called backsight. By Foresight. The transit is set over the point at A, and the line of sight directed to a flag held at B; if the point C is to be set at a given distance from B, the chainmen measure the required distance, the head chainman being kept in line by the transitman. When the required distance has been measured, the point C, which evidently lies in the prolongation of AB, is marked by a stake or otherwise. By Backsight.-The transit is set over the point at B and a sight taken on a flag held at A. The telescope is then plunged so that it is directed along the prolongation of AB. Any re quired distance BC may then be measured from B in the direction indicated by the line of sight. 2. To Run a Line Over a Hill When the Ends of the Line Are Invisible From Each Other.-The points A and B, Fig. 2, are supposed to be on the opposite sides of a hill, and to be FIG. 2 invisible from each other. It is desired to run a line between them, or to locate some intermediate points. Having set two poles at A and B, two flagmen with poles station themselves at C and D, approximately in line with A and B, and in such positions 50 30 FIG. 3 40 B that the poles at B and D are visible from C, and those at C and A are visible from D. The flagman at C lines in the pole at D between C and B, and then the flagman at D lines in that at C between D and A. Then the flagman at C again lines in that at D, and so on, until C is in the line between D and A at the same time that D is in line between C and B. The points C and D will then be in line with A and B. 3. To Erect a Perpendicular to a Line at a Given Point. Let it be required to erect a perpendicular to the line AB at the point B, Fig. 3. A triangle whose sides are in the proportion of 3, 4, and 5 is a right triangle, the longest side being the hypotenuse; for 52-42+32. The following method is based on this principle: Lay off on BA a distance BC of 30 ft. (or li.). Fix one end of the chain at one of the extremities as C, and the end of the ninetieth link at the other extremity B. Hold the end of the fiftieth link and draw the chain until both parts are taut. The point D where the end of the fiftieth link is held will then be a point in the perpendicular, and the direction of the latter will therefore be BD. B The distance BC may be any other convenient multiple of 3. In general, if BC is denoted by 3a, BD must be 4a, and CD must be 5a. Thus, BC may be made equal to 21 (=3X7) li.; in which case BD must be 4X7-28, and CD must be 5X7 35 li. As 35+28-63, one end of the chain must be fixed at one of the extremities of BC, the end of the sixty-third link at the other extremity, and the chain pulled from the end of the thirty-fifth link until both parts are taut. 4. To Determine the Angle Between Two Lines.-Let AD and AE, C FIG. 4 Fig. 4, be two lines on the ground. To determine the angle DAE, measure off from A on AD and AE equal distances AB and AC. Then the angle DAE is calculated from the relation Measure the distance BC. EXAMPLE.-If AB and AC are each 100 ft. and BC is 57.6 ft., what is the value of the angle DAE? SOLUTION. Substituting the values of BC and AB in the formula, sin DAE= X57.6 = .28800; whence, DAE= 16° 44′, nearly; and, therefore, DAE= 16° 44′X2=33° 28'. 5. To Find the Distance of an Inaccessible Point.-Case I.-Let it be required to determine the distance from the point B to an inaccessible point P, FIG. 5 B Fig. 5. Measure BC in any convenient direction and run a line A'D' parallel to BC. Measure AD, the distance between the points where the lines PB and PC intersect A'D'. Measure also AB. Then, 124.2-100 =216.5 ft. Case II.-Measure a horizontal base line AB, Fig. 6, and take the angles formed by the lines BAC and ABC, which gives two angles and the included side. Assuming the angle A to be 60°; the angle B 50°, and the side AB 500 ft., angle C=180°- (60°+50°) =70° Then, sin 70°: AB= sin A: BC, and sin 70° AB = sin B: AC; or, .939693: 500 =.866025: BC, or 460.8+, and .939693: 500 -.766044 AC, or 407.6+. By logarithms: B 6. To Determine the Distance Between Two Points Invisible From Each Other or Separated by an Impassable Barrier.-Case 1.-Let it be required to find the distance between two points A and B, Fig. 7, that are invisible from each other. First run a random line AD' in such a manner that it will pass as near B as can be estimated. From B drop FIG. 7 D D' a perpendicular BD on AD and compute the required distance A B by the formula AB=√AD2+BD2 EXAMPLE.-If, in Fig. 7, the distance AD is 206.1 ft. and the distance BD is 35.1 ft., what is the distance from A to B? SOLUTION.-Here AD=206.1 and BD=35.1; therefore, substituting in the formula, AB= √206.12+35.12=209.1 ft. Case II.-Select any convenient station, as C, Fig. 8, measure the lines CA and CB, and the angle included between these sides, so as to obtain two sides and the included angle. Assuming the angle C to be 60°, the side CA, 600 ft., and the side CB, 500 ft., the following formula is obtained: B-A B = A+B : tan 2 C 68° 57', or angle B, and 60°-8° 57' 51° 03', or angle A, Then, 1,100 Having found the angles, find the third side by the same method as case II, of problem 5. The foregoing formula, worked out by logarithms, is as follows: 60°+8° 57'- 68° 57′, or angle B, and 60°-8° 57' 51° 03', or angle A NOTE. The greater angle is always opposite the greater side. 7. To Find the Distance Between Two Inaccessible Objects When Points Can Be Found From Which Both Objects May Be Seen. Let AB, Fig. 9, be the line, the ends A and B of which are inaccessible. Select two points P and Q from which both ends of the line can be seen, and at a distance from each other of about 300 or 400 ft. Measure the line PQ, and the angles K, L, M, and N. FIG. 9 in which R = 180° − (K+L) – M. From triangle BPQ, AP= PQ sin M sin (X-Y) EXAMPLE.-If, in Fig. 9, the distance PQ is 400 ft., and the angles, as measured, are K=37° 10′, L=36° 30', M=52° 15', N=32° 55', what is the distance AB? SOLUTION.-In the triangle APQ, R=180°- (37° 10′+36° 30'+52° 15') =54° 05', and 400 sin 52° 15' AP sin 54° 05' =390.53 ft. In the triangle BPQ, S=180°- (36° 30'+52° 15′+32° 55′) = 58° 20′, M+N =52° 15'+32° 55′ = 85° 10', and BP = 400 sin 85° 10' Also, K=37° 10', K= 18° 35', and =468.30 ft. sin 58° 20' tan } (X-Y) = (468.30-390.53) cot 18° 35' =283.58 ft. whence, (X-Y) = 15° 04', and therefore AB= (468.30-390.53) cos 18° 35' sin 15° 04' 8. To Determine the Angle Between Two Lines AB and CD, Whose Point of Intersection P is Inaccessible, Also, the Distances BP and DP.—This problem is of frequent occurrence in railroad work, the two given lines being the center lines of two straight tracks that are to be connected by a curve Measure the distance BD, Fig. 10, and the angles K and L. Then, M = 180° are measured on the ground. The area of each triangle may then be determined by the formula A = √s(s—a) (s—b) (s—c) in which a, b, and c represent the three sides and s represents one-half of their EXAMPLE.-In Fig. 11, let the lengths of the sides be as follows: AB =320 ft., BC=217 ft., CD=196 ft., DE=285 ft., and EA304 ft. It is required to calculate the length of the diagonal BE by means of a tie-line. SOLUTION.-Let the line BA be prolonged 100 ft. beyond A; that is, make AFXAE AF 100 ft. Then AG must be equal to AB the length of GF, as found by measurement, be 125 ft. |