Figure 8-6.—Ground elevations: (A) Telescope raised and (B) telescope depressed.DISTANCE AND ELEVATION FORINCLINED SIGHTS.— The following example willdescribe the use of the stadia reduction formulas forinclined sights. Assume you have a stadia interval of8.45 and an angle of elevation of 25014’, as shown infigure 8-6, view A. Let the instrument constant be 1.0.Substituting the known values in the stadia formulafor the horizontal distance, you haveh=kscos2a+(f+c)cosah = 100 (8.45) (0.90458)2 + (1) (0.90458) = 692.34The horizontal distance is 692 feet.Substituting the known values in the formula for thevertical distance, you havev = 50 (8.45) (0.77125) + (1) (0.42631)v = 326.28.The vertical distance to the middle-hair reading on therod is 326.28 feet.To find the elevation of the ground at the base of therod, subtract the center-hair rod reading from thisvertical distance and add the height of instrument (HI).(See fig. 8-6, view A). If the HI is 384.20 feet and thecenter-hair rod reading is 4.50 feet, then the groundelevation is326.28 - 4.5 + 384.20 = 705.98 feetIf the angle of inclination were depressed, then youwould have to add the center-hair rod reading to thevertical distance and subtract this sum from the HI. Asyou see from figure 8-6, view B, the ground elevationwould be384.2- (326.28 + 4.5) = 53.42 feet.STADIA TABLES.— You may save time in findingthe horizontal distance and the vertical distance(difference in elevation between two points) by usingthe stadia reduction tables in appendix II. Here the8-6
Integrated Publishing, Inc. - A (SDVOSB) Service Disabled Veteran Owned Small Business
