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DISTANCE AND ELEVATION FOR INCLINED SIGHTS. The following example will describe the use of the stadia reduction formulas for inclined sights. Assume you have a stadia interval of 8.45 and an angle of elevation of 25014, as shown in figure 8-6, view A. Let the instrument constant be 1.0. Substituting the known values in the stadia formula for the horizontal distance, you have

h = 100 (8.45) (0.90458)2 + (1) (0.90458) = 692.34

The horizontal distance is 692 feet.

Substituting the known values in the formula for the vertical distance, you have

v = 50 (8.45) (0.77125) + (1) (0.42631)

v = 326.28.

The vertical distance to the middle-hair reading on the rod is 326.28 feet.

To find the elevation of the ground at the base of the rod, subtract the center-hair rod reading from this vertical distance and add the height of instrument (HI). (See fig. 8-6, view A). If the HI is 384.20 feet and the center-hair rod reading is 4.50 feet, then the ground elevation is

326.28 - 4.5 + 384.20 = 705.98 feet

If the angle of inclination were depressed, then you would have to add the center-hair rod reading to the vertical distance and subtract this sum from the HI. As you see from figure 8-6, view B, the ground elevation would be

384.2- (326.28 + 4.5) = 53.42 feet.

STADIA TABLES. You may save time in finding the horizontal distance and the vertical distance (difference in elevation between two points) by using the stadia reduction tables in appendix II. Here the values of 100 cos2a and 1/2(100) sin 2a are already computed at 2-minute intervals for angles up to 30. You need to multiply the values in the table by the stadia reading, then add the value of the instrument constant given at the bottom of the page.

To find the values from the stadia table, for the example that we have been discussing, read under 25 and opposite 14. Under Hor. Dist. you find that

100 COS22514 = 81.83.

Under Diff. Elev. you see that

1/2 (100) sin 2 (2514) = 38.56.

The values of the term containing the instrument constant are given at the bottom of the page.

For

You find

Therefore

Using these values in the formulas, you have

and







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