Metric Problems

Problem: Determine the pace factor for an individual that counted 132, 134, and 133 paces for a distance of 100 meters.

Solution:

Distance 100 m 100 m m PF =-= , , =-= 0.7518 or0.75

Problem: An individual with a pace factor of 0.65 m/pace counts 380 paces for the unknown distance. What is the distance?

Solution:

Problem: A wheel with a diameter of 70 cm is used to measure a distance by rolling it along the ground. It was rolled 124 revolutions. What is the distance?

n x dia

D =-x revolutions revolution n x 70 cm 1m

1 rev 100 cm

Problem: Use Table 12.2 to determine the horizontal distance (HD) when the slope distance (SD) is 350 m and the slope is 12.0%.

Table 12.2. Correction factors for converting slope distance to horizontal distance.

True horizontal

Correction factor

distance

Slope

(m/100 m)

(100 m)

1

0.01

99.99

2

0.02

99.98

3

0.05

99.95

4

0.08

99.92

5

0.13

99.87

6

0.18

99.82

7

0.25

99.75

8

0.32

99.68

9

0.41

99.59

10

0.50

99.50

11

0.61

99.39

12

0.72

99.28

13

1.13

99.15

14

1.63

99.02

15

1.13

98.87

18

1.63

98.37

25

3.18

96.82

30

4.61

95.39

= 350m -ii x 0.72 j = 350 m - 2.52 = 347.48 or 47.5 m

Problem: Determine the horizontal distance when the stadia method is used with a metric rod. The top stadia reading is 1.23 m and the bottom stadia reading is 0.82 m.

Solution:

Problem: Determine the corrected distance for a measured distance of 750.5 m when the instrument measured distances of 49.8,49.8,49.7, and 49.8 meters for a standard distance of 50 m.

Solution: The first step is to determine the correction factor. Before determining the correction factor the average measurement must be calculated.

m \ Known distance - Measurement m

50 m m of distance

The last step is to use the correction factor to determine the corrected distance.

Distance = Measurement + Correction

0 0

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