## 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%.

 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