## 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

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