Transcript Document 7452197

Theory of Errors in
Observations
Chapter 3
(continued)
50, 90 and 95 Percent Errors
The 50% error or probable error establishes the limits within an
observation has the same chance of falling within the limits or
outside of them.
The 90 and 95% errors are used to specify precisions required for
surveying projects.
E50  0.6745
E90  1.6449
E95  1.9599
Error of a Sum

Independently observed observations
–
Measurements made using different equipment,
under different environmental conditions, etc.
ESum   Ea2  Eb2  Ec2  ...
Where E represents any specified percentage error
And a, b and c represent separate, independent
observations
Example:


A line is observed in three sections with the lengths
1086.23 ± 0.05 ft, 569.08 ± 0.03 ft and 863.19 ± 0.04 ft.
Compute the total length and standard deviation for the
three sections.
Solution:
ESum  (0.05) 2  (0.03) 2  (0.04) 2  0.07 ft
Length  1086.23  569.08  863.19  2,518.50 ft
Probable length = 2,518.50 ± 0.07 ft
Error of a Series

Similar Quantities (Like Measurements)
–
Measurements taken by the same equipment and
under the same environmental conditions
ESeries   E n
Where E represents the error in each individual
observation and n is the number of observations
Example:

A field party is capable of making taping observations with
a standard deviation of ± 0.015 ft per 100-ft tape length.
What total standard deviation would be expected in a
distance of 500 ft taped by this party?
n = number of tape applications = 500’/100’ = 5
ESeries  0.015 5  0.03 ft
Probable length = 500.00 ± 0.03 ft
Error in a Product
E prod   A2 Eb2  B2 Ea2
Ea and Eb are the respective errors in the sides A and B.
A
-Ea
B
+Ea
- Eb
+Eb
Example:

A rectangular plot of land is surveyed and the following
measurements are recorded: 2245.68 ± 0.12 ft by 664.21 ±
0.06 ft. What is the area of the plot in acres and its expected
error in square feet?
Area = (2245.68)(664.21) = 1,491,603.11 sf
Area in acres = 1,491,603.11 ÷ 43,560 = 34.24 acres
E prod  (2245.68)2 (0.06)2  (664.21)2 (0.12)2  156.55sf
Error of the Mean
E
Em 
n
E is the specified percentage error of a single observation and n is
the number of observations
This equation shows that the error of the mean varies inversely as
the square root of the number of repetitions. In order to double the
accuracy of a set of measurements you must take four times as
many observations.
Weights of Observations
Precise observations should be weighted more heavily than less precise
observations.
____
MW
WM


W
MW is the weighted mean, W is the Weight assigned to each
measurement and M is the value of each measurement.
Wa
1
 a2
An equation can be deduced from this proportionality
which computes the relative weight of measurements
based on their precision.
Weights of Observations (con.)
Wa 
L
a
L
Wb 
b
L
Wc 
c
These equations can be used to determine
the relative weights of measurements
based on their standard errors.