Units and errors


  • SI units
  • imperial units
For measurements and computations we use SI units.
For our purposes is most important 1 meter. 1 m is defined as a length of the path traveled by light in a vacuum during a time interval of 1/299,792,458 s.
It is also good idea to have some background about imperial units, which are non-standard and used mostly in British Empire since they are still widely used there:
Imperial units
Angles are measured in degrees. 360 ° = 2 π. It is also needed skill to know how to convert
between A) decimal notation in degrees and B) notation when degrees are written with minutes and seconds behind them. For example,
12°15'18" = 12.255 °.

Probability basics and errors of measurements

Surveying is based on measurements of, usually, lengths and angles. It is important to understand errors, precision and accuracy.

If we make a measurement, we only find an approximation of the true value. We can get closer to the true value by

and we still have only approximation of unknown value.


Precision is a degree of perfection used in the survey.
For example, length can be observed as 22 m when taken by counting paces. With tape, length can be estimated at 1) 22.1 m, 2) 22.08 m or 3) 22.085 m. In general, more experienced surveyor will reach better precision. We have to have sense how many significant digits to use, otherwise other will consider that we don't understand what we are doing.
The difference between precision and accuracy. Precision is usually linked to a crew, accuracy to an instrument.

When we take distance by counting paces, we can not state the result at 4 significant digits.

If the distance from A to C is computed as distance AB + BC and BC was observed with poor precision or accuracy, the result AC has poor precision as well. Eg. if AB = 120 m, BC = 13.37 m, then the result should be 133 or 133.5 m. If AB = 120.00 m, BC = 13.37 m, it is justified to assume result distance is 133.37 m.

Coastline paradox

If the coastline of Great Britain is measured using fractal units 100 km long, then the length of the coastline is approximately 2,800 km. With 50 km units, the total length is approximately 3,400 km, approximately 600 km longer. More concretely, the length of the coastline depends on the method used to measure it (and can be even infinite).

Coastline paradox


We rather should write "Mistakes and errors" and we can divide them into a few kinds:

Normal distribution

We use so called normal distribution to describe measured variable and its expected errors.
Such distribution is helpful and convenient for further work with random variables. Normal distribution describing measurements is determined by

(1.1)   μ = avg(x0, x1, ... xn),

(1.2)   σn2 = avg(Δ02, Δ12, ... Δn2) = ∑ Δi2 / n, where

(1.3)   Δi = xi - xavg is error of a particular value from its mean value.

Gauss distribution for random errors

Both σ and μ have the same units as original data measured. If we look at the graph above, we see that standard deviation determines the width of the distribution. The smaller deviation is, the smaller is the error of final value μ.

It was explained that the distribution of random value is described by two parameters μ and σ.
For practical results of measurements we are not interested into standard deviation of Gaussian distribution. We are rather finally interested into the numeric value of probable error of our measurements.
Error depends on count of measurements conducted. If only one measurement is conducted, the error is undefined. The more measurements we have, the better idea we have about an error. The more measurements we have, the more confidence we also have into trueness of the value. Standard error of a mean value is calculated as

(1.4)   SE(xavg) = σn-1 / √n,

Above, n is the number of measurements and standard deviation is computed as

(1.5)   σn-12 = ∑ Δi2 / (n - 1).

Because we have only limited sample of measurements and because error (or deviation) on mere one sample is undefined, σn-1 formula slightly differs from (1.2) and provides better estimation. Around 67 % of samples are expected to belong to interval marked by the standard error. If standard error is widened to double, 95 % of samples fall into the interval.


Let us say that our 5 teams conducted a survey and values of distance measured are x = { 1534 m, 1423 m, 1671 m, 1637 m, 1417 m }. Mean value μ is 1536.4 m, ∑ Δi2 = 55,369 m2, σn-1 = 117.6 m, SE = 52.6 m. We can write that distance measured is 1536.4 m ± 52.6 m. Considering poor accuracy, there is no justification to use so many digits and the result should be written as 1540 m ± 50 m.

Adding up errors

The errors are random, some of them negative, other positive. If there are more random errors, they have a tendency to cancel each other.

  1. If a distance from A to B is measured as several separate measurements of distances X0, X1, ... Xn whose sum determines the distance A to B, then error distance A to B is not sum of errors, but SEAB = √n SE1, where SE1 is an error of one measurement. If the error of measurement is 1 cm and we have 4 measurements to collect final distance, then final error is expected at 2 cm.

    The errors are random, some of them negative, other positive. If there are more random errors, they have a tendency to cancel each other.

  2. The formula above is simplification of adding up errors of measured values. When two or more measured values are added, the uncertainty associated with the sum is computed by taking the sum of the squares of the uncertainty associated with each measured value, and then taking the square root of the sum. This process is called summing in quadrature.

    For example if x = 3.0 ± 1.0 and y = 11.0 ± 0.5, the expected result x + y = avg(x, y) ± SEresult = 14.0 ± 1.1, where SEresult = √(1.02 + 0.52) = 1.1

Field notes

The topic is included here because improper recording in a field will most likely cause errors which can be very expensive. At least most important rules:

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