CogoKit

Error propagation

Error propagated into latitude and departure from errors in an angle and a distance

One course, one distance uncertainty and one azimuth uncertainty, carried through the general propagation formula with partial derivatives — and a rotation argument that checks the answer without repeating it.

Analyze· about 26 minutes by hand· 5 steps

Given

  • A single traverse course was observed as: distance D = 1284.62 ft, azimuth α = 63°14′30″ clockwise from north.
  • The distance has a standard error of σ_D = ±0.025 ft.
  • The azimuth has a standard error of σ_α = ±8″.
  • The two errors are independent: the distance came from an EDM observation and the azimuth from an angular observation on an independently oriented backsight.
  • Distances in US survey feet.

Required

  • The latitude and departure of the course.
  • The standard error of each, propagated from both sources.
  • The resultant positional uncertainty of the far end of the course.

Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.

Worked solution

The functions, and the angular unit trap

Latitude and departure are D·cos α and D·sin α. Each is a function of two measured quantities, so the general propagation formula applies: the variance of the result is the sum of the squared products of each partial derivative with its own standard error.

The trap is the unit of the angular standard error. The partial derivative of D·cos α with respect to α is −D·sin α per radian, so σ_α must be in radians before it is multiplied by anything. Divide the seconds by ρ = 206,264.81, the number of arc seconds in a radian.

Leaving σ_α in seconds inflates the answer by a factor of two hundred thousand, which is at least obvious. Converting to degrees instead of radians inflates it by 57, which is not.

lat = D cos α dep = D sin α σ_y = √( Σ ( ∂f/∂xᵢ · σᵢ )² ) for independent xᵢ ρ = 180 × 3600 / π = 206,264.81 arc seconds per radian σ_α = 8″ / 206,264.81 = 3.87851 × 10⁻⁵ rad

The course itself and its four partial derivatives

Evaluate the trigonometric functions once and reuse them everywhere. The course runs largely east: the departure is nearly twice the latitude, which will matter when the two error components come out.

Write all four partials down before combining anything. The distance partials are dimensionless — they are just the cosine and sine — while the angular partials carry the length of the course and are therefore large numbers.

cos α = cos 63°14′30″ = 0.4502283 sin α = 0.8929135 lat = 1284.62 × 0.4502283 = +578.372 ft dep = 1284.62 × 0.8929135 = +1147.055 ft ∂lat/∂D = cos α = 0.4502283 ∂lat/∂α = −D sin α = −1147.055 ft/rad ∂dep/∂D = sin α = 0.8929135 ∂dep/∂α = +D cos α = +578.372 ft/rad

Propagate into the latitude

The latitude picks up 0.0113 ft from the distance and 0.0445 ft from the azimuth. The angular contribution is four times the distance contribution, and the reason is visible in the partial: the angular partial for latitude carries D·sin α = 1147 ft, the full departure of the course.

The physical picture is worth holding on to. Swinging the course by 8″ moves the far end sideways, perpendicular to the course. Because this course runs mostly east, sideways is mostly north–south — so an angular error lands mainly in the latitude, not the departure.

The combined standard error of the latitude is ±0.046 ft, essentially all of it angular.

distance term: | cos α | σ_D = 0.4502283 × 0.025 = 0.01126 ft angular term: | D sin α | σ_α = 1147.055 × 3.87851 × 10⁻⁵ = 0.04449 ft σ_lat = √( 0.01126² + 0.04449² ) = √( 0.0001268 + 0.0019794 ) = √0.0021062 = 0.0459 ft

Propagate into the departure

The departure's two contributions are 0.0223 ft from the distance and 0.0224 ft from the azimuth — almost exactly equal, by coincidence of this particular geometry. Combining them gives ±0.032 ft.

So the departure, despite being the larger of the two components by a factor of two, is the better known of the two by a factor of 1.45. That is not a paradox: the departure runs along the course, where distance error lives, and distance is the better-observed quantity here. The latitude runs across the course, where angular error lives.

Anyone reporting a single positional uncertainty for this course without noticing that σ_lat and σ_dep differ by 45 per cent is throwing away the most useful thing the propagation just told them.

distance term: | sin α | σ_D = 0.8929135 × 0.025 = 0.02232 ft angular term: | D cos α | σ_α = 578.372 × 3.87851 × 10⁻⁵ = 0.02243 ft σ_dep = √( 0.02232² + 0.02243² ) = √( 0.0004983 + 0.0005031 ) = √0.0010014 = 0.0316 ft

Resultant positional uncertainty

The uncertainty of the far end of the course, taken as a single number, is the resultant of the two components. It comes to ±0.056 ft, or about 1:23,000 of the length of the course.

Report the two components as well as the resultant. The far end of this course is uncertain by 0.046 ft north–south and 0.032 ft east–west, and those two numbers are the semi-axes of the error ellipse projected onto the coordinate axes. Which of them matters depends on what the point is going to be used for.

In its natural frame the ellipse is simpler still. Its semi-axes are σ_D = 0.025 ft along the course and D·σ_α = 0.0498 ft across it — the second being twice the first, so the angular observation is the limiting one for this course by a factor of two.

σ_position = √( σ_lat² + σ_dep² ) = √( 0.0459² + 0.0316² ) = √0.0031074 = 0.0557 ft along-course semi-axis = σ_D = 0.0250 ft across-course semi-axis = D σ_α = 1284.62 × 3.87851 × 10⁻⁵ = 0.0498 ft relative precision = 1284.62 / 0.0557 = 1 : 23,000

Answer

  • Latitude = +578.372 ft, departure = +1147.055 ft.
  • σ_lat = ±0.046 ft, of which 0.0445 ft comes from the 8″ azimuth error and only 0.0113 ft from the 0.025 ft distance error.
  • σ_dep = ±0.032 ft, made up of 0.0223 ft from the distance and 0.0224 ft from the azimuth in almost equal parts.
  • Resultant positional uncertainty of the far end of the course = ±0.056 ft, about 1 : 23,000 of the course length.
  • The latitude is the less well determined of the two components even though it is the smaller one, because the course runs mostly east and an angular error therefore acts mostly north–south.

Check

Check by rotation invariance instead of by repeating the partials. The error ellipse of the far end has semi-axes σ_D along the course and D·σ_α across it; σ_lat and σ_dep are that same ellipse projected onto north and east. The sum of the squares of the projections onto any two perpendicular axes is invariant under rotation, so σ_lat² + σ_dep² must equal σ_D² + (D σ_α)².

Left side: 0.0021062 + 0.0010014 = 0.0031076. Right side: 0.025² + 0.0498² = 0.000625 + 0.0024800 = 0.0031050. The two agree to three significant figures, the difference being rounding in the intermediate terms. Nothing in this check reuses the partial derivatives, so it catches a sign or a factor lost in either propagation.

Check the angular contribution by its physical meaning: 8″ at 1284.62 ft subtends 1284.62 × 8 / 206,264.81 = 0.0498 ft. That is the across-course displacement, and it must be the larger of the two ellipse axes here — it is, by a factor of two over the 0.025 ft distance error.

Check the trigonometry: lat² + dep² = 578.372² + 1147.055² = 334,514 + 1,315,735 = 1,650,249, and √1,650,249 = 1284.62 ft, recovering the distance.

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