The hook
Least squares is what Bowditch wishes it were. When you have more measurements than unknowns(a redundant network), least squares finds the set of coordinates that minimizes the sum of squared residuals — weighted by how reliable each measurement is. Modern survey software runs it on every GPS network and total-station traverse.
Memorize these
Concepts that show up on the exam
Residual (v)
Difference between an observed value and its adjusted (best-fit) value: v = obs − adj. Least squares minimizes Σv².
Weight
Inverse of variance: w = 1/σ². A measurement with σ = 0.01 ft has 100× the weight of one with σ = 0.1 ft. The adjustment "trusts" high-weight observations more.
Redundancy (degrees of freedom)
Number of observations minus number of unknowns. df = n − u. With df = 0, no adjustment possible. df > 0 lets least squares estimate accuracy.
Standard error a posteriori (σ₀)
Scaling factor from the adjustment. σ₀ ≈ 1.0 means weights were realistic; 2.0 means observations were noisier than you said.
Error ellipse
2D analogue of a 1σ confidence interval. Each adjusted point gets one — orientation and size show the direction of greatest uncertainty.
Constrained vs. free network
Free = nothing fixed; assesses internal precision. Constrained = fixes one or more coordinates to known values; gives final published coordinates.
Keep these in muscle memory
Formulas to know cold
Normal equations (matrix form)
(AᵀWA) x = AᵀWℓA = design matrixW = diagonal weight matrixℓ = observation - approximationx = corrections to unknowns
Standard error a posteriori
σ₀ = √( vᵀWv / df )Variance of adjusted unknowns
Σ_xx = σ₀² · (AᵀWA)⁻¹Diagonal entries are variances of each adjusted coordinate.
Try it before you peek
Worked example
The problem
A line is measured 4 times: 100.02, 100.05, 100.01, 100.04 ft. All measurements have equal precision. What is the least-squares estimate of the line length, and what is the standard error of the mean?
Don't fall for these
What trips people up
Equal-weight assumption when it isn't true
Distance and angle observations have very different precisions (mm/ft vs. arcseconds). Treating them as equal-weight means angles get drowned out by distances. Always weight by 1/σ².
Interpreting σ₀ above 1.0 as a problem
σ₀ > 1 means residuals are bigger than your weights predicted. Could be: (a) blunder in data, (b) optimistic σ values, (c) systematic error not modeled. Investigate before publishing.
Confusing accuracy with precision
Least squares estimates PRECISION (internal consistency). It cannot detect systematic shifts that affect every observation equally. A traverse can have σ₀ = 1.0 and still be 5 m from where it should be.
Test yourself
How well did it stick?
A quick 5-question check on Least Squares Adjustments. See where you stand and what to review.