CODEWITHSUNDEEP
mathgeometrytrilateration
padawan
padawan

Reputation: 1315

TULIP Trilateration in 2-D algorithm in simpler form

I found an algorithm to do trilateration in 2-D this link. But the formulas are too complicated. What is happening here? Can you please break it down to terms like dot product, cross product, distance etc?

Upvotes: 0

Views: 492

Answers (1)

user1196549
user1196549

Reputation:

Let P be the unknown point. (Bold for 2D vectors.)

Write the implicit equations of circles 1 and 2:

(P - P1)² = d1²

(P - P2)² = d2²

Substract memberwise and rearrange:

2.(P2 - P1).P = d1² - d2² + P2² - P1²

Similarly with circles 1 and 3:

2.(P3 - P1).P = d1² - d3² + P3² - P1²

Looking closely, you will notice that this forms a system of two linear equations in two unknowns:

2.(X2 - X1).X + 2.(Y2 - Y1).Y = d1² - d2² + P2² - P1²

2.(X3 - X1).X + 2.(Y3 - Y1).Y = d1² - d3² + P3² - P1²

Use Cramer's rule, or if you insist on using vector calculus, work it out as follows.

Rewrite the system as:

A.P = a

B.P = b

Compute vectors perpendicular to A and B in the xy plane, using cross products A' = A /\ 1z and B' = B /\ 1z, and express P as a linear combination of these:

P = u . A' + v . B'

Performing a dot product with A and B gives, after simplification:

A.P = a = v . A.B'

B.P = b = u . B.A'

Note that A.B' = A.(B /\ 1z) = 1z.(A /\ B) = -1z.(B /\ A) = -B.(A /\ 1z) = -B.A' (mixed product).

All in all:

P = [ (- b.A + a.B) /\ 1z ] / [ 1z.(A /\ B) ]

(which is a rewrite of Cramer's result.)

Upvotes: 1

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