Surveying I– coordinate account

Page takes part in the competition "Notes on the Internet" organized by the Rector of Academy of Science and Technology in Cracow.
V o t e !

Back>>

Analytical method for calculating the surface area of the polar coordinates

Calculate the surface area of the polar coordinates is dependent on the number of positions, from which measurements were made.

⇒ For one position to designate the area to be on the ground from his position to make a minimum of two points with known coordinates, measure the length and directions to points of the contour bends. Then you need to make a reduction of measured directions*.

*If you do not remember how to reduce the measured directions, click

Figure 1. The siting of points 1-2-3 in the collapse of a contour in coordinate system

Search:
P


Measured
length: d₁, d₂, d₃


Directions reduced:
k₁, k₂, k₃

Field P the polygon 1-2-3 calculate from formula:

2P = ∑₁ⁿ d_id_i+1sin(k_i+1 – k_i)

where:
n – number of contour points of collapse,
i – the number of point.

The control is to determine whether:
∑₁ⁿ (k_i+1 – k_i) = 0

Field measurements are given in hectares, remembering that 1ha = 100a = 10 000m².

The following is a table that facilitates the calculation of surface area.

⇒ If the lengths and directions to points of the contour collapse were measured from several observation stations (Figure 2) we need to know the coordinates of the observation stations. Then we calculate coordinates of a collapse contour points:
X_i = X_S + d_icosA_Si
Y_i = Y_S + d_isinA_Si

Continuation of the calculations performed of the Gauss's formulae*.

*If you do not remember how to calculate the surface area of the Gauss's formulae, click

Figure 2. Location points 1-2-3-4-5-6 in the collapse of a contour in coordinate system

Note: To be able to calculate the surface area of the polar coordinates measured from several observation stations, the observation stations must be in one coordinate system.

Example

Calculate the surface area of a polygon 1-2-3-4.

Measured length:
d₁ = 123,32m
d₂ = 153,74m
d₃ = 147,35m
d₄ = 98,56m


Directions reduced:
k₁ = 40^g,0300
k₂ = 62^g,1193
k₃ = 89^g,2873
k₄ = 111^g,4289

SOLUTION

Point number	ki	di [m]	didi+1 [m2]	ki+1 – ki
1	40g,0300	123,32	18959,217	22g,0893
2	62g,1193	153,74	22653,589	27g,1680
3	89g,2873	147,35	14522,816	22g,1416
4	111g,4289	98,56	12154,419	-71g,3989
1	40g,0300	123,32	_	∑ = 0

2P = ∑₁ⁿ d_id_i+1sin(k_i+1 – k_i)

2P = 18959,217•sin(22^g,0893) + 22653,589•sin(27^g,1680) + 14522,816•sin(22^g,1416) + 12154,419•sin(–71^g,3989)

2P = 9825,466m²


P = 4912,73m² ≈ 49,13a ≈ 0,4913ha