Week 3 Outcomes Area Determination One of the questions we frequently want to answer with aerial photos isβ¦ How much area is contained within a certain image? Problem is this: You see a field on an aerial photo at a scale of 1:12000. The field measures 0.5 inches by 1 inch. What steps you would use to determine the area of the field in square feet? When measuring rectangular areas, we can simply use an engineers scale or MPI, to measure length & width, and use the formula area = length * width. PRS * PD = GD 12,000 β 0.5 ππ. β 6000 ππ. β 1 ππ‘. = 500ππ‘. 12 ππ. 1 ππ‘. 12,000 β 1 ππ. β 12,000 ππ. β = 1,000ππ‘. 12 ππ. (500 ππ‘. β 1000 ππ‘. ) β 1 ππππ 500,000 ππ‘. 2 β 43,560 ππ‘.2 = 11.478 Acres β¦Area Determination Except in some agricultural applications, the areas of interest are rarely rectangular is shape, so we need other techniques. Several techniques are available for determining the area of irregular shapes. A) Counting squares on graph paper 1) Fairly accurate if squares are small enough 2) Somewhat tedious 3) Not sure how to handle partial squares? 4) Gives area in square inches on the photo Photo at 1:20,000 scale 1 ππ2 = 20,0002 = 400,000,000 ππ2 1 ππ2 = 400,000,000 ππ2 / (144 ππ2 /ft.) 1 ππ2 = 2,777,777.8 ππ‘ 2 1 ππ2 = 2,777,777.8 ππ‘ 2 / (43560 ππ‘ 2 /ac.) 1 ππ2 = 63.77 ac. B) Transect line method 1) Parallel lines are laid out over the object at a known distance apart 2) Total length of the lines within the polygon are determined 3) Total length times the distance apart gives area in square inches 4) The closer the lines, the more accurate the method 5) Less tedious than counting squares C) Polar planimeter: specifically designed to find areas (Swiss mathematician J.A. Laffon, 1854) 1) Consists of a weight, pivot arm, & a recording wheel. 2) Perimeter of polygon boundary is traced ο starting/ending at same point. 3) Tracing done in clockwise direction 4) Repeated a few times, then calculate the average 5) Must be sure to set bar so planimeter reports square inches 6) Somewhat laborious if one has to do numerous polygons β¦Area Determination D) Electronic polar planimeters 1) usually have a small processor to facilitate other measurementsβ¦ a) Lengths b) Rescaling of axis so lengths & areas can be read in ground units c) Transmit information to a host computer for additional processing E) Dot grids 1) Dot grid reduces problem of partial counting around edges of an area that is present when counting squares. 2) Dots set up so they fall in the center of squares 3) By knowing the number of dots per inch, can calculate the number of square inches 4) Can and should be used in a statistical procedure ο do it 3 times & take an average. # acres = (πππ )2 β Or # acres = PSR β 1 ππππ 6,272,640 ππ.2 1 ππ‘. 12 ππ. 2 β β( 1 ππππ 43,560 ππ‘.2 π·πΆ ) π·πΌ β( π·πΆ ) π·πΌ OR DI = dots/inch on grid used DC = dot count within for some polygon F) If interested in a single area, can convert: sq. inches on photo to sq. inches on ground 1) For repeated area measurements, though, itβs easier to calculate acres/sq. inch of photo Acres/square inch β¦.for any photo given the PSR 3 Ways to obtain acres/square inch: Formula 1: Acres/Sq.In. = π·πΊπΉ ππ,πππ ππ/ππ 640 / 63,360 πππ 2 OR ο 640 β πππ 63,360 2 β’ Note that β’ If this number is squared, it represents ο square miles / square inch β¦at a given PSR β’ Since there are 640 acres / 1 square mile, β¦. β¦multiplying ο gives a number that represents mile/inch β¦at a given PSR πππππ 1 π π. ππππ β π π. πππππ 1 π π. πππβ = πππππ 1 π π. πππβ β¦at a given PSR Acres/square inch β¦.for any photo given the PSR β¦3 Ways to obtain acres/square inch: Formula 2: Acres/Sq.In. = πππ 12 ππ/ππ‘ 2 / 43560 ft./ac. Or πππ 2 (12)2 β 43,560 ο¨ π·πΊπΉπ πππ β ππ,πππ 43560 sq.in. / acre β’ 1) ππππ’ππ π π.ππ. πβππ‘π π π.ππ. Note that π·πΊπΉπ gives ο ππππ’ππ π π.ππ. πβππ‘π π π.ππ. β 1 π π.ππ‘. 144 π π.ππ. β¦.gives ο square feet on the ground per square inch on the photo 2) (ππ‘.)2 πππ ππ. 2 (ππ‘.)2 πππ ππππ = (ππ‘.)2 (ππ.)π β ππππ (ππ‘.)2 β¦.gives ο acres on the ground per square inch on the photo Acres/square inch β¦.for any photo given the PSR 3 Ways to obtain acres/square inch: Formula 3: Acres/Sq.In. = 2 πππ β 10000 4 or πππ 2 (10000/4)2 The logic behind this βrule of thumbβ is a bit more difficult to see: β’ From formula 2, recall that (144 sq.in./ft) * (43,560 sq.ft./ac.) = 6,272,640 sq.in/ac. β’ Above, note that (10000/4)2 = 6,250,000 β 6,272,640 β’ Error associated with this formula (rule of thumb) is approximately 0.36% Acres/square inch β¦.for any photo given the PSR An example: Suppose you have a photo with a scale of 1:15840 ο PSR = 15840 63360 πππππ By formula 1: 640 / 2 = 640/16 = 40 acres/sq.in. 2 By formula 2: πππππ 12 By Formula 3: πππππ 10000 / 43,560 = 1,742,440/43,560 = 40 acres/sq.in. 2 β4 = (6.336)2 = 40.14 acres/sq.in. Once any of these is calculated, then β¦. (πππππ π π. πππβ) β (π π. πππβππ ππ π πππ ππππ ππ π πβππ‘π) β π¨πππ ππ π¨ππππ FYI: 40.14 β 40 40.14 = 0.349% πππππ Use of an Appropriate Dot Grid May be able to avoid the β3 formulasβ aboveβ¦ A) If a dot grid were set up so there were 64 dots per square inch (DI),a count of dots would give ο number of acres directly on 1:20,000 photos. B) Such is the case with dot grids we will use. C) What about using this dot grid for other photo scales? 1) Need a way to correct area estimates when photo scale β dot grid scale 2) Must do something similar to the Distance Correction Factor: DCF = a) Can think of a process a correcting two distancesβ¦ β’ A length distance β’ A width distance π·πΉπΊ π·ππππ π·πΉπΊ π«πππππ b) Could simply use DCF twice c) Area Correction factor: ACF = (π·πΆπΉ)2 d) An example: 1:20,000 β’ β’ 64 center dots/in. If a 1:20,000 dot grid is used on a 1:10,000 photo & 20 dots counted (DC = 20) in a polygon, the correct acreage is: π·πΆ β 10000 20000 2 = 20 β 0.5 2 = 5 πππππ Area Determination Area measurements made from aerial photos are subject to errors caused by distortion & displacement. Mainly from tilt and changes in scale due topography Zoom Transfer Scopeβ¦ Corrects tilt due to aircraft pitch or roll Topographic effectsβ¦a bit tougher to correctβ¦ Mathematical transformation/modeling is needed. Things closer to the camera are bigger
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