Week 3 Outcomes

Week 3 Outcomes
Area Determination
One of the questions we frequently want to answer with aerial photos is…
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 βˆ— 0.5 𝑖𝑛. β†’ 6000 𝑖𝑛. βˆ—
1 𝑓𝑑.
= 500𝑓𝑑.
12 𝑖𝑛.
1 𝑓𝑑.
12,000 βˆ— 1 𝑖𝑛. β†’ 12,000 𝑖𝑛. βˆ—
= 1,000𝑓𝑑.
12 𝑖𝑛.
(500 𝑓𝑑. βˆ— 1000 𝑓𝑑. ) β†’
1 π‘Žπ‘π‘Ÿπ‘’
500,000 𝑓𝑑. 2 βˆ— 43,560 𝑓𝑑.2 = 11.478 Acres
…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 𝑖𝑛2 = 20,0002 = 400,000,000 𝑖𝑛2
1 𝑖𝑛2 = 400,000,000 𝑖𝑛2 / (144 𝑖𝑛2 /ft.)
1 𝑖𝑛2 = 2,777,777.8 𝑓𝑑 2
1 𝑖𝑛2 = 2,777,777.8 𝑓𝑑 2 / (43560 𝑓𝑑 2 /ac.)
1 𝑖𝑛2 = 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 οƒ  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
…Area Determination
D) Electronic polar planimeters
1) usually have a small processor to facilitate other measurements…
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 οƒ  do it 3 times & take an average.
# acres = (𝑃𝑆𝑅)2 βˆ—
Or
# acres = PSR βˆ—
1 π‘Žπ‘π‘Ÿπ‘’
6,272,640 𝑖𝑛.2
1 𝑓𝑑.
12 𝑖𝑛.
2
βˆ—
βˆ—(
1 π‘Žπ‘π‘Ÿπ‘’
43,560 𝑓𝑑.2
𝐷𝐢
)
𝐷𝐼
βˆ—(
𝐷𝐢
)
𝐷𝐼
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, it’s easier to calculate acres/sq. inch of photo
Acres/square inch ….for any photo given the PSR
3 Ways to obtain acres/square inch:
Formula 1:
Acres/Sq.In. =
𝑷𝑺𝑹
πŸ”πŸ‘,πŸ‘πŸ”πŸŽ π’Šπ’/π’Žπ’Š
640 /
63,360
𝑃𝑆𝑅
2
OR οƒ 
640 βˆ—
𝑃𝑆𝑅
63,360
2
β€’
Note that
β€’
If this number is squared, it represents οƒ  square miles / square inch …at a given PSR
β€’
Since there are 640 acres / 1 square mile, ….
…multiplying οƒ 
gives a number that represents mile/inch …at a given PSR
π‘Žπ‘π‘Ÿπ‘’π‘ 
1 π‘ π‘ž. π‘šπ‘–π‘™π‘’
βˆ—
π‘ π‘ž. π‘šπ‘–π‘™π‘’π‘ 
1 π‘ π‘ž. π‘–π‘›π‘β„Ž
=
π‘Žπ‘π‘Ÿπ‘’π‘ 
1 π‘ π‘ž. π‘–π‘›π‘β„Ž
…at a given PSR
Acres/square inch ….for any photo given the PSR
…3 Ways to obtain acres/square inch:
Formula 2:
Acres/Sq.In. =
𝑃𝑆𝑅
12 𝑖𝑛/𝑓𝑑
2
/ 43560 ft./ac.
Or
𝑃𝑆𝑅2
(12)2 βˆ— 43,560

π‘·π‘Ίπ‘ΉπŸ
πŸπŸ’πŸ’ βˆ— πŸ’πŸ‘,πŸ“πŸ”πŸŽ
43560 sq.in. / acre
β€’
1)
π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ π‘ π‘ž.𝑖𝑛.
π‘β„Žπ‘œπ‘‘π‘œ π‘ π‘ž.𝑖𝑛.
Note that π‘·π‘Ίπ‘ΉπŸ gives οƒ 
π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ π‘ π‘ž.𝑖𝑛.
π‘β„Žπ‘œπ‘‘π‘œ π‘ π‘ž.𝑖𝑛.
βˆ—
1 π‘ π‘ž.𝑓𝑑.
144 π‘ π‘ž.𝑖𝑛.
….gives οƒ  square feet on the ground per square inch on the photo
2)
(𝑓𝑑.)2 π‘π‘’π‘Ÿ 𝑖𝑛. 2
(𝑓𝑑.)2 π‘π‘’π‘Ÿ π‘Žπ‘π‘Ÿπ‘’
=
(𝑓𝑑.)2
(π’Šπ’.)𝟐
βˆ—
𝒂𝒄𝒓𝒆
(𝑓𝑑.)2
….gives οƒ  acres on the ground per square inch on the photo
Acres/square inch ….for any photo given the PSR
3 Ways to obtain acres/square inch:
Formula 3:
Acres/Sq.In. =
2
𝑃𝑆𝑅
βˆ—
10000
4
or
𝑃𝑆𝑅2
(10000/4)2
The logic behind this β€œrule of thumb” 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‰ˆ 6,272,640
β€’
Error associated with this formula (rule of thumb) is approximately 0.36%
Acres/square inch ….for any photo given the PSR
An example:
Suppose you have a photo with a scale of 1:15840 οƒ  PSR = 15840
63360
πŸπŸ“πŸ–πŸ’πŸŽ
By formula 1: 640 /
2
= 640/16
= 40 acres/sq.in.
2
By formula 2:
πŸπŸ“πŸ–πŸ’πŸŽ
12
By Formula 3:
πŸπŸ“πŸ–πŸ’πŸŽ
10000
/ 43,560 = 1,742,440/43,560 = 40 acres/sq.in.
2
βˆ—4
=
(6.336)2
= 40.14 acres/sq.in.
Once any of these is calculated, then ….
(π‘Žπ‘π‘Ÿπ‘’π‘  π‘ π‘ž. π‘–π‘›π‘β„Ž) βˆ— (π‘ π‘ž. π‘–π‘›π‘β„Žπ‘’π‘  π‘œπ‘“ π‘ π‘œπ‘šπ‘’ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘› π‘Ž π‘β„Žπ‘œπ‘‘π‘œ) β†’ 𝑨𝒓𝒆𝒂 π’Šπ’ 𝑨𝒄𝒓𝒆𝒔
FYI:
40.14 βˆ’ 40
40.14
= 0.349% π‘’π‘Ÿπ‘Ÿπ‘œπ‘Ÿ
Use of an Appropriate Dot Grid
May be able to avoid the β€œ3 formulas” above…
A) If a dot grid were set up so there were 64 dots per square inch (DI),a count of dots would give
οƒ  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 distances…
β€’ A length distance
β€’ A width distance
𝑷𝑹𝑺 𝑷𝒉𝒐𝒕𝒐
𝑷𝑹𝑺 π‘«π’†π’—π’Šπ’„π’†
b) Could simply use DCF twice
c) Area Correction factor: ACF = (𝐷𝐢𝐹)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:
𝐷𝐢 βˆ—
10000
20000
2
=
20 βˆ—
0.5
2
=
5 π‘Žπ‘π‘Ÿπ‘’π‘ 
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 Scope…
Corrects tilt due to aircraft pitch or roll
Topographic effects…a bit tougher to correct…
Mathematical transformation/modeling is
needed.
Things closer to the camera are bigger