1. Flight planning

2. # photos/ flight line
One part of price estimation is calculating the number of photos in a project
Let’s assume 9x9 in. format, 6 in. focal length, 60% end lap (along the flight line)
The 2nd photo edge must extend to the start of the project for stereo coverage
The 2nd exposure must thus start 50% of 9 in. converted to ground from the project edge

3. # photos/ flight line
Since 60% overlap is 40% advance, the 1st exposure will thus be 10% inwards from the project edge
Flying is not an exact science so it makes more sense to take the first exposure at the project edge.
Usually the pilot takes 1-4 photos before and after the project edge just to be sure of coverage
Photos not needed are simply not processed

4. # photos/ flight line
Starting 1st exposure on project edge, how many photos are required on a 3 mile flight line flown at 1200 ft. flying height?
{3mi*(5280 ft/mi) / [0.4*9in*1200ft/6in]} +1
One is added because dividing computes # of intervals between exposures, so adding one “includes” the 1st exposure
The answer is 23 photos

5. # photos/ flight line
If the computation results in a decimal number such as photos, one has to round up to 28 photos.
This is because 27 photos would not provide the desired coverage.
Note it was assumed 1st and last exposures will in a worst case situation be on the project exterior not 10% inwards.

6. # flight lines/ project
Typical sidelap (across flight lines) is 15-30%. We will assume 20% for our computations.
20% sidelap is 80% advance between flight lines.
A typical “safety” factor is to fly the first flight line inside the project exterior by the sidelap.
Note mathematically the first flight line could be flown 50% inside the project exterior.

7. # flight lines / project
Assume the previous project is 2.5 miles wide (usually the flight lines are in the longer direction of the project) and at least 20% sidelap is desired
Assume the first and last flight lines must be within 20% format of the project exterior
Total width is 5280*2.5-2(.2)(9)(1200/6) =
= 12480

8. # flight lines / project
Note the distance inside the project exterior is subtracted twice for both first and last flight lines
# flight lines = ft./ (.8*9in*1200ft/6in)+1
# flight lines = 9.67
So ten flight lines are required unless you want to “move” the first and last flight lines inward more than 20%

9. # flight lines / project
It is up to the photogrammetrist, due to rounding up, to stay at 20% sidelap and move exterior flight lines outward, to fly at more sidelap
Try changing the flying height and recomputing # of photos/flight line and # of flight lines
Fewer photos means less cost but higher flying height so less accuracy

10. Flying height limits
Fixed wing aircraft are limited to 1200 ft. in urban areas and 1000 ft. in rural areas.
A helicopter can be used to obtain lower flying heights for higher accuracy.
A helicopter is more expensive to use than a fixed wing aircraft.
Lower flying height means less coverage on one 9 x 9 in. format exposure.

11. (Forward or End Overlap)

13. B = D – E
W = D – S
PE = [(D - B)/D] × 100
PS = [(D - W)/D] × 100

14. example F = 153 mm E= 60% Zavg = 2 km Photo format (d) = 230 mm
Calculate air base (B).
B = D – E
Savg = c / Zavg = 153mm / 2000 m
D = (230mm) × (2000 m / 153 mm) = 3007 m
E = (60/100) × (230mm) × (2000 m / 153 mm) = 1804 m
B = D – E = 3007 m – 1804 m = 1203 m

15. EXAMPLE 2 Savg = 1 : 5000 B = 500 m Photo format (d) = 230 mm
Calculate end lap percentage (PE)?
PE = [(D – B) / D] × 100
D = d × (1 / Savg)
D = (230 mm) × (5000) = 1150 m
PE = [( )/1150] × 100 = 56.5 %

16. EXAMPLE 3 Savg = 1 : 5000 B = 500 m Photo format (d) = 230 mm
The distance between adjacent flying lines is 900 m.
Calculate side lap percentage (Ps)?
PS = [(D - W)/D] × 100
PS = [( ) / 1150 ] × 100
PS = 21.7 %

17. Example 4 Zavg = 1530 m f = 153 mm d × d = 230 × 230 mm PE = 60 %
PS = 25 %
Calculate D and W?
S = c / Zavg = 153mm/1530m = 1/10000
DXD= (230 mm × 10000) ( 230 mm × ) = 5.29 sq.km
b (on photo)= d – e = 230 mm – 60 × (230 mm / 100) = 92 mm
W = d – s = 230mm – 25(230mm/100) = mm
B × w = 92 mm × mm
AN (net) = B × W = b (10000) × w (10000)
AN = B × W = 92mm (10000)×172.5mm(10000)
AN = sq.km

18. Example 5
It is desired to generate topographical map for an area based on stereo photo model, calculate the no. of photos needed, the time between two successive photos, neat diagram to show the flying photos. Given the following:
Area = (6 × 4.5 km)
haverage = 250 m
S= 1:10000
c = f = 153 mm
d × d = 230 × 230 mm
PE = 60 %
PS = 20 %
Plan speed = 150 km / hr
Flying direction with along with largest dimension of the area

19. One photo coverage in long direction
=[230 mm (10000)] / 1000 = 2.3 km
actual coverage = ( ) % × 2.3 km = 0.92 km
No. of photos in long direction for one line=(6 km / 0.92) = 6.5
In practical = 8.5 say 9 photos
One photo coverage in short direction
actual coverage = (100 % - 20 %) × 2.3 km = 1.84 km
No. of lines (strips): Ns = (4.5 km / 1.84 ) = 2.4
In practical 2.4 = 3 lines
No. of photos = 3 x 9 = 27 photos
B= (100 % - 60 %) × (230 mm) × (10000) / 1000 = 920 m
V = (150 × 1000 m) / (3600)sec. = m /sec
t = B / V = (920m) / (41.67 m /sec) = sec