Part I

Question 1

F/STOP is the relative ___ or lens ___.

Answer 1

Aperture or lens opening

Question 2

As F/STOP increases, d ___, E ___, t ___. (Increases/Decreases).

Answer 2

d decreases, E decreases, t increases

Question 3

Scale = ___/___ or ___/___.

Answer 3

f/H’ or d/D

Question 4

Resolution is ___ to grain size of film.

Answer 4

Inversely proportional

Question 5

Slow speed equates to ___ (higher/lower) resolution.

Answer 5

Higher

Question 6

Resolution is expressed in ___ mm.

Answer 6

lines per mm

Question 7

In metric, kilo is 10^x, x=?

Answer 7

10³

Question 8

In metric, nano is 10^x, x=?

Answer 8

10^-9

Question 9

In metric, micro is 10^x, x=?

Answer 9

10^-6

Question 10

In metric, mega is 10^x, x=?

Answer 10

10⁶

Question 11

In metric, centi is 10^x, x=?

Answer 11

10^-2

Question 12

In metric, milli is 10^x, x=?

Answer 12

10^-3

Question 13

In metric, giga is 10^x, x=?

Answer 13

10⁹

Question 14

Ground Resoltuion, denoted R_g =(a*b)/c where a, b and c are, (symbolicaly and numerically), ___, ___, and ___, respectively. Fill in the blanks.

Answer 14

R_s =System resolution in line pairs/m

f=Camera focal length in mm

H’=Flying height in meters

Question 15

If my camera focal length is 300mm, the system resolution is 40 line pairs/mm, and flying height is 6000m, what is the ground resolution?

Answer 15

R_g=(40ln prs/mm*300mm)/6000m

Question 16

If R_g=2.0 ln prs/m, then the Ground Resolution Distance (GRD) is = ?

Answer 16

The reciprocal, 1/R_g , and thus 1.0 ln prs / 2 ln prs/m=0.5m

Question 17

If the image scale is RF = 1:50,000 and the system resolution is 35 line prs/mm, then the GRD is ___.

Answer 17

GRD(m)=(Reciprocal of Image Scale)/(SystemResx1000mm/m)=(50000)/(35x1000)=1.4m line pr^-1

Question 18

If the array element size is 0.009mm, f=28mm, and H’=1800mm, find the equivalent ground distance.

Answer 18

GSD(m)=(0.009mmx1800mm)/28m=0.6m

Question 19

If the array element size 6.5mm, f=62.5mm, and H’=2880mm, find the equivalent ground distance.

Answer 19

GSD(m)=(6.5x10⁶x2880m)/(0.0625m)=0.30m=30cm

Question 20

The dimension D of the Ground Resolution Cell (GRC) is the ___.

Answer 20

IFOV, Instanteneous Field of View

Question 21

The ability of the system to resolve detail is limited by the ratio ___, where the larger the lens ___, d, for a given wavelength, the finer the detail which can be distinguished from the image.

Answer 21

The ratio (lambda)/d, lens aperture

Question 22

While not always true the GSD (G. Sample Distance) and GRC (Ground Resolution Cell) are equated with the ___ of the image projected onto the ground.

Answer 22

Pixel Size;

No guarantee that you will be able to discriminate objects that are the same size as the GSD – depend where the pixels fall in relation to the object of interest. Figure: Object same size as GSD butmay or may not dominate any one pixel

Question 23

The rule of thumb is that the GRC should be less than ___ the size of the smallest object of interest.

Answer 23

less than one half the size of the smallest object of interest (which at a minimum equals to 4 pixels for a simple square object)

Question 24

If I am using a Kodak film with resolving power of 80 ln prs/mm, focal length of 152mm, what flying height do I need to resolve individual cars in a parking lot (if the approximate distance between cars is 1m)?

Answer 24

We require GRD/2=1m, thus GRD=2m/1pr. Now R_g=1/GRD. Thus R_g=0.5lpr/m. Now reverse-engineer the formula process. If R_g=(R_s*f)/H’, Rg=0.5=(80 ln prs/mm*152)/H’.

Now H’=12160lpr/0.5lpr=24320m, which is quite high. (See this slide #49)

Question 25

The Landsat Multispectral Scanner has an angular IFOV of 0.086 mrad and is flying at a height of 915km, what is the GRC?

Answer 25

ARP(IFOV)=L/r rad

0.086mrad=L/915km rad=

L=0.086mrad/915km=

L=0.086mrad/915000m

L=78.69m

Question 26

If I were using Kodak 460 digital camera that has a pixel detector size of 9mm, a focal length of 28mm and a flying height of 3,110 feet, what is the GSD?

Answer 26

Scale=f/H=PSD/GSD

GSD=(H/f)*PSD=3110ft/28mm*9mm

GSD=999.6ft…=0.9996, GSD=1 foot

Question 27

GRD(m)= …?

Answer 27

(Reciprocal of Image Scale)/(System Res*1000mm/m)

Question 28

The minimum separation you can recognize on the ground given scale and system resolution is …

Answer 28

GRD/2

Question 29

Minimum Ground Separation is the same as ___.

Answer 29

Ground Sample Distance

Question 30

Derrive the Ground Sample Distance from the scale formula. What is PD?

Answer 30

Scale=(f/H’)=(PD/GD). Now solve for GD in terms of PD, H’ and f. The result is GD=(PD*H’)/f. PD is the size of the pixel array element.

Question 31

Yellow is _ + _ and is the absence of _

Answer 31

R+G, blue

Question 32

Cyan is _ + _ and is the absence of _

Answer 32

G+B, red

Question 33

Magenta is _ + _ and is the absence of _

Answer 33

R+B, green

Question 34

The PP is __ while the CPP is ___.

Answer 34

PP Principal Point, center of photo 1; CPP=Conjugate Principal Point is the adjacent photo’s PP

Question 35

Absolute steroscopic parallax is the ___, which is the ___.

Answer 35

Average photo base length which is the distance between PP and CPP.

Question 36

The differential parallax is the __.

Answer 36

Difference between the steroscopic parallax at the top and base of the object.

Question 37

To get height h using stereoscopic parallax, the equation is h=(a)*b/(c+d) … find a, b, c, d.

Answer 37

h=(H’)*dP/(P+dP)

Question 38

If you are given the flying height and focal lenth of the camera, to calculate the object height: (1) calculate ___ by and then (2) use this formula h=a*b/c. Fill in the unknown.

Answer 38

First calculate flight altitute H’ by multiplying the RF denominator by the focal length of the cam (hence use f/H’=PD/GD) and then the object height is h=(d*H’)/r where d=length of object from base to top and r=distance from PP to top of object

Question 39

Assume the photo relief displacement for a tank is d=2.0mm, the radial distance from PP to top of tank is r=71.5 and the flying height above terrain is H’=918m. Find the height of the tank.

Answer 39

h=d*H’/r=(2mm*918m)/71.5mm=25.7m=26m

Question 40

You measure the displacement of the Statue of Liberty (to the top of the torch) using a single photo as 13mm, and the distance from the PP to the top as 140mm. The flying height of the mission was 1000 m. What is the height of the Statue of Liberty?

Answer 40

Use formula h=d*H’/(r). Now H’=1000m, d=relief displacement from base to top=13mm and r=distance from PP to top of object=140m. Thus (h=1000m*13mm)/(140mm)=93.0m

Question 41

If you didn’t know the flying height of the aircraft or the focal length of the camera but you did know the height of a single object in the photo, how could you estimate the heights of other objects in the photo?

Answer 41

For the known object, measure d and r, then solve for H’. Thus from h=d*H’/r obtain H’=(h*r)/d. Then use H’ in h=d*H’/r to solve for other unknown objects

Question 42

What flying height is needed to resolve individual SAV beds of 1m wide x 10m long (0.001 ha in size)? If the array element size is 0.00465mm, f=12mm, and GSD=0.5m, H’=? What will be the image width if FOV=28.1 deg?

Answer 42

The general rule of thumb is a GSD at a minimum of 1/2 the size of the smallest feature. In this case a GSD of 0.5m. Recall GSD=(array element size*H’)/focal length. H’=0.5m*12mm/0.00465mm=1290m. Tan(FOV/2)=(1/2 image width)/H’ thus =645m

Question 43

Assume that relief displacement for a tower at A is 2.01mm, and the radial distance from the center of the photo PP to the top of the tower is 56.43mm. If the flying height is 1220m above the base of the tower, find the height of the tower.

Answer 43

h=dH/r, so (2.01mm*1220m)/56.43mm=43.4m

Question 44

The GSD=(a*b)/c where a, b, and c are ?

Answer 44

GSD=(ArrayElementSize*H’)/f