Part I Flashcards
F/STOP is the relative ___ or lens ___.
Aperture or lens opening
As F/STOP increases, d ___, E ___, t ___. (Increases/Decreases).
d decreases, E decreases, t increases
Scale = ___/___ or ___/___.
f/H’ or d/D
Resolution is ___ to grain size of film.
Inversely proportional
Slow speed equates to ___ (higher/lower) resolution.
Higher
Resolution is expressed in ___ mm.
lines per mm
In metric, kilo is 10x, x=?
103
In metric, nano is 10x, x=?
10-9
In metric, micro is 10x, x=?
10-6
In metric, mega is 10x, x=?
106
In metric, centi is 10x, x=?
10-2
In metric, milli is 10x, x=?
10-3
In metric, giga is 10x, x=?
109
Ground Resoltuion, denoted Rg =(a*b)/c where a, b and c are, (symbolicaly and numerically), ___, ___, and ___, respectively. Fill in the blanks.
Rs =System resolution in line pairs/m
f=Camera focal length in mm
H’=Flying height in meters
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?
Rg=(40ln prs/mm*300mm)/6000m
If Rg=2.0 ln prs/m, then the Ground Resolution Distance (GRD) is = ?
The reciprocal, 1/Rg , and thus 1.0 ln prs / 2 ln prs/m=0.5m
If the image scale is RF = 1:50,000 and the system resolution is 35 line prs/mm, then the GRD is ___.
GRD(m)=(Reciprocal of Image Scale)/(SystemResx1000mm/m)=(50000)/(35x1000)=1.4m line pr-1
If the array element size is 0.009mm, f=28mm, and H’=1800mm, find the equivalent ground distance.
GSD(m)=(0.009mmx1800mm)/28m=0.6m
If the array element size 6.5mm, f=62.5mm, and H’=2880mm, find the equivalent ground distance.
GSD(m)=(6.5x106x2880m)/(0.0625m)=0.30m=30cm
The dimension D of the Ground Resolution Cell (GRC) is the ___.
IFOV, Instanteneous Field of View
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.
The ratio (lambda)/d, lens aperture
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.
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
The rule of thumb is that the GRC should be less than ___ the size of the smallest object of interest.
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)
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)?
We require GRD/2=1m, thus GRD=2m/1pr. Now Rg=1/GRD. Thus Rg=0.5lpr/m. Now reverse-engineer the formula process. If Rg=(Rs*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)
The Landsat Multispectral Scanner has an angular IFOV of 0.086 mrad and is flying at a height of 915km, what is the GRC?
ARP(IFOV)=L/r rad
0.086mrad=L/915km rad=
L=0.086mrad/915km=
L=0.086mrad/915000m
L=78.69m
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?
Scale=f/H=PSD/GSD
GSD=(H/f)*PSD=3110ft/28mm*9mm
GSD=999.6ft…=0.9996, GSD=1 foot
GRD(m)= …?
(Reciprocal of Image Scale)/(System Res*1000mm/m)
The minimum separation you can recognize on the ground given scale and system resolution is …
GRD/2
Minimum Ground Separation is the same as ___.
Ground Sample Distance
Derrive the Ground Sample Distance from the scale formula. What is PD?
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.
Yellow is _ + _ and is the absence of _
R+G, blue
Cyan is _ + _ and is the absence of _
G+B, red
Magenta is _ + _ and is the absence of _
R+B, green
The PP is __ while the CPP is ___.
PP Principal Point, center of photo 1; CPP=Conjugate Principal Point is the adjacent photo’s PP
Absolute steroscopic parallax is the ___, which is the ___.
Average photo base length which is the distance between PP and CPP.
The differential parallax is the __.
Difference between the steroscopic parallax at the top and base of the object.
To get height h using stereoscopic parallax, the equation is h=(a)*b/(c+d) … find a, b, c, d.
h=(H’)*dP/(P+dP)
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.
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
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.
h=d*H’/r=(2mm*918m)/71.5mm=25.7m=26m
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?
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
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?
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
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?
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
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.
h=dH/r, so (2.01mm*1220m)/56.43mm=43.4m
The GSD=(a*b)/c where a, b, and c are ?
GSD=(ArrayElementSize*H’)/f
