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Topic Series 07Scale and Measurements from Aerial Photography

Photogrammetry is the science, art, and technology of obtaining reliable information aboutphysical objects and the environment through processes of recording, measuring, and interpretingphotographic images and patterns of recorded radiant electromagnetic energy, and other phenomena.Included in the definition of photogrammetry are two distinct areas:

1. Metric photogrammetry - making precise measurements from photos and other imagemedia to determine relative locations of points, distances, scales, angles, areas, volumes,elevations, and the sizes and shapes of objects.

2. Interpretative photogrammetry - recognizing and identifying objects on aerial imagery andjudging their significance.

This topic series deals with the geometry and metric applications of photogrammetry. Recall that anaerial photograph cannot be considered a map because: (see Figure 1-1 in Topic Series 1)

Map 1. map is a two-dimensional scale representation of a portion of the earths surface.

2. all points appear as if they are viewed from above, straight down; i.e. orthogonalprojection.

3. the scale is constant across the entire map and all points have been located, orthogonally,on the datum plane.

Aerial Photo

1. aerial photo is a perspective or central projection.

2. a photo is a map for all points that are on the datum plane.

3. distortion results for all points not on the datum plane.

4. the scale is not constant across the entire photo.

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I. Geometry of Aerial Imagery

An aerial image that has been exposed in a precision "mapping" camera can be treatedanalytically because the camera has been calibrated to certain precise metrical characteristics -primarily the format and the focal length.

A camera is an angle-recording instrument. The angle Theta between object and opticalaxis in object space is equal to Theta Prime in image space. (see Figure 7-1 below)

Recall the geometry of a single photo relative to the fiducial marks, principal point, conjugateprincipal point, nadir point, and isocenter.

Fiducial Marks - an indicator of the center of the film plane in the aerial camera. The marks appearin the middle of each photograph border.

Principal Point (PP) - the optical (and geometric) center of the film plane of the photograph. Theprincipal point of each photograph is located by connecting the fiducial marks, at the center of thephotograph. There is only one PP per photo.

Conjugate Principal Point (CPP) - the PP of an adjoining photograph for photos with at least 55%end overlap. Each photo of a stereo pair will have at least 2 CPP's. If sidelap is sufficient, there willbe 3 CPP's along each side of the photo.

NADIR POINT (NP) - a plumb point perpendicular to the 0 datum passing through the lens at theinstant of exposure.

On a tilted image, there are three (3) possible photo centers:

PRINCIPLE POINT (PP): a geometric or optical center, found by connecting fiducialmarks.

NADIR POINT (NP): a plumb point perpendicular to the 0 datum passing throughthe lens at the instant of exposure.

ISOCENTER (IC): the intersection of the axis of tilt and the line between NP andPP.

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Figure 7-1: Basic camera optics

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II. Scale

The objectives of this series of lectures are:

1. determine the attributes of SCALE - photos or maps;

2. determine the differences between point scale and average scale;

3. understand the causes of variation in scale.

Scale defines the relationship between a linear distance on a vertical photo and thecorresponding actual distance on the ground.

Scale = the ratio of a distance on an image or map to the corresponding distance on the ground.

Figure 7-2: Geometry of similar triangles on aerial imagery

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The geometry of similar triangles in Figure 7-2 above is used to derive the following relationships:

Since scale is the relationship between image distance (d) and ground distance (D), the scalerelationship of an aerial photograph can be expressed as:

Exercise: Explain/show how the standard scale equation RF = 1/S = d/D is expanded to includethe geometry of scale on an aerial photo.

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Expressions of Scale

1. Equivalent Scale - the most common form is known as an equivalent scale where 1 unitof map measure is equal to X units of ground scale.

1 inch on map/photo = X ft./chains/meters on the ground

2. Graphic Scale - seen on highway maps where a graphic bar scale is drawn to indicatespecific distances on the map. The advantage of a graphic scale is that if you copy the mapand reproduce it, the "bar" will increase or decrease accordingly.

Figure 7-3: Graphic bar scale

3. Representative Fraction (RF) - where 1 unit of photo/map measure is equal to X units ofsame measure on ground. The forms most commonly used in photogrammetry are either aREPRESENTATIVE FRACTION, OR A RATIO; They are written as 1/24,000 or 1:24,000.

The characteristics of an R.F. are

1. the numerator is always 1; 1 over something.

2. it is unitless, until you assign units.

3. it indicates 1 unit of horizontal distance on the image or map is equal to Xnumber of the same units on the ground.

That is, 1/24,000 means that 1 photo unit equals 24,000 ground units. You can look atit as 1 photo inch = 24,000 ground inches. OR, 1 photo mm = 24,000 ground mm.; or 1of anything to 24,000 of the same thing. ft/ft., m./m., chains/chains.

RF is the most commonly used expression of scale and we will use it extensively inphotogrammetry.

Conversion of RF to Photo Equivalent Scale (PES)

1. RF to Feet: From 1/24,000 to ft./in.? If 1 in. = 24,000 inches, then 24,000"/12" per ft. =2000 ft. and 1"=2000'

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2. RF to Chains: How about chains/in.? 24,000"/12in. per ft./66 ft per chain = 30.0303chains per inch

3. RF to Meters: (24,000"*2.54 cm per inch)/100 cm per meter = 609.6 meters per inch

4. RF to Miles per Inch: 24,000/12 = feet per inch/5,280 ft. per mile = miles per inch or24,000/12/66 ft. per chain/80 = miles per inch

5. RF to Area Measure:

a. acres: (30.0303)2/10 = 91.827 acres/sq. in.

b. hectares: (609.6)2/10,000 = 37.16 hectares/sq. inch

check on hectares: 91.809/2.471 acres/hectare = 37.17 hectares

Conversion of Ground Distance to PES

Example 1. - we have one corner of a 40 ac. block located on an aerial image, but we can seethe 40-lines and we need to draft them on the photo. If the scale of the image is 1:15,840,how long are the sides of a forty?? 40 ac. = 400 sq. ch. = 20 ch. on side.

Convert RF to ES; 1 in. = 15,840/12 = 1320/66 = 20ch So ... we have a ES of20ch/in, and a

Ground Distance = 20ch Therefore, PES = GD/ES = 20 ch/20ch/in = 1 inch on the photo.

Example 2. Photo R.F. of l:40,000. What is the length (on the photo) of one side of a 40 acreblock?

1:40,000 === 1 inch = 40000/12 = 3,333.33 ft. 1 inch = 3,333.33/66 = 50.51 chains

thus, PES = 20 chns/50.51 chns = 0.396 inches on photoor,

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if we define R.F. = 1/S = d/D (from definition) then,

1/40,000 = d/20 chainsd = .0005 chains = 0.396 inches

Example 3. Let's try metric for the dimensions of 1 hectare on a photo with an R.F. of1:25,000:

Since, 1 hectare = 10,000 square meters thus 1 hectare = 100m on a side

RF: 1mm=25,000mm/1000mm per m = 25m per mm on photoGD: 100m on side for hectarePES = GD/ES = 100m/25m = 4 mm on photo

or, 1/25,000 = d/100 metersd = 0.004 meters = 4 mm on photo

Questions??????

Scale Determination

There are only two (2) basic formulae for scale determination:

1. With photo/map and ground distances:

2. With camera focal length and flying height: , but combining these.....

3. Combination of RF with d, D, f, and H:

Equation 3 is the most used equation in Photogrammetry. Be sure that you understandeach component of the equation and how it is used.

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Figure 7-4 Geometry of scale determination on aerial imagery

NOTE: The platform height, H, is height of camera above mean sea level (m.s.l.). The object height,h, is the elevation above average ground level (a.g.l.) of the object above m.s.l. -- either at a point(point scale) or at the average terrain for the photo. The (H-h) is flying height of the camera platformabove average ground level (a.g.l.).

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Figure 7-5: Scale determination on aerial imagery

Example 1: What is the point scale of an aerial photo if the focal length of the camera is152.4mm, the aircraft altitude is 5,000 ft. and the object is located at 200 ft. above m.s.l.?

R.F.= 1/S = (152.4mm/25.4/12)/(5000 - 200) = .5/4,800 = 1/9,600

Thus, the R.F. = 1:9,600

Example 2: If an object measures 800 ft. on the ground and 1 inch on an aerial photo, whatis the scale of the photo?

R.F. = 1/S = d/D 1/S = (1/12)/800 1/S = 1/9,600

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Sources of Scale Variation: Scale Determination:

1. Terrain 1. Point scale2. Topographic Displacement a. using f and H at h3. Focal length and flying Height b. using d and D4. Tip and Tilt 2. Average Scale

a. using f and H at haverageb. using �di and �Di

Scale Variation Attributable to Terrain

Only if a truly vertical image is acquired over perfectly flat terrain is the scale of an aerialimage uniform across the image!!

Figure 7-6: Variation in photo scale from terrain

THIS IS A VERY GRAPHIC ILLUSTRATION OF WHY A PHOTO IS NOT, ANDNEVER WILL BE A MAP!!! Even if you have a perfectly vertical photo (the OPTICAL AXISIS PERPENDICULAR TO M..S.L.) and perfectly flat terrain at exactly 0 datum (sea level), youmay still have small camera or lens defects that distort scale. We use photos as maps all the time--BUT, we know that distances we measure or areas we calculate from measurement are NOTCORRECT!!

So, do NOT, under any circumstances, EVER REFER TO A PHOTO AS A MAP!

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Now, having gotten that out of the way, lets look at POINT SCALES, AVERAGESCALES AND SCALE VARIATION. We have two lines laid out on the ground, each exactly1,000 ft, long. One line (AB) is located at 1,000 ft elevation and Line CD is at a 500 ft. elevation.Given a Focal Length of 6" and a plane ALTITUDE of 6,000 ft., determine the appropriate scale foreach line.

As a check on yourself, which topo position will have the smallest scale? The one closestto the camera? or the one furthest away? A small scale means a large denominator; conversely,a large scale means a smaller denominator. Is 1/12,000 larger or smaller than 1/24,000??LARGER - COVERS FEWER FEET ON THE GROUND PER INCH OF IMAGE!! Or look at itthis way -- is 1/12 larger than 1/24? 0.083 vs. 0.042.

Point Scales:

1. Using f and H at a known h: (i.e. a known/specific elevation)

Scale AB: 1/S = 0.5/(6000 - 1000) = 1/10,000 or 833 ft/in (using f and H-h) 1/S = 1.20"/12/1000 ft = 1/10,000 or 833 ft/in (using d and D)

Scale CD 1/S = 0.5/(6000 - 500) = 1/11,000 or 916.67 ft/in. (using f and H-h) 1/S = 1.09"/12/1000 ft = 1/11,000 or 916.67 ft/in (using d and D)

Thus, on the photo image, the 1,000 ft. ground distance would be represented by line ab as1000/833 = 1.20 inches and line cd would be 1000/916.67 = 1.0909 inches. If we hadmeasured the image and ground distances, we could also compute a point scale.

2. Using d and D: (i.e. assuming both ends of line are at the same elevation)

Scale AB = 1/S = 1.2/1000 = 1/10,000 Scale CD = 1/S = 1.0909/1000 = 1/11,000

Average Scale: The average scale is obtained by averaging all of the terrain elevationsimaged on a frame or averaging the measured photo and ground distances:

1. Using average elevation: Average elevation = (1,000 + 500)/2 = 750 ft.

Thus the scale is: 1/S = .5/(6,000-750) = 1/10,500.

2. Using the sum of photo and ground distances: (gives a weighted average not the scale atthe average elevation!)

Thus the scale is: 1/S = d/D [(1.09+1.2)/12]/ (1,000+1,000) = 1/10,480 (weighted average)

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How much error would be incurred if you applied the scale computed at Line AB to the LineCD? Line cd is measured on the image and found to be exactly 1.09." Applying the PES of 833ft/in, we would calculate the line length to be 908 ft., but we know it is 1000 ft.; thus the error wouldbe:

Error% = (X - True)/True x 100%

1. using lengths: (908 - 1000)/1000 x 100% = -9.2%

2. using scales: (833 - 917)/ 917 x 100% = -9.2%

Point Scale = scale at a point, assuming same elevation

Average Scale = scale at average elevation

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Scale Variation Attributable to Topographic Displacement

Figure 7-7: Topographic displacement on an aerial image.

ANOTHER effect of terrain variation if DISPLACEMENT! Objects are displaced radiallyfrom the CENTER of the image; this is the basic reason why we are able to measure heights ofobjects.

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Other Sources of Scale Variation

1. Obviously, as you change flying height, you will change Scale and changing focal lengthwill also change scale

Figure 7-8: Relationship between focal length and flying height

What is the relationship between f and (H-h) in the scale expression 1/S = f/(H-h)?

a. Scale is inversely proportional to H-h: with f constant, as (H-h) increases in value, the SCALE DECREASES -

GETS SMALLER!!

b. Scale is directly proportional to f:with (H-h) constant, as f increases, the scale becomes larger.

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2. tip and tilt.

Figure 7-9: Relationship of Tip and Tilt on an aerial image

Figure 7-10: Change in scale on a tilted photo

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On a tipped or tilted photo, the scale is uniform on any line parallel to the axis of tilt; it doeshowever, from line to line perpendicular to the axis. NOTE that on a tilted image, there arethree (3) possible photo centers:

1. the PRINCIPLE POINT (PP): a geometric or optical center, found by connectingfiducial marks.

2. the NADIR POINT (NP): a plumb point - a point perpendicular to the 0 datumpassing through the lens at the instant of exposure.

3. the ISOCENTER (IC); the intersection of the axis of tilt and the line between NPand PP.

Summary - The major sources of scale variation on a single photo are terrain and tip/tilt; betweenphotos we have to add changes in flying height. There are three possible photo centers on a tiltedphoto - all three coincide on a vertical photo.

Helpful Hints for Problem Solving!!

A. Read or listen to the problem to gather information;anything that has to do with focal length, elevation, altitude, photo or ground distance:

PROBLEM: If the flying height of a plane is 5,000 ft. and a camera with a focal length of152.4mm is used, WHAT IS THE SCALE OF THE IMAGERY??

1. Read it again and write down important clues:H-h = 5000 ft. (always carry units) f = 304.8mm

2. State the problem: 1/S = ?

3. Convert known values to common units (if required): convert f to feet; 152.4/25.4/12 = 0.5 ft.

4. Determine solution equation: 1/S = f/(H-h)

5. Show the SETUP: 1/S = 0.5/5,000

6. Derive/compute the SOLUTION: 1/S = 0.5/5,000 = 1/10,000

B. Estimate! square root of 146? Between 12 (144) and 13 (169), but closer to 12.

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C. Diagram to help visualize.

D. When we get into "compound" problems involving two or more formulas, BREAK THEPROBLEM DOWN INTO COMPONENT PARTS.

FOR EXAMPLE: combining a scale and a height problem. Determine the scale of a photoif you have the following information - f = 0.5'; a tree on the image is known to be 100 ft. tallwith a displacement of 0.2" and a radial distance of 2.00". Your first inclination is to try thescale formula:

1/S = f/(H-h), but you only have f = 0.5'So then try the Displacement formula:

dh = d (H-h)/r100ft. = 0.2" (H-h)/2.00" 200 = 0.2 (H-h) and H-h = 1,000 ft.

NOW solve scale 1/S = 0.5/1000 1/S = 1/2000

Sample Scale Problems

1. Given: camera with a f = 3", a photo scale = 1:30,000 Question: What is the altitude of aircraftif the average terrain elevation was 1,000 ft.?

Solution: 1/S = f/(H-h)

1/30,000 = (3"/12')/(H - 1,000)

(H-1,000)=0.25 x 30,000

H = 7,500 + 1,000 = 8,500 ft.

2. Given: flying height of 12,000 ft., scale of 1:18,000 Question: Determine the focal length in MILLIMETERS

Solution: Get H-h in Millimeters H-h = 12,000ft x 12in/ft x 25.4mm/in

= 3,657,600 mm

1/S = f/(H-h)

1/18,000 = f/3,657,600

f = 3,657,600/18,000 = 203.2 mm

3. Given: a section-line road (5,280 feet) appears on an image with a photo distance of 1.5", and focal length = 6 inches.

Questions: What is the scale of the photograph? What was the platform height?

Solution: 1/S = d/D = f/(H-h)

f = 6/12 = 0.5 ft.

d = 1.5"/12" = 0.125 ft.D = 5,280 ft

1/S = 0.125/5,280 = 1/42,240 (scale)

1/S = f/(H-h) 1/42,240 = .5/(H-h)

(H-h) = 0.5(42,240) = 21,120 ft.

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III. Lengths and Linear Distance

Ground Distance - Equal Elevation

Recall that ground distance for all points at the same elevation is

Note that d and f are in the same units of measure (i.e. inches) and that the unit of measure for D willbe the same as the unit of measure for H and h. This assumes that all points along the grounddistance D are at the same elevation.

Ground Distance - End Points at Different Elevations

If the end points of a ground distance (line) are at different elevations, you could computethe ground distance using:

1. Average Scale (i.e. elevation) along the line, or

2. End Point Scales (i.e. elevations) of the line.

For example, if f=6 inches, H=6,500 ft., the elevation at endpoint a was 500 ft. (i.e. ha=500) and theelevation at endpoint b was 1,000 ft. (i.e. hb=1,000); then......

Average Scale - Approximation

The average elevation have would be: have = (ha + hb ) / 2 = 750 ft.

and the average scale would be 1/S = .5/(6,500-750) 1/S = 1/11,500 or,PES = (6,500-750)'/6" = 958.33 ft./inch1 inch on the photo = 958.33 ft. on the ground.

Thus, if photo d=2.81 inches then ground distance D=2,692.91 ft. using the average scale.

NOTE: This is only an approximation because the scale is not weighed proportionally to thedisplacement position on the photograph. This is a reasonable method for quick approximations, butit will not yield the correct scale because topographic displacement is radial from the PP and varieswith distance.

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Figure 7-11: Computation of end point scale on a single aerial photograph.

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End Point Scale: (Refer to Figure 7-11 above)

The RF scale at endpoint a would be .5/(6,500-500) = 1/12,000 or the PES = (6,500'-500')/6"= 1,000 ft. on ground per photo inch.

The RF scale at endpoint b would be .5/(6,500-1000) = 1/11,000 or the PES =(6,500'-1,000')/6" = 916.67 ft. on ground per photo inch.

The only way to weight the scales in proportion to their displacement positions relative tothe x and y axis on the photograph is to use the following procedure:

The PP is used as the origin and the endpoints are referenced in x,y coordinate distances fromthe PP. The photo coordinate distances for each endpoint are converted to the grounddistance coordinates using the appropriate scale at each endpoint.

Example: Photo Coordinates are:xa = +0.79" , ya = +1.50", ha = 500 ft.xb = -1.28" , yb = -0.40", hb = 1,000 ft.

Photo Values are: H = 6,500 ft., f = 6 inches

The x,y coordinates in ground distance scale are:

Di = (H-hi)/f * di for i= points a,bdi is in inches (i.e. photo distance)

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The PES for points a and b are:

PESi = (H-hi)/f for i = points a,b

H and hi are in feet; f is in inchesPESi will be in ground feet per inch of photo scale

PESa = (6,500'-500')/6' = 1,000 ft. per inch Xa = 1,000 ft. * +0.79 inches = + 790 ft.

Ya = 1,000 ft. * +1.50 inches = +1,500 ft.

PESb = (6,500'-1,000')/6' = 916.67 ft. per inch Xb = 916.67 ft. * -1.28 inches = -1,173.3 ft.

Yb = 916.67 ft. * -0.40 inches = - 366.7 ft.

Using the Pythagorean theorem: c2 = a2 + b2

D = {[790 - (-1173.2)]2 + [1500- (-366.7)]2}0.5

= 2,709.08 ft.

Note: The ground length of each side, a=1500+366.7 and b=790+1173.2, is composed of 2different scale values.

DO NOT USE PHOTO LENGTHS AND AVERAGE SCALE. You must convert the compute thephoto scale at the desired point.

Magnetic Bearing of N-S Fiducial: Assume the N-S Fiducial line has a True Bearing of N 05o 15'W and the current declination is 02o 15' E. What is the magnetic bearing of the N-S Fiducial line?

Answer: N 07o 30' W How??

What is the Bearing of the Line in Figure 7-11?

Tan � = (1173.3 + 790)/(1500 + 366.7) � = 46.44o Line bearing = N 07o 30' W + (46.44o to right) = N 39.944o E

Acres of Rectangle: (assuming the line is one diagonal of a rectangle) = 84.13 acres How?

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Scale Problems for Study

1. A flight mission is scheduled to obtain imagery at a scale of 4 in. = 1 mi. with a camera of focallength 6 in. The average terrain elevation is 2000 ft. (above mean sealevel - a.s.l.) What must bethe flying height to achieve the desired scale?

2. A satellite overpass is scheduled for a given date; it is desired to obtain "ground truth" with colorinfrared imagery at a scale of 20 ch = 1 in. If the average terrain elevation is 800 ft., and the camerafocal length = 6 in., what should be the correct platform altitude ?

3. The Space Shuttle Columbia carried aboard a mapping camera with a 12 in. focal length. Atapogee, an image was acquired of a section of the Midwest which illustrated the spacing of "section-line" roads. These roads were spaced at intervals of 50/60 in.. Assuming an average terrain elevationof 1500 ft. a.s.l., what was the maximum orbital height achieved by Columbia?

4. An oil storage tank is photographed at scale of 1/16,000; if the tank is 120 feet long and 60 feetin diameter - what are the equivalent photo measurements?

5. How many square miles are covered by a 9" x 9" aerial image at the following scales?

a. 1/6000 b. 1/12,000 c. 1/30,000

6. What is the flying height required to achieve a scale of 2000 ft/in. with a camera focal length of:

a. 6 in? b. 8.25 in? c. 12 in?

7. What should be the flying height with f = 6 in. to acquire imagery where an 80 ft. road right-of-way appears to be at least 0.4 in. on the image?

8. Two points along a highway are known to be exactly one mile apart. If the image measurementbetween these two points is 0.42 feet:

a. what is the scale of the image? b. if the plane flying height was 11,000 ft., what was the focal length of the camera?

9. A clearcut measures 1.5 in. by 3.0 in. on an aerial image taken at a scale of 1/11,500.

a. how many acres are in the clearcut? b. how many hectares?

10. Determine the scale of an image taken at 12,000 ft. over a land surface with an average elevationof 1500 ft. f = 12 in.

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11. The objective of an aerial mission is to acquire imagery that will permit the interpreter tocalculate plantation survival. Assume that this interpreter can consistently distinguish imageobjects having a diameter of 1/100 in. If the camera has a focal length of 6 in., what is the maximumflying height at which young pine with a minimum crown diameter of 4 ft. can be discriminated?

12. It is desired to lay out a photo cruise using one-half ac. plots at a spacing of 20 chains.Assuming an image scale of 1/15,840, what would be the photo measurements of plot diameter andspacing?

13. Given a scale of 1/12,000, a focal length of 6 in. and a platform altitude ( a.s.l.) of 7000 ft., whatwas the average terrain elevation?

14. A study of photos at an average scale of 1/7920, taken at an average terrain elevation of 1200ft., indicates that one square inch at an elevation of 1500 ft. contains 150 trees, while one square inchat an elevation of 300 ft. contains 100 trees. Assuming a focal length of 6 in., determine which standcontains the largest number of trees/acre.

15. You are attempting to verify the average scale using two road intersections on a singlephotograph. If the elevation of the NE intersection is 165 ft. (above m.s.l.), the elevation of the SWintersection is 385 ft. (m.s.l.), the camera had a focal length of 209.55 mm, and the aircraft altitudewas 8,250 ft (m.s.l.);

The average scale between the NE and SW intersections is calculated to be: 1:___________

16. You don't have a quad sheet for this area, but you need to determine the true ground distance andmagnetic bearing between the two road intersections described in question 15 above. Using thefiducial lines as a coordinate system, your obtain/compute the following measurements for thephotograph using the aircraft altitude and camera focal length from question 15 above and thediagram of a single photo below: (note: PES is photo equivalent scale)

Photo Coords Ground Distance Point Elev PES X Y X Y

NE 165' ___._' +1.5" +3.0" 1470.0' ____._'SW 385' 953.3' -2.5" -3.5" ____._' 3336.7'

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17. The true horizontal distance between the two road intersections (in question 15 & 16 above) iscalculated to be _____________ ft.

18. If the N-S Fiducial is N4o15'W, True and the Magnetic Decl is N1o15'E, the Magnetic bearingof the N-S fiducial in question 16 above is ________________, and

19. The Magnetic Bearing (in degrees and minutes only) of a straight line between the two roadintersections in questions 16 - 18 above is ______________.

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IV. Height Determination

Displacement - Height Measurement on Single Photos

The major reason why a photo is not a map - is displacement. On a map, all objects arepictured as they would appear at sea level - 0 datum.

Figure 7-12: Displacement of aerial image

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Note that objects themselves are displaced outward - that is, the top lies away from the photocenter and the base toward. Likewise, the base of the object appears outside of the true horizontalmap position "C". Top "A" outside of "B" outside of "C". If you carefully examine an verticalphoto, you will note that everything radiates in a 360 degree circle form the Photo Center. Thisphenomenon gives us the mathematical basis for measurement of Height of Objects.

Figure 7-13: Profile of height on a single photo

Again we see the phenomenon of radial displacement of the top of the object out from thebase: but note the relationship of height of object and the radial distance from the Nadir Point to theBase and Top of the object. Both trees are the same height, but one is located three times furtherfrom the NP than the other - it is obvious from the sketch that the apparent height (delta h) of the

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fartherest tree will be greater than the one closer to the NP. It is also apparent that the flying height(H-h) will a directly proportional influence on delta h. As H-h increases, delta h decreases.

Figure 7-14: Two dimensional view of height determination on single photo

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The basic relationship for determination of height of objects on a single, vertical aerial photo is:

Thus,

and the height of object is directly proportional to H-h and d, and inversely proportional to r. Youshould also note that at a given Photo SCALE, a longer focal length will increase displacement -because the flying height is less. For example, to achieve a photo scale of 1/6000, a 6" focal lengthwill be flown at 3000 ft., but a 12" f will be flown at 6000 ft.. So, if your flight objective is toacquire stand height, you are better off using a short focal length at lower altitudes.

Also note that d and r have to be in the same units - whatever, mm, inches; H-h and deltah also have to be in the same units, feet or meters - BUT NOT NECESSARILY THE SAMEUNITS AS "D" AND "R." You normally measure the photo distances in inches or millimeters,but structures or trees are given in feet or meters.NOTE you have to be able to see both top and bottom of object for this to work!

Example 1: Given a 6000 ft., H-h, a measured "d" of 3/60", and a measured "r" of 193/60" - what is the height of object?

dh = d(H-h)/r = 0.05" x 6,000 ft./3.22" = 93 ft.

Example 2: Given that H-h = 3000 ft., measured "d" = 6/50", and the distance from the NPto the base of the object is 194/50", what is the HEIGHT?

d = 6/50 = 0.12"r = 194/50 + 6/50" = 200/50 = 4.00" dh = (3000 x 0.12)/4.00 = 90 ft.

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Example 3: Determine the average terrain elevation given the following information: d =0.08", r = 3.20", H = 5,500 ft., dH = 50 ft.

0.08/3.20 = 50/5,500 - h3.20 (50) = 0.08 (5,500 - h) 160 = 440 - 0.08h0.08h = 280h = 3500 ft.

Example 4: Given that the point on top of a 1,000 ft. hill appears on a radial line 3.00 in.from nadir. Assuming a flying height of 6,000 ft.. What would be the displacement of thehill above its sea level position?

d/r = dh/(H-h) SKETCH d/3.00 = 1,000'/6,000' 3(1,000) = d(6,000)d = 3,000/6,000 = 0.5 inches

This only holds on vertical images - tilted photos require rectification!!! Or location of theISOCENTER INSTEAD OF THE NADIR POINT.

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Stereoscopic Parallax - Height Determination on Stereopairs

Figure 7-15: Parallax relationships on stereo pair of aerial photographs.

Parallax IS the reason why we see 3-D; without PARALLAX ... NO STEREO. This is why youcannot take two centerfolds from the same issue and see 3-D. Remember - two images of the sameobject -TAKEN AT DIFFERENT ANGLES!!

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Parallax = the apparent displacement of position of a body (object), relative to some referencepoint or coordinate system, caused by a shift in the point of observation.

Three Components of parallax:

ABSOLUTE PARALLAX of the Base = APb= lxb1 - xb2l = the absolute value of thealgebraic difference, parallel to the line of flight, of the distances from nadir to the photo baseof objects; also called ABSOLUTE PARALLAX.

ABSOLUTE PARALLAX of the Top = APt = lxt1 - xt2l = the absolute value of thealgebraic difference, parallel to the line of flight, of the distances from nadir to the photo topof objects.

PARALLAX DIFFERENCE = lAPt - APbl = the absolute value of the algebraic differenceof the distance from photo base to photo top of the object. Also called differential parallax,dp.

Parallax Formula - this is the basis for all topographic mapping instruments. All topo mapscompiled from aerial imagery use this basic formula:

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Note the similarity between the two formulas: (single vs stereo)

NOTE - the average stereobase "b" [avg. b = (b1 + b2)/2] can be substituted for "APb" when:

1. the elev. diff. between PP & of the base of the object is within 3% of (H-h).

2. photo tilt is less than 3 degrees

3. both PP's are near the same elevation

4. Plane altitude is essentially the same for both exposures

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Manual Measurement of Parallax on Stereo Pair

Since the parallax height equation involves only dp and APb, we can manually obtain theparallax measurements on a stereo pair and compute the height of an object.

Figure 7-16: Manual measurement of parallax on a stereo pair

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Two Components of the manual application method are:

ABSOLUTE PARALLAX = APb = lZ - Z'l = the absolute value of the algebraicdifference, parallel to the line of flight, of the distances from nadir to the photo base ofobjects.

PARALLAX DIFFERENCE = dp = lX - X'l = the absolute value of the algebraicdifference of the distance from photo base to photo top of the object. Also called differentialparallax, dp.

Figure 7-17: Manual measurement of tree height on stereopair with parallax

Direct Parallax Measurement of Height - Optical devices

Delta p is usually measured by some form of a stereoscopic "floating dot" instrument suchas the Harvard or Michigan Parallax Wedges, an Abrams Height Finder (both inexpensive), or themore sophisticated mapping instruments such as the Multiplex plotter - multiple projectors; Balplex- dual projectors. Either type uses the anaglyph principle of obtaining stereo - more modern versionsuse POLARIZED light.

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Example - given an ABRAMS HEIGHT FINDER (a low cost version of a floating dot instrument),a reading is taken at the base of a tree and at the top of the same tree. The difference in reading is0.55 mm = dp . The average stereobase "b" = 3.22 inches, the flying height 3,00 ft. - what is theheight of object?

dh = 0.55mm (3000 ft)/(3.22 x 25.4 + 0.55)1650 = 82.34 dhdh = 20 ft.

Height Measurement with Shadow Height

Figure 7-18: Geometry of shadow height measurement

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Height Measurement by Ocular Estimate

Summary: Methods of height determination1. Single Photos - d/r = dh/(H-h)2. Stereopairs - dp/(APb + dp)= dh/(H-h)3. Shadow height4. Ocular estimates

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Practice Problems HEIGHT DETERMINATION

I. Height Determination on Single Photo

Parallax formula

Shadow Height

1. What is the elevation of a point that is located 3.75 inches from the nadir and has a reliefdisplacement of 0.80 inches if the platform flying height is 10,500 feet?

Elevation = ________ ft.

2. What is the elevation of the following points on the side a mountain if the base of themountain (i.e. datum point) is located 3.06 inches from the nadir point, the platform altitudewas 18,000 feet above the datum point, and the relief displacement from the datum point tothe elevation point is:

a. 0.1800 inches; elevation = _________ ft.

b. 0.3825 inches; elevation = _________ ft.

c. 1.1770 inches; elevation = _________ ft.

3. On a single aerial image, a radial line is selected that passes over a coastal redwood treewhose base is located at 350 feet of elevation. What is the height of the tree if the photoscale is 1 inch equals 400 feet at the tree base elevation, the camera focal length is 6 inches,and the 50 scale on the engineer's scale is used to obtain the following measurements parallaxmeasurements?

Nadir to base of tree = 145 incrementsNadir to top of tree = 165 increments

Height of tree = __________ ft.

4. The image of a point with an elevation of 950 ft. is located 54.67 mm from the nadir pointon a photo where the platform altitude was 12,000 ft.. What would the radial distance be ifthe base of the point was located at m.s.l.?

d = _______ mm

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5. What is the height of a tree whose shadow measures 0.1 inches on single photo which hasa scale of 1 inch equals 660 ft. and where the sun's rays tangent to the top of the tree crownmeet the ground at a 50.47 degree angle?

a. Shadow length, ground distance = ____ ft.

b. Tree height = ______ ft.

6. You have obtained a single aerial photo that has a scale of 1:12,000 at an average elevationof 1,000 ft. taken with a camera focal length of 6 inches. If the measured distance along aradial line to the base of a loblolly pine tree is 3.50 inches and to the top of the tree is 3.566inches and the elevation of the tree base is 500 ft.:

a. The flying height of the aircraft was _______ ft.

b. The tree height is ___________ ft.

II. Height Determination on Stereo PairsDirect Measurement

Abrams Parallax Bar

7. The stereo pair has a photo scale of 1:12,000 at an average elevation of 175 feet and thecamera had a 6 inch focal length. The following direct, coordinate measurements were takenfrom the stereo pair:

X = + 0.75 mm X' = + 0.70 mm (quadrant I)Z = +28.00 mm Z' = +31.55 mm (quadrant I)

a. Calculate the height of the object, assuming the base of the object was at theaverage elevation.

Height = _________ ft.

b. Calculate the height of the object, assuming the base of the object was at 500 ft.elevation.

Height = _________ ft.

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8. Using the parallax bar (floating dot) instrument on a stereo pair where the flying heightwas 3000 ft. and the absolute parallax of the base was 80.00 mm, the following parallaxmeasurements were obtained:

Reading at top of object = 12.76 mmReading at bottom of object = 10.00 mm

The height of the object is _______ ft.

9. Using the Abrams Height Finder (parallax bar) on a stereo pair where the absolute parallax(of the base) is 3.20 inches and the platform altitude is 3,000 ft. (m.s.l.), you obtain a readingof 12.15 mm (dhtop) on a vertical control point P1 that is located at elevation 250 ft..

a. The base parallax reading (dhbot) for P1 = ______ mm.

b. The differential parallax for P1 = ______ mm.

If subsequent measurements (dhtop) on Points 2, 3, and 4 are obtained as:

dhtopP2 = 13.75 mm P3 = 9.71 mm P4 = 10.55 mm

determine the elevations of points 2, 3, and 4.

c. P2 = _______ ft.

d. P3 = _______ ft.

e. P4 = _______ ft.

III. Bearing Determination

10. The bearing of a road (as determined from a quad sheet) is N65015'E (true) and currentmagnetic declination is N5015'W. If the angle between the N-S fiducial and the road on anaerial photo, is 630:

a. The True Bearing of the Fiducial is __________.

b. The Magnetic Bearing of the Fiducial is ________.

c. The Magnetic Bearing of the Road is _________.

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V. Area Measurement

Measurement of area (i.e. acres, etc.) on a photograph or map begins with the establishmentof the appropriate scale. On a map, the scale is constant across the entire map surface, but on anaerial photograph, the scale varies in accordance with topographic displacement, tilt, and otherdistortions. Ideally, areas should not be measured directly on a photograph. The area should betransferred to a planimetric base map or digitized into a GIS or CAD system (with control points)so that the scale is rectified. In real life however, we seldom have the time to perform these desirabletasks prior to measuring area.

Preparations for Photo Area Measurement

1. All Points within Area are at Same Elevation: Compute the average scale for the portionof the photograph that contains the desired area to be measured by measuring distances between roadintersections (or identified points) and correlating the map to photo distance; i.e. 1/S = d/D.

2. Elevation Varies across Area: If elevations vary substantially across the area to bemeasured, the average scale procedure above will not be adequate. You should use the endpoint scaleprocedure to establish the correct ground distance along each side of the area.

Methods of Area Measurement

Dot Grid:

1. Place the dot grid over the area to be measured. Count the number of complete dots thatfall within the area being measured;

D = dots that fall with area being measured = 20

2. Determine the number of acres per dot for the scale of the map being measured byconverting the scale expression to acres per square inch.

a. Let N = number of dots per square inch on dot grid= 100 dots per square inch

b. Let S = map scale in acres per square inch, thus

1) If scale is in feet as 1 inch = 2000 feet thenS = (2000)2/43,560 square feet per acre = 91.827 acres per square inch

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2) If scale is in chains as 1 inch = 20 chains thenS = 202/10 square chains per acre = 40 acres per square inch

c. Let A = acres per dot = S/N

1) for S = 91.827 acres/sq. in. A = 91.827/100

= 0.91827 acres per dot

2) for S = 40 acres per square inch A = 40/l00

= 0.40 acres per dot

3. Compute area in acres as:

Acres = Dots * Acres/dot = D * A

1) A = 20 * 0.91827= 18.37 acres

2) A = 20 * 0.40= 8.00 acres

Polar Planimeter:

1. Connect the pivot arm to the main body of the planimeter.

2. Place the planimeter and pivot arm on the map to be measured so that all wheels of theplanimeter will remain on the map surface as the pointer is moved around the tract boundaryin a clockwise direction.

3. Position the pointer at a beginning point, on the tract to be measured, and read the dialsfor the starting number. Do not try to "zero" the dials; read the numbers as is. There are 2dials and a vernier.

a. The large dial reads in 10's of square inches, as 10, 20, 30, etc.,

b. The small dial reads in square inches and tenths of square inches, and

c. The vernier reads in hundredths of square inches. Find the vernier marks that lineup completely, top and bottom, and read the hundredths.

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4. Move the pointer in a clockwise direction around the boundary of the tract and trace theboundary 2 times back to the starting point.

5. Read the dials for the ending number. This is 2 times the square inches for the tract. Halfthe square inches.

6. Multiply the measured square inches times the acres per square inch for the map (see step2.b. above) to obtain the acres for the tract.

Geometric Figures and Polygons:

An area can be subdivided into geometric figures and/or polygons and the area computedwith the appropriate area equation:

Square: area = L x WTriangle: area = 1/2 B HPolygon:

Computer Assisted Mapping/Drafting (CAM/CAD) Systems:

Most CAM/CAD systems permit area measurement of digitized polygons, but most of thesesystems compute the area in square inches only and the user must convert the square measure todesired area units. The area to be measured is placed on a digitizing table/tablet and either digitizedinto the system with a digitizing puck or the puck is used to "trace" or "click-on" the boundary.

The ACRES Measurement Software System developed by Dr. Robert C. Parker runs ona microcomputer and utilizes a digitizing table or tablet and a puck to trace the area. The userspecifies the scale of the map/photo in terms of RF, feet per inch, chains per inch, or acres per squareinch and the measured area and perimeter are computed at the specified scale to yield area in acresand perimeter in the specified linear measure.

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Sample Problems on Area Measurement:

1. For a photo scale of 1:40,000 and a dot grid with 100 dots per square inch, the acres perdot will be: ________________.

2. You are using a planimeter on a photo of scale 1:15,840 and got the following 1-passreadings. The acreage of each tract is:

a. 8.46 square inches = ___________ acresb. 10.96 square inches = ___________ acres