Scale of Vertical Photograph Numerical

1. Advance Surveying
Aerial Photogrammetry
Prof. Rajguru R.S.
Civil Engineering Department
(rajgururajeshcivil@sanjivani.org.in)
Sanjivani College of Engineering,
Kopargaon,MH,India

2.  Lecture Outline
 Scale of Vertical Photograph
 Numerical

3.  Scale of Vertical Photograph:
• Scale in Aerial Photography is the ratio of the
distance between two points on an image to the
actual distance between the same two points on
the ground.
• Scale is not uniform and it vary from point to
point depending upon the ground profile whether
tilt or tip. If the terrain to be photographed is flat,
the photographs taken are vertical.
• If the altitude of the airplane is changed, then
scale of photograph is also changed.

4.  Scale of Vertical Photograph:
• O=Exposure Center
• f=Focal length
• Ok=Optical axis/camera axis
• H=Altitude of airplane
• 𝒉 𝑨=Height of pt. A above datum
• Scale=
𝑝ℎ𝑜𝑡𝑜 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑔𝑟𝑜𝑢𝑛𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
=
𝑘𝑎
𝐾𝐴
• From similar triangle, ΔOka &
ΔOKA
• s=
𝑘𝑎
𝐾𝐴
=
Ok
OK
=
𝑓
H−ℎ 𝐴
Scale at pt. A on ground

5.  Scale of Vertical Photograph:
• Datum Scale: Scale at datum
S =
𝑓
𝐻
• Point Scale: Scale at any point on ground
𝑆 𝑝 =
𝑓
H−ℎ 𝑝
, Scale at pt. p on ground.
• Average Scale: Average ground variation
𝑆 𝑎𝑣𝑔 =
𝑓
𝐻−ℎ 𝑎𝑣𝑔

6.  Scale of Vertical Photograph:
• True length of line on ground
Let say PQ is the line on ground
PQ= (𝑋 𝑃 − 𝑋 𝑄)2+(𝑌𝑃 − 𝑌𝑄)2
𝑿 𝑷=
𝐻−ℎ 𝑃
𝑓
𝑥 𝑃 , 𝒀 𝑷=
𝐻−ℎ 𝑃
𝑓
𝑦 𝑃 & 𝑿 𝑸=
𝐻−ℎ 𝑄
𝑓
𝑥 𝑄 , 𝒀 𝑸=
𝐻−ℎ 𝑄
𝑓
𝑦 𝑄
H = Flying Height of airplane
𝒉 𝒑 , 𝒉 𝑸 , = Ground elevation of P & Q above datum
𝒇 = Focal length
𝑿 𝑷 , 𝒀 𝑷 & 𝑿 𝑸 , 𝒀 𝑸 = Ground coordinate of P & Q resp.
𝒙 𝑷 , 𝒚 𝑷 & 𝒙 𝑸 , 𝒚 𝑸 = Photograph coordinate of P & Q resp.

7.  Numerical:
Ex.1: A line AB 2000m long lying at an elevation of 500m measures 8.65 cm
on a vertical photograph for which focal length is 20 cm. Determine the scale
of the photograph in an area the average elevation of which is about 800m
Solution:
Scale=
𝑝ℎ𝑜𝑡𝑜 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑔𝑟𝑜𝑢𝑛𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
=
𝑓
H−ℎ 𝐴𝐵
(1m=100cm & 1m=1000mm)
Let find out H first and then scale,
𝑝ℎ𝑜𝑡𝑜 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑔𝑟𝑜𝑢𝑛𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
=
𝑓
H−ℎ 𝐴𝐵
,
0.0865
2000
=
0.20
H−500
, H=5124.027m
Scale of Photograph at Avg. elevation of 800m,
𝑆800 =
20𝑐𝑚
5124.027−800
=
20𝑐𝑚
4324.027𝑚
=
1𝑐𝑚
21620.135𝑚
Therefore , 𝑆800 is 1cm =216.20cm

8.  Numerical:
Ex.2: Two point A and B which appear in a vertical photograph taken from a
camera having focal length of 220mm and from an altitude of 3000m have
their elevation as 400m and 600m respectively Their corrected photo
coordinate are as under, Determine the length of ground line AB.
Solution:
Given-
H = Flying Height of airplane/altitude =3000m
𝒉 𝑨 , 𝒉 𝑩 , = Ground elevation of A & B above datum = 400m ,600m resp.
𝒇 = Focal length =220mm=0.220m
𝑿 𝑨 , 𝒀 𝑨 & 𝑿 𝑩 , 𝒀 𝑩 = Ground coordinate of A & B resp.= Unknown
𝒙 𝑨 , 𝒚 𝑨 & 𝒙 𝑩 , 𝒚 𝑩 = Photograph coordinate of A & B resp.= As per above
Point Photo coordinates (mm)
𝒙 𝒚
A 23.8 16.4
B -13.6 -29.7

9.  Numerical:
First let find out Ground coordinate of A point
• 𝑿 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑥 𝐴 =
3000−400
.220
0.0238 = +281.27 m
• 𝒀 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑦 𝐴 =
3000−400
.220
0.0164 = +193.82 m &
Let find out Ground coordinate of B point
• 𝑿 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑥 𝐵 =
3000−600
.220
(−0.0136) = - 148.37 m
• 𝒀 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑦 𝐵 =
3000−600
.220
(-0.0297) = - 324 m &
Let find out length of line AB on ground
• AB = (𝑋 𝐴 − 𝑋 𝐵)2+(𝑌𝐴 − 𝑌𝐵)2 = (281.27 − (−148.37))2+(193.82 − (−324))2 =
672.85 m

10.  Numerical:
Ex.3: The ground length of a line AB is known to be 545m & the elevation of
A & B are respectively 500m & 300m above m.s.l. On a verical photograph
taken with a camera having focal length of 20 cm .The distance ab scaled
directly from photograph is 5.112 cm. Calculate flying height above msl.
Solution:
Given-
H = Flying Height of airplane/altitude =Unknown
𝒉 𝑨 , 𝒉 𝑩 , = Ground elevation of A & B above datum = 500m , 300m resp.
𝒇 = Focal length =20 cm = 0.20 m
𝑿 𝑨 , 𝒀 𝑨 & 𝑿 𝑩 , 𝒀 𝑩 = Ground coordinate of A & B resp.= Unknown
𝒙 𝑨 , 𝒚 𝑨 & 𝒙 𝑩 , 𝒚 𝑩 = Photograph coordinate of A & B resp.= As per above
Point Photo coordinates (cm)
𝒙 𝒚
A 2.65 1.36
B -1.92 3.65

11.  Numerical:
First let find out the 𝑯 𝒂𝒑𝒑𝒓𝒐𝒙
photo distance
ground distance
=
f
Happ−havg
, here A & B having different elevation
havg =
𝟏
𝟐
(hA + hB) =
1
2
( 500 + 300 ) = 400 m
photo distance
ground distance
=
f
Happ−havg
=
𝟓.𝟏𝟏𝟐 𝒄𝒎
𝟓𝟒𝟓 𝒎
=
𝟐𝟎 𝒄𝒎
(Happ −𝟒𝟎𝟎 ) 𝒎
Happ= 2532.23 m
• Now find out ground coordinate A & B from Happ= 2532.23 m

12.  Numerical:
First let find out Ground coordinate of A point
• 𝑿 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑥 𝐴 =
2532.23−500 𝑚
20 𝑐𝑚
2.65 𝑐𝑚 = 269.27 m
• 𝒀 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑦 𝐴 =
2532.23−500
20
1.36 = + 138.19 m &
Let find out Ground coordinate of B point
• 𝑿 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑥 𝐵 =
2532.23−300
20
(−1.92) = - 214.3 m
• 𝒀 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑦 𝐵 =
2532.23−300
20
3.64 = 406.3 m &
Let find out length of line AB on ground
• AB = (𝑋 𝐴 − 𝑋 𝐵)2+(𝑌𝐴 − 𝑌𝐵)2 = (269.27 − (−214.3))2+(138.19 − 406.3)2 =
552.92 m based on approximate height
• Actual length of AB given in the problem is 545 m.

13.  Numerical:
let find out second approximate height & ground coordinate
Happ−havg
Happ−havg
= 𝑮𝒊𝒗𝒆𝒏 𝑨𝑩
𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅 𝑨𝑩
,
Happ−400
2532.23−400
=
𝟓𝟒𝟓 𝒎
𝟓𝟓𝟐.𝟗𝟐 𝒎
,
Second approximate height , Happ = 2501.68 m
let find out Ground coordinate of A point
• 𝑿 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑥 𝐴 =
2501.68−500 𝑚
20 𝑐𝑚
2.65 𝑐𝑚 = 265.22m
• 𝒀 𝑨=
𝐻−ℎ 𝐴
𝑓
𝑦 𝐴 =
2501.68−500
20
1.36 = + 136.11 m &

14.  Numerical:
Let find out Ground coordinate of B point
• 𝑿 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑥 𝐵 =
2501.68−300
20
(−1.92) = - 211.36 m
• 𝒀 𝑩=
𝐻−ℎ 𝐵
𝑓
𝑦 𝐵 =
2501.68−300
20
3.64 = 400.7 m &
Let find out length of line AB on ground
• AB = (𝑋𝐴 − 𝑋 𝐵)2+(𝑌𝐴 − 𝑌𝐵)2
= (265.22 − (−211.36))2+(136.11 − 400.7)2 = 545 m
This agree with given AB , Hence flying height = 2501.68 m

15. THANK YOU