Survey Iv

TECHNICAL UNIVERSITY OF MOMBASA

FACULTY OF ENGINEERING AND TECHNOLOGY

DEPARTMENT OF BUILDING AND CIVL

CLASS: BCE/12JA/MA

REG NO: BEBC/023J/2012

NAME: OPWORA N HARUN

UNIT: SURVEY IV

TITLE: ASSIGNMENT 1

LEC: MADAM NDUATI

1.An area of 150 km² at datum level is to be photographed at a scale of 1:10000 using a camera whose focal length is 152.4mm.The photo format is 23*23cm.Fore overlap and after overlap are 60% and 25% respectively. Assuming an aircraft speeds of 250km/hr compute the following.

i. Flying height of the aircraft

S=f/H-h but h=o then

S=f/H 1/10000=0.1524/H

H=1524m

ii. Time interval between exposure

Find base line first

BL=s’*W*(1-WO) where

W=23cm=width of negative photo plate

WO=width of percentage of fore overlap

Therefore

BL=10000*0.23*(1-0.6) =920m

Given

Flight speed=250km rep 1hr

0.92 reps?

Time interval between exposure=0.92*1/250*3600=13sec

iii. Number of photos

Given: fore overlap=60% and after overlap =30%

BL=S’*W*(1-0.6)

10000*0.23*(1-0.6) =920m /0.92km

Length of study area given = 15km

Number of photos longitudinally

=study length/base line

=15/0.92 =17 photos

But normally there is plus 4 photos, hence

Total photos=17+1= 18photos

In the lateral direction, base line is given as

BL=S’*W*(1-0.25)

=10000*0.23*(1-0.25) =1725m

Number of flight lines=perpendicular study distance/base line

=10*1000/1725 =6lines

Total number of flight lines=6+1 =7 slides

The total number of photos in the study area=number of photos*number of slides

=18*7=126 photos

2. using an appropriate diagram explain relief displacement and derive the expression for the relief displacement relationship.

Relief displacement is the shift in an object’s image position caused by its elevation above a particular datum. For vertical or near vertical photography the shift occurs radialy from the nadir point.

This...