ENGI 3703 Surveying and Geomatics

Transcription

1 Paralla Measurement: (Section 27.10) We have seen that by rules of simle geometry that we can determine the elevation of vertical objects in a vertical air hoto. This is due to the fact that the base of the object and to are both visible in the hoto due to the effect of relief dislacement. Equations develoed from similar triangles allow the elevation to be determined because the to and bottom of the object share a common distance from the rincial oint. We can also determine the difference in height of objects that are distant from one another. However, to do this we require more information from a second air hoto and use of the aralla henomena. Paralla is the aarent dislacement of objects due the changing location of the observer. Paralla Demonstration - Hold u a finger at arms length, close one eye and focus on a distant object. Keeing on eye closed and your finger stationary, turn your head from side to side. You ll notice that your finger aarently moves a greater distance than the distant object. This is aralla! In aerial hotograhy the camera moves with the aircraft and images that are closer (higher elevation objects) aear to farther than more distant ones (lower elevation). If we have overlaing aerial hotos, we can use this knowledge to etract elevation data from aerial hotograhs. Toic Lect 23 - Nov 14/07 Slide 1 of 6

2 Aerial Photo Image Pairs - St. John s, NL July 1995 (National Air Photo Library) X- X+ X- X+ Princial Point Corresonding Princial Point Corresonding Princial Point 1 Princial Point 1 Left Image Right Image Toic Lect 23 - Nov 14/07 Slide 2 of 6

3 Paralla Equations Difference in between a & a 1 is the aralla measure () = 1 L,L 1 Here we have a common rincial ais and can develo similar triangles LA 1 A and La 1 a H-h f a 1 o,o 1 a f = B H h Solving for H-h H h = Bf A 1 O,O A 1 B X Also we have triangles LOA and Loa Toic X = f H h Solving for H-h H h = Xf Equating both Bf Xf = H h = X = B similarly Y = B y Lect 23 - Nov 14/07 Slide 3 of 6

4 Paralla Eamle - Find the Elevation of St John s Harbour and Windsor Lake X a X a1 y a1 b Corresonding Princial Point y b1 X b Xb1 Left Image Measurements a =4.28 in b =2.53 in b =3.63 in Paralla Measurements i = i - i1 a = = 3.48 in b =2.53-(-1.05)= 3.58 in Right Image a1 =0.80 in b1 =-1.05 in y a1 =3.85 in y b1 =-2.91 in Toic Lect 23 - Nov 14/07 Slide 4 of 6

5 Find Air Base (B): we must find the distance from the Princial Point (PP) to the Corresonding Princial Point (CPP) on the air hoto (b). This distance must be adjusted for relief dislacement and must be used to determine the flying height (H) at the same time. For this we can utilize a toograhic ma to determine the airbase (B) and from this etract the flying height (H). We measure B from a toograhic ma as 9500 ft and the elevation of the CPP is h B = 670 ft (you would be given this [see ma net age]). We can determine the flying height (H) if we know the the elevation of the Corresonding Princial Point from: 2 L 2 = (H h ) B b (H h ) A a + (H h )y B b (H h )y A a f f f f along the rincial ais all y measurements are zero. In addition, the measurement at the rincial oint is zero this leaves: L 2 = (H h ) 2 B b f 2 (H 670.0) = 6 H =16372 ft 2 Elevation of Harbour (actual 0 ft) f = B H h h = H Bf h = (6) 3.48 h = 7 ft Elevation of Windsor Lake (actual 460 ft) f = B H h h = H Bf h = (6) 3.58 h = 450 ft Toic Lect 23 - Nov 14/07 Slide 5 of 6

6 PP 1 km CPP Toic Lect 23 - Nov 14/07 Slide 6 of 6