G373: Lec 8, Tue. Jan. 29 Rev. 1.4, revised January 27, 2002 p. 1/7

Geog 373: Lecture Summary: Tue., Jan 29

• New material: p. 125-153 (scale and relief displacement), p. 154-164 (image parallax)

• Before you begin lab 3, it will lessen the confusion to fix two typos:

On p. 1 of Lab 3:

Change the equation (which we will show below is equivalent to text equation 3.13):

Ho = H∆P/(b+ ∆P ) (1)

to read:Ho = H ′∆p/(b+ ∆p) (2)

where Ho is the height of the object (also denoted as ∆h = ha − hb below, where ha, hb aredefined on p. 160 as the heights of the higher and lower points, respectively), H ′ is the flyingheight above the object (H −h in the text Fig. 3.17, p. 158), ∆p = pa− pb is the differentialparallax (discussed below), and b is the photo airbase. We will follow the book and denotethe airbase B, measured in ground units (m), and the photo airbase b, measured in photounits (mm). They are related by the scale, defined on p. 140:

b =f

H − hB = SB (3)

• On the next page of the lab, make the following changes:

∆h =∆pH

(b+ ∆p)(4)

to read:

∆h =H ′∆p

(b+ ∆p)(5)

(i.e. Eq. (5) is identical to Eq. (2)).

• Now, begin the discussion of relief displacment with Figure 1, a modified version of the textfigure 3.8, showing why relief causes differences between the airphoto location of points andtheir map location:

• Note in Figure 1 that if the ground was flat, the tree base would show up at position A onthe film. Instead, the relief moves the tree’s location on the file to position B.

• The fact that a vertical object appears displaced radially away from the principal point onan airphoto can be used to measure its height. Figure 2 (Fig. 3.13 (p. 152) of the book)shows how:

• As the book shows on p. 149, the fact that AA′A′′ is similar to LOA′ implies that h = dHr

.

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Figure 1: My version of text figure 3.8

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Figure 2: Figure 3.13, p. 152 of text

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• I think that Figure 3 on page 5 is clearer version of text Figure 3.17. It shows a pair ofstereophotos overlying each other, centered on a house at height ha. The principal andconjugate principal points are given by O and O′ (L and L′ on the focal plane), just as inFigure 3.17, and the airbase B, height above datum H and flying height H ′ = H − ha arealso shown. As the Figure shows, is you can measure the parallax p = x − x′, you can theheight of the object above datam, h, via the relationship:

h = H − Bf

p(6)

• Next we need to show that the lab equation (5) is identical to the text equation 3.13 on p.160, i.e. that:

∆h =H ′∆p

(b+ ∆p)=

∆pH ′

pa(7)

• To prove this, from the similar triangles on Figure 3 we get Equation 3.10, p158:

ha = H − Bf

pa(8)

for a single object (say the top of a tree), because of similar triangles (as explained in lecture).Convince yourself, by moving ha up and down in Figure 3, that the higher theobject A is above datum, the larger the parallax pa, which agrees with Eq. (8)).

For a second object b (say the base of the same tree)::

hb = H − Bf

pb(9)

We want the height of the tree, which is ha − hb, so subtract Eq. (9) from Eq. (8):

∆h = ha − hb =

(H − Bf

pa

)−(H − Bf

pb

)(10a)

or rearranging

∆h = fB

(papapb

)−(

pbpapb

)(10b)

∆h =fB

papb∆p (10c)

(10d)

where ∆p = pa − pb.But using Equation (8) we can write the combination:

fB

pa= H − ha ≡ H ′a (11)

where H ′a is the vertical distance of the plane above point a.

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Figure 3: My version of Figure 3.17, p. 158, showing that ha = H − fB/pa

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and

fB

pb= H − hb ≡ H ′b (12)

Which gives two different expressions for the height difference ∆h = ha − hb:

∆h =H ′apb

∆p (13a)

∆h =H ′bpa

∆p (13b)

• Now how do we get (5) from (13)? To show that these agree we need to prove that b+∆p = pa.To see that this is true, Figure 3, which shows that Xa−X ′a is just the airbase B (rememberX ′a < 0 in this example. Since the photo measurements and the ground measurements differonly by the scale S, this means also that xa − x′a = b. This will also be true for a secondpoint b:

xb − x′b = b (14)

To make sure you understand this, take a look at Figure 4 (Figure 3.18 of the text), whichshows the relationship between the parallax and other distances on the photo:

Figure 4: Text Figure 3.18 showing parallax

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• So use Eq. (14) to expand b+ ∆p:

b+ ∆p = b+ pa − pb = b+ (xa − x′a)− (xb − x′b)= xb − x′b + (xa − x′a)− (xb − x′b) = pa (15)

So we now have what we need:

∆h =H ′bpa

∆p =H ′∆p

(b+ ∆p)(16)

as long as we define H ′ = H ′b.