Chapter 14, Problem 10P

To determine

The scale of the first vertical aerial photograph and the flying height of both the photographs.

Answer to Problem 10P

  $\underset{\_}{1:19,070},\underset{\_}{{\text{H}}_{\text{1}}=9835\text{ft}\text{,}}\text{and}\underset{\_}{{\text{H}}_{\text{2}}=12300\text{ft}}\text{.}$

Explanation of Solution

Given information:

${\text{Scale of the first photograph S}}_1=1:24,200$, $\text{focal length of the camera, }\text{f}\text{=}\text{6}\text{inches}$ and the average elevation, $\text{h}\text{=}\text{300}\text{ft}\text{.}$

Calculation:

Take two control points on a vertical aerial photograph:

Let the length of DE $\text{=}\text{X}\text{ft}\text{.}$

We have been given the value of BC $\text{=}4.9\text{inch}\text{.}$ for one photograph.

For scale two the value of X is found as follows:

  $\frac{\text{BC}}{\text{DE}}=\frac{\text{AB}}{\text{AD}}$

Substituting the values, we get

  $\text{Put}\frac{\text{AB}}{\text{AD}}=1:24,000\text{and}\text{BC}=4.9\text{inch =}\text{0}\text{.41}\text{ft}$

  $\begin{array}{l}\frac{0.41}{\text{X}}=1:24,000\\ \frac{0.41}{\text{X}}=\frac{1}{24,000}\\ \text{X}\text{=}0.41\times 24,000\\ \text{X}\text{=}9840\text{ft}\text{.}\end{array}$

Now, the value of BC $\text{=}6.2\text{inch}\text{.}\text{=}\text{0}\text{.516}\text{ft}$ for another photograph.

We have the following formula, for scale first:

  $\frac{\text{BC}}{\text{DE}}=\frac{\text{AB}}{\text{AD}}$

Substituting the values, we get

  $\text{ Put}\text{BC}=6.2\text{inch}\text{.}\text{=}\text{0}\text{.516}\text{ft}\text{and}\text{X}\text{=}9840\text{ft}\text{. }$

  $\begin{array}{l}\frac{0.516}{\text{X}}=\frac{\text{AB}}{\text{AD}}\\ \frac{0.516}{9840\text{ft}\text{. }}=\frac{\text{AB}}{\text{AD}}\\ \frac{\text{AB}}{\text{AD}}\text{=}19,069.76\\ \frac{\text{AB}}{\text{AD}}\text{=}19,070\\ \frac{\text{AB}}{\text{AD}}\text{=}\frac{1}{19,070}={\text{S}}_{\text{1}}\end{array}$

Implying that the scale of the first vertical photograph is $1:19,070$

Now, the flying height $({\text{H}}_{\text{1}})$ of the first photograph using formula

  ${\text{S}}_{\text{1}}=\frac{\text{f}}{{\text{H}}_{\text{1}}-\text{h}}$

Where, ${\text{Scale of the first photograph S}}_1=1:24,200$, $\text{focal length of the camera, }\text{f}\text{=}\text{6}\text{inches}$ and the average elevation, $\text{h}\text{=}\text{300}\text{ft}\text{.}$

  $\begin{array}{l}\frac{1}{19,070}=\frac{0.5\text{ft}}{{\text{H}}_{\text{1}}-3\text{00}\text{ft}}\\ {\text{H}}_{\text{1}}-3\text{00}\text{ft}=0.5\text{ft}\times 19,070\\ {\text{H}}_{\text{1}}-3\text{00}\text{ft}=9535\text{ft}\\ {\text{H}}_{\text{1}}=9535\text{ft}\text{+}3\text{00}\text{ft}\\ {\text{H}}_{\text{1}}=9835\text{ft}.\end{array}$

The flying height of the first photograph is, ${\text{H}}_{\text{1}}=9835\text{ft}\text{.}$

Now, the flying height $({\text{H}}_{\text{2}})$ of the first photograph using formula

  ${\text{S}}_{\text{2}}=\frac{\text{f}}{{\text{H}}_{\text{2}}-\text{h}}$

Where, ${\text{Scale of the first photograph S}}_2=1:24,000$, $\text{focal length of the camera, }\text{f}\text{=}\text{6}\text{inches}$ and the average elevation, $\text{h}\text{=}3\text{00}\text{ft}\text{.}$

  $\begin{array}{l}\frac{1}{24,000}=\frac{0.5\text{ft}}{{\text{H}}_{\text{2}}-3\text{00}\text{ft}}\\ {\text{H}}_{\text{2}}-3\text{00}\text{ft}=0.5\text{ft}\times 24,000\\ {\text{H}}_{\text{2}}-3\text{00}\text{ft}=12000\text{ft}\\ {\text{H}}_{\text{2}}=12000\text{ft}\text{+}3\text{00}\text{ft}\\ {\text{H}}_{\text{2}}=12300\text{ft}.\end{array}$

The flying height of the first photograph is, ${\text{H}}_{\text{2}}=12300\text{ft}\text{.}$

Conclusion:

Therefore,the scale of the first vertical photograph is $1:19,070$, the flying height of the first photograph is, ${\text{H}}_{\text{1}}=9835\text{ft}\text{,}$ the flying height of the first photograph is, ${\text{H}}_{\text{2}}=12300\text{ft}\text{.}$