Altitude question?

Question

You need to obtain photographs at a scale of 4 inches per mile with a camera focal length of 6 inches. Average ground elevation of the area to be photographed is 1,500 feet above mean sea level. What flight altitude, in feet, above mean sea level must be maintained to obtain the desired scale.

 

So I know the scale, the camera focal length, and the ground elevation, but I'm confused as to how to solve for flight altitude.

Answer 1

The formula in optics relating a lens's focal length f to the object distance u and image distance v is as follows:

 

1/f = 1/u + 1/v  --Equation 1

 

The image size to object size ratio should be 4 inches to 1 mile or 1:15840

 

By the property of similar triangles, the ratio of object size to image size equals the ratio of object distance to image distance.  This gives us

 

u/v = 15840or 1/v=15840/u

 

Applying this relation in Equation 1 above, we have

 

1/f = 15840/u + 1/u

 

or 15841/u = 1/f.  But we know that f=6"

 

Therefore 15841/u=1/6 or u = 6*15841=95046" = 7,290' from the area to be photographed.  Since the area to be photographed is at 1,500' above sea level, the flight altitude must be 7,290 + 1,500 = 8,790'