Jonathan,
Agreed that there is much math in photography or viceversa - the challenge you
posed was to somehow make the library building a part of the project! And then
there was the matter of getting students to "engage" rather than simply do
something they might call "boring".
Anyway, I think this suggestion was made by others and is connected with the
little size/distance relationship situation. A student of mine suggested you
make a photograph at right angle to the building wall and with the camera level
encompassing the whole width of the building locating the building's base in the
along a line across the middle of the frame. Then, without changing camera
location you aim the camera so the top of the building is in the middle of the
frame. Now you pace off the distance to the building wall.
So now you can compare the width of the image of the building in case 1 with the
width of the building top in Case 2. This should give you an idea of how much
further the top is from the camera position than the bottom. Let's say the top
appears to be 1/2 the length of the building side. That means the top is twice
as far as the bottom. So, if the camera was 25 meters from the building base the
top is 50 meters away.
At this point I'd resort to a graphical solution to determine the height of the
wall. Draw a right angle and call the horizontal ground level. Identify a point
as being located 25 m away from the wall. Use a compass set to an "opening" that
is twice that distance. Draw an arc and where that intercepts the vertical part
of the right angle is an indication of the height of the wall.
To determine that in actual dimensions compare the height of that segment to the
segment you called 25 m. If it is 1.5 times as long as the 25 m dimension the
height of the wall is 25 x 1.5 or 37.5 meters.
In the case mentioned above actually the height would be 1.73 times the distance
to the building or 25 m. And thus the height would be about 43 m.
Or, you could square the 25 and subtract that from squaring the 50 and then find
the square root of the difference and that would be the height of the wall. In a
right triangle the sum of the squares of each side is equal to the square of the
hypotenuse (side opposite the right angle).
All this assumes a square or rectangular wall.
I hope the logic above is OK - if not hit me with a wet noodle!
Andy
PS: I would not suggest having students at ground level photographing something
like a falling refrigerator!
