Nono, I see what he means... it's not just the horizontal distance, but the distance added by the drop to the lower plateau. There is probably solid math to determine the length of the gap (pythagoras? if a=across and b=down, then a²+b²=c², where c is the length of the bridge needed to settle on both sides
But unless the drop distance is more than a few feet I don't think it should dramatically affect the numbers.
EXAMPLE: Bridge Four comes to a chasm 20' wide, where the other side is 5' lower than the side they're on (a good sized drop). The length of bridge they would need is about 20.6' (25² + 5² = 425, square root of 425 is 20.6155blahblahblah)
But your point was entirely valid, you have to consider the vertical drop in addition to the horizontal distance.
Exactly. It's not just the horizontal distance we're concerned with here, but the diagonal distance created where the bridge will be going. The bridge needs to be able to be pushed out far enough that the arc of the bridge falling lands solidly and in a stable position to continue on.
And yes, we would need to use the Pythagorean theorem to figured this out (a and b are the horizontal and vertical, in no particular order, and C is the length out bridge needs to be able to reach to)
It's true that a 5 foot or so drop isnt a big deal, but it all depends on how all this is actually measured out. We don't really have dimensions for anything, so its kinda hard to say, but both horizontal and vertical distance are going to play into how the bridges need to be worked. If it's a huge drop, even though it isnt more than say 15 feet on average, we're gonna need to go out pretty far before we can let the bridge drop.
edit: also, inkthinker, your math is wrong. 25^2 is 625, plus 25 (5*5) added to it is 650, which is a diagonal distance of about 25.5 feet after rounding. When dealing with right triangles, your hypotenuse (The diagonal distance) is always gonna be greater than each side. So if there's a drop involved, you're always gonna need more length than your horizontal distance.
Further edit: wait, i see what happened. In the equation you put 25, in the text you said 20. your math is correct with 20 feet, just not 25 feet