| Bradley Mickle |
For stairs (or likely ramps even):
"Difficult Terrain: Difficult terrain, such as heavy undergrowth, broken ground, or steep stairs, hampers movement. Each square of difficult terrain counts as 2 squares of movement. Each diagonal move into a difficult terrain square counts as 3 squares. You can’t run or charge across difficult terrain."
| Lakesidefantasy |
Difficult terrain aside, I guess you could think about this by considering the equivalent orthogonal movement to get to neighboring 5 ft. cubes in a three dimensional grid.
From any 5 ft. cube there are 26 adjacent cubes. Six of those adjacent cubes are connected at one of the cubes faces, 12 are connected at its edges, and 8 at its vertices.
It takes 5 ft. of orthogonal movement to get to a cube connected at a face, 10 ft. to get to one connected at an edge, and 15 ft. for one connected at a vertex.
Considering squares in a two dimensional grid, the 15 ft. rule for a second diagonal requires 20 ft. of orthogonal movement.
| Bradley Mickle |
I'd say half speed as if it were on a 2D grid. 15 ft diagonal and up 15 feet, assuming a 30 feet move speed.
And maneuverability doesn't automatically grant you the ability to fly straight up in PF. It grants you a bonus +8 to fly skill checks. Flying at an angle greater than 45 degrees is a 20 DC fly check.
| Lakesidefantasy |
I think I agree with Bradley if you're flying up. You would be moving diagonally so every other cube would count as two, and you are moving up so you would be moving at half speed. So, the first cube would be 10 ft. of movement while the second would be 30 ft. The third would be 40 ft. of movement and the fifth would be 60 ft.
However, flying down would be different, as would moving around without the effects of gravity.
| DM_Blake |
Assuming perfect maneuverability or no gravity or a downward diagonal, isn't there just a basic mathematical formula here?
Yes there is, but we need to understand the math for the 2D grid first.
On a 2D grid, the math is easy, we're just calculating the hypotenuse of a triangle with sides of 5' each: sqrt(5^2 + 5^2) = sqrt(50) = 7.07 feet, or nearly 15 feet when you move two of them which is why we have the rule that the first diagonal counts as 5 feet (7.07 feet rounds down to 5') and the second diagonal is 10 feet (14.14 rounds up to 15').
For the third dimension, you just do it again, using the 5' and 7.07' dimensions for the two sides: sqrt(5^2 + 7.07^2) = sqrt(25+50) = sqrt(75) = 8.66 feet. (not for math geeks, there is a faster way but the formula is not as intuitively obvious).
That would equate to a 10' first move (8.66 rounds to 10) and a 5' second move (17.32 rounds to 15'), then 10' again (25.98 rounds to 25') then another 10' (34.64 rounds to 35'), then repeat the pattern (the next move is 10' (43.3 rounds to 45'), the next would be 5' (51.96 rounds to 50'), etc.)
Or just to be simpler, why not say each 3D diagonal amounts to 10'?
| Bradley Mickle |
DM_Blake:
What if you double move?
You could go 15 feet then bank 90 degrees and do 15 more feet, you'd be 30 feet in the air. That same move at a 10 foot speed would be 10 feet, insufficient movement left, then 10 feet and insufficient movement stop, so only 20 feet. You have to compensate for that remainder, which is why, I think, for simplicity, just picture it on a 2D board and then elevate it up to amount moved that round.