Long jump apex height

I can't seem to find any rule for how high you rise while making a long jump in Pathfinder. I believe 3.5? had it as you rise 1 quarter of the total distance traveled horizontally. i.e.: Long jump 40ft = rising 10 ft. This becomes important when your monk is trying one of his crazy distance long jumps in a chamber with a ceiling that is only about 10 or so feet high.

An informatic would be great for both jumping and the flying skill. As it stands we have one member of the party do Pythagoras Theorem on his smartphone while the rest of us wait for him and his precious science.

I would very much like to have an official answer for this too, as it came up in one of my PFS games this weekend. We settled on the 3.5 rule as Mark described above, but if there is no height restriction for long jumps of 20' or more. it could have made an already strong PC a super munchkin for that encounter.

No pythagoeans necessary... unless gravity works funny in your world and things don't follow a parabola.

I'm almost sure there's no rules answer, so here's the google+math answer.

This suggests that 22 degrees is the optimum take off angle for a jumper. Assuming your players don't want to land prone we'll treat the landing height as the same as the takeoff height...

Quick answer: a little over 1/5 the jump distance.

v is velocity of jump
vy is vertical velocity at start
xy is horizontal velocity (assuming constant ignoring wind resistance)
h is vertical distance traveled
s is seconds in air
d is horizontal distance traveled

vy = sin22 * v
vx = cos22 * v (so v = vx/cos22)
d = vx * s (so vx = d / s)

h = vy * s / 2 = sin22 * v * s / 2 = sin 22 * vx * s / 2 / cos22 = sin 22 * (d / s) * s / 2 / cos22 = tan22 * d / 2 = .202 * d

No pythagoeans necessary... unless gravity works funny in your world and things don't follow a parabola.

The falling damage rule (linear damage) strongly suggests that gravity works funny in the Pathfinder world.

Speed when you hit the ground is proportional to the square root of the height you fell from. Kinetic energy when you hit the ground is proportional to the square of the speed you are going. So kinetic energy is directly (linearly) proportional to the height you fall from (still ignoring air resistance).

h = v^2 / 2 g
KE = mv^2 /2

I have generally followed the rule of thumb that your height is 1/4 of your distance, based on the fact that the DC for a high jump is 4 times the height, and the DC for a long jump is 1 times the distance. That takes you a bit higher than oneplus' math indicates it should, but is easy to calculate and has at least a tangential rules basis.

I have generally followed the rule of thumb that your height is 1/4 of your distance, based on the fact that the DC for a high jump is 4 times the height, and the DC for a long jump is 1 times the distance. That takes you a bit higher than oneplus' math indicates it should, but is easy to calculate and has at least a tangential rules basis.

That's also the way it was calculated in 3.X, and it's what I use, too.

The trajectory math is reasonably easy to work out. I did the figures for a 50-foot-long, 1-foot-high longjump a while back. You need to be going something like 70 miles an hour on launch to make it happen. Doing that inside of ten feet is something like 16g. For a 200 pound adventurer, that's 3200 pounds. Presuming that translates neatly to drag capacity, you could do it with 24 strength.

It's kinda silly once you start trying to stuff reality into it.