Reverse Gravity - the sky is the limit ?

[Castellan]

Actually, of course, the character won't just "stop" at the top of the range, but will keep "falling" upwards until his speed reaches 0, and would then fall back downwards again until he hit the Reverse Gravity. He will thus oscillate between the two until he eventually comes to rest at the top of the spell area. Ignoring air resistance, and assuming that terminal velocity is not reached, then I am fairly sure that his initial oscillation will take him to twice the height of the spell (please correct me, I don't have the time to work it out properly!); alternately, if he falls upwards more than 200' (at which point DnD assumes terminal velocity is reached since damage is constant) he will fall an additional 200' from the top of the spell.

You certainly are correct, if you choose to apply physics to the situation. In the 3.5 SRD, it explicitly states that targets "fall upward and reach the top of the area in 1 round." Depending on the height of the 10-foot cubes, this could mean a strong or weak gravitational field. The description also states that "f an object or creature reaches the top of the area without striking anything, it remains there, oscillating slightly, until the spell ends."

So, the designers realized that D&D has lots of geeks who play, and that some of them would want to apply real physics to the situation. They countered those geeks by imposing certain rules.

However, that's not fun, so let's take the gloves off and do some math. For purposes of this experiment, there'll be no air resistance, and we'll assume that there are no wind or Coriolis effects that would cause a target to "drift" out of the affected area (something that is legitimately a problem, though probably only on super windy days).

We'll presume a wizard/sorcerer is casting it at the first opportunity he can, so he's 13th level. We'll round down for the number of 10-foot cubes he gets, so he has 6, which he'll stack, one on top of the other, for a 60-foot high area of effect. The target is assumed to have failed his saving throw.

Finally, we'll also assume that gravity is earth-like, at 32-feet-per-second-squared.

To find the velocity of the target at the top of the effect, we'll use the following equation:

1. v(f)^2 = v(i)^2 + 2ad.

The final velocity (v(f)), squared, is equal to the initial velocity (v(i)), squared, plus two times the product of acceleration (a) and distance (d). Since the initial velocity is zero (the target was standing on the ground), this is reduced to:

2. v(f)^2 = 2ad,

or, plugging in,

3. v(f)^2 = 2 * 32 ft/(s^2) * 60 ft = 3840 (ft^2)/(s^2),

and,

4. v(f) = (3840 (ft^2)/(s^2))^(1/2) = 62 ft/s [rounding, slightly].

So, at the top of the spell effect, the target has an upward velocity of 62 feet-per-second. This will be bled off by the now-active downward gravity. the bleed-off will take exactly the same amount of time, so we can assume that the target will max out at 120-feet in height, at which point, he'll have a velocity of zero. Gravity will begin pulling him downward again, and at the barrier of the spell area, he'll begin decelerating, once more.

With no air resistance, he'll end up with a velocity of zero, just as his feet touch the ground. Now, if you include air resistance, he won't quite make it back down to the ground, but he'll get really, really close (air resistance isn't that great, actually). So, he'll oscillate in a slightly decaying manner, up and down, up and down, until the end of the spell. Let's see what happens if we let the spell run out. It's on round per level, so we're going for 13 rounds, or 78 seconds. But, how long did it take for our target to do his one complete oscillation?

We'll use the equation,

5. v(f) = v(i) + at,

where (t) stands for time, in seconds. Solving for (t), we get,

6. t = (v(f) - v(i)) / a.

During the first half of the target's upward trip, he begins with an initial velocity of zero, so we get:

7. t = (62 - 0) / 32 = 1.9 seconds

to reach the top of the spell effect. There are four segments for this motion: (1) from rest on the ground to the top of the spell effect, (2) from the top of the spell effect to rest in mid-air, (3) from rest in mid-air down to the top of the spell effect, and (4) from the top of the spell effect to rest on, or near, the ground. So, we'll get our total time for round-trip of:

8. t = 1.9 * 4 = 7.6 seconds,

or slightly longer than one round. We need to figure out what happens after 78 seconds, though. What if the target is miraculously lighting on the ground as the spell ends? Useful for friends, but not for enemies. What if the target is hovering motionless in the sky? Useful for enemies, but not for friends.

The easiest thing to do is divide the 78 seconds by the 1.9-second trip through one of the segments. Since there are four segments, if we get a number that has no remainder (remember that?) then we know the target has landed on the ground safely at the end of the spell duration. Otherwise, the target is in motion.

9. 78 seconds / 1.9 seconds = 41, or 40 remainder 1.

This means that the target goes through 40 complete segments, or 10 complete oscillations, with one left over. Which one is that? The one sending him back up into the sky. So, at the end of our spell's duration, the target is being flung a final 120-feet into the sky, for a rather unpleasant ending.

Given the various number of 10-foot cubes that can be used, and the spell duration based on level, the plethora of possibilities are left as an excercise for the reader (I've always wanted to say that!).

Enjoy!

 

[Piratecat]

I love this place.

 

[Hand of Evil]

Castellan should not the rotation of the planet be taken into affect, most worlds are still globes and at some point based on the speed of rotation the object would move out of the field. Not that it matters...in the end.

 

[Taloras]

Castellan.......thank you. I will now use this against the group im running for.

I have one question...if a monk finds a way to get out of the reverse gravity area, and begins falling to the ground, can he use the reverse gravity area to slow fall? reaching in at key times to slow his descent?

 

[shilsen]

I have one question...if a monk finds a way to get out of the reverse gravity area, and begins falling to the ground, can he use the reverse gravity area to slow fall? reaching in at key times to slow his descent?

It's a creative idea and not really covered by the rules, so it's up to DM fiat. Incidenatally, why specifically a monk? I presume you're thinking of the monk's Slow Fall ability in some way, but that is irrelevant here, since it only applies if the monk is within touching distance from a wall.

 

[Castellan]

Regarding planetary rotation (see Coriolis Effect, briefly mentioned in my post above) would certainly have an effect. However, it wouldn't have a significant effect over the area this spell covers. Well... unless you're playing an Epic-level campaign (and even then, I wouldn't worry about it until you reach into the hundreds- or even thousands-of-levels campaign).

As for Taloras' question, I'd go with shilsen on this one: DM's call. Under most circumstances, I probably wouldn't allow it, though I'd be more inclined to if it involved a cinematic/creative solution to a problem.

Take all of this a step further: Imagine a wizard teamed up with some sort of ranged marksman (rogue, ranger, whatever). The wizard casts Reverse Gravity and stacks the cubes for maximum height. He targets the area where his sharp-shooter friend is located. Sharp-shooter flings upward (maybe with a seriously high DC on a Balance check or some such), at the uppermost limit (where velocity is zero) shoots someone at great range, or behind a wall, or whatever else you can think of. Sharp-shooter falls back down to the ground, and wizard Dispels the magic before he can rocket back up again. Or, he can lather, rinse, and repeat until the spell runs out (or almost runs out, as the case may be). The sharp-shooter is always popping up just in time to get a shot off, but "ducks" back down into foliage/obscuring mist/behind a wall before anyone can get a bead on him. Of course, that would require DM's permission... You might want to bring something to bribe said DM, too...

Take this a step further. Imagine some sort of clever castle-breaching design. You create two "trampolines" with two magic items. Each has a Reverse Gravity spell on them. One is positioned outside the castle walls. The other is snuck inside the walls and positioned appropriately (Knowledge (Physics) check, or something). Attacking forces run into the outer Reverse Gravity, maybe make some sort of Jump, Tumble, or Balance check (say, DC 20 or 25, so not for the shock troops among you) and fling themselves quickly into the inner castle, where the other Reverse Gravity slows them down, they run out of the area, and wreak havoc.

And think how much fun it'd be when one of the defenders moves that magic item 10 feet to the left....