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How fast would a Tenser's Floating Disk descend if I pulled it over a hole covered by a plank and then removed the plank so that the disk falls?

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There's no way you can bring Tenser's floating disk down a long drop unless the terrain is, itself, moving. Ask your GM what it means when the terrain moves out from under your floating disk, because the game has no mechanism to let you know that before you try it.

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    \$\begingroup\$ @Scanicus Please do not be rude to people on this site. We expect everyone to Be Nice to one another. That means that you should not be telling people to delete their answers or to go away. If you don't like it because you think they misunderstood the question, feel free to kindly help them by explaining what they missed. Otherwise, when you have enough reputation, you can downvote and move on. \$\endgroup\$ – Rubiksmoose Apr 5 at 20:52
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It's unclear, because the rules as written (RAW) don't address this.

In this situation, a Tenser's floating disk floats above a surface, and the surface suddenly becomes a hole. Unfortunately, the rules provide no guidance on what exactly happens next, since the disk is not expected to be above a hole that is 10+ feet deep.

The spell's mechanics suggest that it wasn't intended for this scenario, since it normally "floats 3 feet above the ground" and "can move across uneven terrain, up or down stairs, slopes and the like, but it can't cross an elevation change of 10 feet or more".

The spell description doesn't specify a speed for how quickly the disk moves, either horizontally or vertically. The disk itself is a "horizontal plane of force", which may or may not mean the plane is fixed in place.

So there are various possible outcomes, depending on interpretation:

  1. The disk is not stationary, and therefore it would fall like any other object affected by gravity. It falls until it floats 3 feet above a surface, or exceeds the maximum distance from the caster.

  2. The disk is affected by gravity, but "can't cross an elevation change of 10 feet or more". If the hole is 10+ feet deep, then the disk stays in place. Otherwise the disk falls, as described above.

  3. The disk is magic force, and thus gravity doesn't apply, so it remains hovering in place.

Ultimately the DM would need to decide on what happens.

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    \$\begingroup\$ Options 4, the spell ends early. \$\endgroup\$ – Pureferret Apr 4 at 19:13
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Effectively instantaneously.

According to XGtE, the default falling speed in DnD is 500 feet if falling speeds are being used at all. However, Tenser's Floating Disk has a maximum range of 100 feet, and will dissipate if forced beyond that range. As a result, if it were to fall as a result of the a pit trap being opened underneath it, it would fall to the bottom of the pit so that it can continue hovering 3 feet above the floor of the pit, or it will reach its maximum distance and the effect will end.

If the wizard who cast it is also caught by the pit trap, the disk will follow the wizard at the same speed as the wizard, unless something like the Featherfall spell slows down the wizard's fall - in which case the wizard will fall at their reduced speed, and the disc will fall at its normal speed (and the spell will end if the disc goes more than 100 feet away from the wizard at any point during the fall).

If the wizard who cast it is caught in a pit trap and the disc isn't, it will float to the edge of the trap if if the trap is more that ten feet deep and then remain there; if the wizard falls more than 100 feet, or falls and then moves 100 feet away from the disc, the spell will end.

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    \$\begingroup\$ If falling speeds are used at all, xanathars provides one optional rule of 500' per turn, but that's not the default. \$\endgroup\$ – NautArch Apr 5 at 11:26
  • \$\begingroup\$ what is the default? \$\endgroup\$ – Scanicus Apr 5 at 12:44
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    \$\begingroup\$ @Scanicus See this question but it's basically up to the DM. \$\endgroup\$ – NautArch Apr 5 at 16:32

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