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It is known that a Immovable Rod will deactivate if more than 8,000 lbs of weight is put on it. So, prompted by this question and some of the answers on it. I "asked my GM" and he told me to ask here (i swear).

If we assume that the rod is stationary to the ground and not the vehicle, meaning that the rod remains in place while a locomotive moves (as seen on the this answer for that question), so the rod would be pushed against one of the inside doors, how much weight could this door withstand before breaking down? Or would the rod deactivate against the tractive force caused by the locomotive against the rod?

To put it into another perspective (excuse me if it's wrong, not a physics expert), would an object that weights 8,000 lbs at x miles/hour be enough to bring down a wooden/iron door?

For the sake of this question, let's consider both wooden doors from earliest locomotives and iron doors from the modern ones. If it's not possible to figure this out due to lack of rules, answers with backup from previous editions are acceptable.

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    \$\begingroup\$ It's fine to say that it came up in chat and didn't have a satisfactorily obvious answer. It's not necessary to sorta pretend that one's GM was asked. ;) \$\endgroup\$ – SevenSidedDie Jun 28 '17 at 19:19
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    \$\begingroup\$ @SevenSidedDie actually the GM would be myself, I thought the joke would be more obvious by the comment within parenthesis. But it's not that the answers on the chat weren't satisfactory, just that the answers should be rewarded properly if they are good. \$\endgroup\$ – ShadowKras Jun 28 '17 at 19:32
  • \$\begingroup\$ Are you trying to determine the maximum speed a train can be going before the door moves the immovable rod? \$\endgroup\$ – Timi Jun 28 '17 at 20:19
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    \$\begingroup\$ @Timi Is that an African or a European Train? \$\endgroup\$ – KorvinStarmast Jun 28 '17 at 20:39
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    \$\begingroup\$ @KorvinStarmast It must be a European train...African trains don't migrate \$\endgroup\$ – A.B. Jun 29 '17 at 11:39
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If you accept answers from 3.5e

If it's not possible to figure this out due to lack of rules, answers with backup from previous editions are acceptable.

then

in 3.5e the door will break first, in 5e it's up to DM.

From the description of the immovable rod:

An immovable rod can support up to 8,000 pounds before falling to the ground. If a creature pushes against an immovable rod, it must make a DC 30 Strength check to move the rod up to 10 feet in a single round.

(emphasis mine)

Table: DCs to Break or Burst Items:

$$ \begin{array}{|l|r|} \hline \mbox{Strength Check to:} & \mbox{DC} \\\hline \mbox{Break down simple door} & 13 \\\hline \mbox{Break down good door} & 18 \\\hline \mbox{Break down strong door} & 23 \\\hline \mbox{Break down barred door} & 25 \\\hline \mbox{Break down iron door} & 28 \\\hline \end{array} $$

The DC for breaking the strongest door is less than the DC for moving the immovable rod. Both are strength checks. It means that the rod is “stronger” than the door.

Though I've never played 5E, I've looked into the 5e SRD and found that the description of immovable rod is almost the same there:

Immovable Rod

Rod, uncommon

This flat iron rod has a button on one end. You can use an action to press the button, which causes the rod to become magically fixed in place. Until you or another creature uses an action to push the button again, the rod doesn’t move, even if it is defying gravity. The rod can hold up to 8,000 pounds of weight. More weight causes the rod to deactivate and fall. A creature can use an action to make a DC 30 Strength check, moving the fixed rod up to 10 feet on a success.

Unfortunately, I haven't found the same table for breaking objects, the SRD only says:

A character can also attempt a Strength check to break an object. The GM sets the DC for any such check.

So, in 5e it seems that designers left the descision to a DM.

This is what we can know from the rules and their interpretation. To get more we need to refer to real world physics, but it is not safe, because real world physics often contradicts DnD physics.

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    \$\begingroup\$ @ keithcurtis 50 "If it's not possible to figure this out due to lack of rules, answers with backup from previous editions are acceptable." \$\endgroup\$ – Ols Jun 29 '17 at 1:07
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    \$\begingroup\$ Well then I plus one'd you. It looks like this is not turning out to be a popular answer, but that may have something to do with my lack of familiarity with 3.5. However, it looks like a good quick "let's roll the dice and move on" answer. \$\endgroup\$ – keithcurtis Jun 29 '17 at 1:10
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    \$\begingroup\$ @Adeptus I've added some quotes from 5e SRD, though they are not as helpful. \$\endgroup\$ – Ols Jun 29 '17 at 10:03
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    \$\begingroup\$ @ShadowKras If the door moves and meets the rod then yes, it can. You need an impact anyway. The DC's in the table are for situations when a creature tries to break a door. That is the creature moves. It may not move in squares, but it needs to press against the door somehow, and that's also movement in physical sense. But the rod is immovable, thus the door has to move. The rod and the door have to be moving relative to each other. If they are moving relative to anything else or not is irrelevant. \$\endgroup\$ – Ols Jun 29 '17 at 14:54
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    \$\begingroup\$ DMG p.238 has rules for setting DCs. DCs of 25 or 30 are nearly impossible tasks that you normally wouldn't even roll for. Unless you're getting a bonus from Bardic Inspiration or magic, you'd need a nat 20 and a Strength score of 20 (the maximum for PCs) to make a DC 25 Strength check, or a Strength score of 30 (the absolute maximum for creatures, e.g. a titan or deity.) I wouldn't expect even an iron door to put up so much resistance that a titan needs a nat 20 to succeed, so I'd say the rod wins. \$\endgroup\$ – Doval Jun 29 '17 at 15:09
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See this BBC website for a really basic primer.

All we have to go on is a weight of 8,000 lbs - assuming they actually mean pound force not pounds this is a mass of 3,628.7 kg or a weight of 35,585.8 Newtons. If they didn't mean pound force then we need to know the acceleration due to gravity at the point where the rod is and since gravity varies with elevation and where you are on the Earth this is problematic.

Now, according to the rules, "The rod can hold up to 8,000 pounds of weight". If we want to be a physics pedant (and who doesn't want to be that?) the weight "of an object is the force of gravity on the object." The force imposed by the locomotive is not weight and neither is the decelerating force the rod imposes on the door. Therefore if the door weighs less than 8,000 lb (35,585.8 N) then the rod does not move - what happens to the door is up to the DM but one imagines it would be spectacular thing to see from a safe distance, particularly with an ultra-slo-mo camera.

However, if you want to assume that the writers of D&D did not actually intend to make a distinction between forces due to gravity (weight) and other forces (not weight) then the question is unanswerable as there is insufficient information. Questions that need to be addressed are:

  • Does the Rod deactivate if there is an instantaneous force above 8,000 lb (35,585.8 N) or must that force be applied for some measurable time and, if so, how long?
  • Does the rod or the door distort under the impact and over what time frame does this occur and what is its shape?
  • At what point will the distortion lead to failure of the rod or the door (or the hinges, or the lock, ...) and what does that look like?
  • How much energey is lost in heat and sound from the impact and what shape is that energy loss over time?

In the real world there are no empirical answers to these questions. Real world physicists would call this an engineering problem, and they would be right. Real world engineers would conduct a series of experiments with different rods and different doors and build a statistical model of rod-door interaction that they could put in a standard for door design to resist immovable rods which everyone would happily use until there was a catastrophic failure when it would need to be revised.

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    \$\begingroup\$ I'd downvote you for not answering the question(I'm not buying: "the question is unanswerable as there is insufficient information"), but this was too entertaining too read, so I'm upvoting. \$\endgroup\$ – godskook Jun 29 '17 at 3:17
  • \$\begingroup\$ Thanks @godskook but how is "the rod does not move" not answer the question? \$\endgroup\$ – Dale M Jun 29 '17 at 3:40
  • \$\begingroup\$ That relied on the tenuous assumption that the authors meant weight as in only gravitational forces, which is, at best, ahistorical to how the magic item has functioned in 3.5. Worse, your answer reads like even you do not believe that answer to be correct; as if you're only covering it for that pedant in the back of the room. \$\endgroup\$ – godskook Jun 29 '17 at 3:58
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    \$\begingroup\$ @godskook answers are not given under oath :-) - just playing Devil's Advocate \$\endgroup\$ – Dale M Jun 29 '17 at 10:08
  • \$\begingroup\$ Which is why I'm not arresting you for perjury :P \$\endgroup\$ – godskook Jun 29 '17 at 13:03
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Assuming the wooden door is 2 inches thick, 6 feet 8 inches high and 3 feet wide, and is perfectly dry, it would have a breaking force of 1250lbs or 8.61844662MPa, assuming the size of the immovable rod is 2cm diameter and 30cm long, is a perfect and regular dodecagonal prism and does not deform under high pressures. This would give a contact area of 0.001554m^2. 8000lbs in Newtons is 35597.9292N, so the door must withstand 22907290.3Pa or 22.9MPa. So under increasing force, the rod would last much longer than the door. This would be useful if one was using the rod as an anchor point and extending something into the door. The door, in this scenario, would give out way before the rod.

However, if the wood was moving, say as a door in a train, then it would provide considerably more force before breaking. F=ma, a=F/m. Assuming the weight of the door is 50kgs and is bends 2cm before breaking, the minimum deceleration a door would have to give to move the rod is 35597.9292/50 = 711.9585ms^-2.

0 = u^2 + 2*(-711.9585)*0.02
v^2 = u^2 + 2as

This works out as u=5.34ms^-1 or roughly 12 miles per hour.

An immovable rod inside a train moving at 12mph, with a door inside that the immovable rod is going to collide with, will have the immovable rod make a hole in the door and then deactivate immediately afterwards.

All in all, decreasing the weight of the wood from 80kgs to 50kgs increased the speed by 2mph. From this we can learn that the weight of the door doesnt matter very much.

Sources
The only outside source I have used is from here and everything else is pure maths.

Wood bares the toleration of roughly 625 pounds per square inch (PSI) of a compression load. Steel can bare 30,000 PSI of a compression load.

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  • \$\begingroup\$ Not sure if a 76kg door is a reasonable weight \$\endgroup\$ – Nat Jun 28 '17 at 21:50
  • \$\begingroup\$ @Nat That is a good point. I am working off the assumption that it is 8 foot by 4 foot and solid for 2 inches. I googled it and it said it was 76kgs. I will look at a different source and recalculate appropriately \$\endgroup\$ – Timi Jun 28 '17 at 22:59
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    \$\begingroup\$ Suggestions: cite a source for your breaking force; don't present 9 digits of precision on a number derived from one with 3; don't equate force and pressure, they're dimensionally inconsistent; describe what support for which edges of the door is already baked into that breaking force measurement; explain why the fact that it's a dodecagonal cross-section matters, or leave that bit out; explain what you mean that a moving door "provides more force" than one that is not; explain your variables (I think u and v are final and initial velocities, not sure?); mark up your equations with... \$\endgroup\$ – nitsua60 Jun 29 '17 at 2:54
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    \$\begingroup\$ All in all, decreasing the weight of the wood from 80kgs to 50kgs increased the speed by 2mph. From this we can learn that the weight of the door doesnt matter very much....... \$\endgroup\$ – Timi Jun 29 '17 at 12:52
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    \$\begingroup\$ @Timi that information is actually really useful if you could include that into your answer. \$\endgroup\$ – ShadowKras Jun 29 '17 at 13:02
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Add 8000 pounds resistance to the door's resistance for one instant effect, then the rod deactivates and any sustained force, such as a continuous stream of moving locomotive car force, would act as if there were no rod at all.

If you are looking for damage values by crossing editions, then you know 8000 lbs doesn't really exist (easily) in AD&D or 5e, but it does exist as a strength value in 3e, and is about 32 Strength, since a cloud giant with a 35 strength is listed as 12,800 lbs.

The door itself probably presents pocket change a small fraction of the 8000lbs resistance, but it's real and won't "deactivate" if your first jolt/hit doesn't destroy the hinges and lock after the rod fails.

A 32 strength is the same as a very strong frost giant, lower than a cloud giant, and higher than a standard frost giant. In 5e, these monsters exist, with the Frost Giant at 23 strength, and the cloud giant at 27 strength. The fire giant in 3e has 31 strength, which is very close to 32, and in 5e the fire giant has 25 strength, which is also midway between the Frost Giant and Cloud Giant.

So I would reference the defense added for 1 attack/sudden burst of force the rod is capable of withstanding, as the max damage you would expect from a Fire Giant, or other 25 strength monster, such as a Goristo (5e MM p. 59) And treat it like the bonus hit points an abjurist wizard gets, in this case, the door is the wizard and the rod is their special ability to have bonus hit points from their 2nd level Arcane Ward.

A door, is probably a medium object (4d8/18 hit points), and worst case scenario 25 strength attacks are around 4d8-7d10 damage, which means a cannon ball or trebuchet force (8d10) has a 100% chance of deactivating the rod and a fair chance of also bashing through the door simultaneously, while in the next round/attack or sustained force, the door doesn't stand a chance.

  • I admit I'm not sure if I understood your question to precision.
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