The RW Physics of the Decantur of Endless Water

[strongbow]

Editing Note: I will be adding some good estimates/figures here shortly after the final, best estimates are given the thumbs up by the people actually doing the computations, or when I get bored. Also, the initial computations were flawed, so read the whole thread before assuming you have the correct estimates. The original post is below.

I have a weak background in physics, so could someone help me please? I know there is someone out there wanting to show off their mad skills. I know that trying to apply real world physics to D&D is dangerous, but this involves set forces so it shouldn't be a problem.

Assume you have a Decantur of Endless Water, set at the geyser setting.
• “Stream” pours out 1 gallon per round.
• “Fountain” produces a 5-foot-long stream at 5 gallons per round.
• “Geyser” produces a 20-foot-long, 1-foot-wide stream at 30 gallons per round.

Before you turn the decantur on, you mount it in a boat of varying size. First a two man canoe, then a 15 ft by 15 raft with 1 ft thick logs, then a 10 ft by 10 ft raft with 1 ft thick logs. Assume the logs are oak if that helps with density calculations.

The decantur opening plays a large role in determining the force of the blast, so if the formula is easy to calculate, assume 1 inch, 1.5 inches, and 2 inches as the opening size. Perhaps you might come up with a more reasonable size; I don't have a decantur in front of me right now.

When mounted underneath a boat, how much force does the decantur output? Assuming the rafts had 1200 lbs of equipment and adventurers in it, how fast would it go in real speed and D&D ft/round. What would be its maneuverability rating, per the DMG? If the canoe had 400 lbs of gear/people, how fast would it go? How many parts to a multi-part series of questions is too many?

For bonus points, calculate the velocity of an unladen swallow (African & European), a swallow with a coconut, and a swallow with a decantur on geyser mode as they move away from the ground.

I really would like the answers to the decantur questions because I have always been curious and I am designing an adventure. If calculating is too much, please give a formula and I'll plug in the numbers myself, although I would like the formula anyway.

Thank you all for your time and busy calculations,

 

[AuraSeer]

It's a jar, not a jet engine.

The geyser mode is the only one that produces any significant recoil, and it's not very much. The Strength check for a wielder to avoid being knocked down is only DC 12, which is equivalent to an opposed check against a character with Strength 14 who takes 10.

If you're using that force to propel a boat, I'd treat it as if the boat were being rowed by one person of Strength 14. So a canoe would go approximately normal speed, but larger boats would go proportionally slower, assuming they had no other means of propulsion.

Maneuverability class of a boat or ship wouldn't get any better, as IRL it depends mostly on the size and shape of the vessel itself. In fact steering might get a whole lot more difficult if the decanter were mounted in a fixed position. (Plus if it you did mount it on the underside, you'd need a water breathing spell if you ever wanted to turn it off.)

A raft is the only thing that would get some maneuverability benefit, since unaided it steers about as easily as a dead cow. Assuming the raft is small enough to be rowed by one person, and that you mount the decanter somewhere that it can be turned and controlled from on board, I'd let it increase the raft's maneuverability from clumsy to poor.

[Edited to add]
I'll work out the RL physical calculations if you really want them, but the values in kilograms and Newtons and whatnot would be pretty useless in game terms.

 

[strongbow]

I would like some RL numbers to work with, thanks. Metric rates of speed/calculations are fine, I can convert them over to English. Your verbal examples are good. I am curious as to how powerful the decantur is. I have looked online a little bit, and have come up with a range of 120-200 gallons per minute of how much water is displaced by a firehose nozzle approximately 1.5 inches wide. Most of the sources I saw placed about 120-150 gallons per minute. After seaching some more I saw a fire department listing that boasted it had a 1500 gallon per minute pump, although this is distinct from a firehose of course.

The decantur has a 300 gallon per minute rate. Even with a 1 ft dispersal, the force that it generates is tremendous at first glance. Instead of flying the natural course of tens of feet, the force magically ends after 20 ft and does enough damage (1d4) to kill a child with one blast. Granted that isn't saying much in D&D (morality aside), but to someone without a background in physics the decantur is amazing.

I defer to your calculations, AuraSeer, but at first glance the decantur looks more powerful than a firehose over its short distance. I look forward to reading your response.

 

[Treebranch]

I'm also working on the calculations at the moment... By the way, D&D is VERY abstract. Fact is, you can actually figure out the size of the opening by the speed of the water. However, given the varying modes of operation, the opening would change size in a "real world" environment I do believe. Since I'm fairly sure that "geyser" is the only relevant one, I'll get you some numbers.

However, the best thing to do, realizing that D&D is a very abstract world, is simply to take my numbers and round off to 5' or whatever is necessary. I'll post 'em when I have 'em.

 

[strongbow]

So to make calculations easier, let's assume the decantur is 1.5 inches wide, which makes the firehose comparison more valid.

Edit: I keep wanting to put firehouse instead of firehose. Maybe I have been getting too hungry for some Firehouse subs...

 

[strongbow]

I don't have AIM, sorry.

 

[Treebranch]

Don't worry about the size of the opening - it's not relevant. It's purely a matter of force. Gimme a few

 

[strongbow]

Exxxxxcellent Smithers! My reign of doom will be complete, as soon as I blot out the sun. After reading through the underwater combat section, a creature at least half-way in the water that is attacked from land gets a +8 cover bonus to AC. For a "difficult" encounter, how about a hydra attacking a boat? Sorry, the body has total cover, but at least the heads don't get the cover bonus to AC...

 

[Coredump]

Okay, here is my math on the topic. It has been quite awhile since I have fooled with Fluid Dynamics, so hopefully someone can check/verify/debunk my assertions.

First:

Geyser” produces a 20-foot-long, 1-foot-wide stream at 30 gallons per round.

Is non-sensical. If it is really a 1 foot diameter stream 20 feet long, it would take 111 gallons to 'fill' this stream. Which is over 22 seconds. It just makes no sense whatsoever.

For purposes of calculating this, I assumed that the relevant info was the 20' long stream, and the 30 gallons/round.

The size of the opening:

First, must define what a 'stream' is, and how you measure the distance. I defined it at the distance from the decanter till the water hits the ground...assuming a normal human is holding it about chest high, and pointing it straight forward. (In math speak, the flow is parallel to the ground, at about 4' above the ground)

Due to gravity, the water will be in the air for about .5 seconds before it hits the ground.

Since it will travel 20', that means it must be traveling 40' per second (note: I have ignored the impact of air resistance on this. It should be negligible)
Now we know we have water traveling at 1200 cm/sec (40') and it must deliver 19000 cc (5 gallons) in one second.
1200*Pi*radius^2=19,000
or radius is 2.25 cm, or diameter is 4.5cm, or 1.75 inches

The above calculations I am quite confident about, the next step, however, I may have missed an assumption...

Pressure of the water

Okay, the velocity of 1200 cm/s is the same as if the stream was caused by squirting out the side of a large bucket that was 733 cm tall. (There is a long calculation for this, but the number is the same as just figuring out freefall speed from the same height)

Now,
Pressure = g*rho*z or force-gravity*density*height
Pressure = 9.81 * 1 * 7.33m = 72 Pa (pascals)

So the velocity would require a pressure of 72 pascals

Force of Water
Pressure = Force/Area or 72=F/.006 (A=Pi*r^2 but for this r=.045m)
Thus, F=.45N

Assuming 500 lb canoe/riders/gear and *NO* friction from the water.... that would mean an acceleration of .002m/s. Or that after one straight minute, about 5 inches a second.
With friction, I would estimate need *at least* 10 of these to make any difference at all on a canoe.

Picture yourself on a raft, and take a pee, see if it speeds anything along...

.

 

[Piratecat]

Coredump, that's just cool. Nice analysis.