# Decanter of Endless water question

### Rules Questions

If you attached a DoEW to each side of a rowboat and turned it on geyser, how fast would it push the boat? :)

All decanters will produce equal force, thus canceling each other out.

If you turn a rowboat on a geyser, the speed at which the geyser will propel the rowboat upwards depends entirely on the size of the rowboat and the strength of the geyser.

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Both of the above suggestions were made by the same person.

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I assume the intended question is, if you attach a couple of decanters to a rowboat and have them both facing backwards, how fast would it push the boat forwards?

A rowboat might weigh 27 kg. If there's a couple of adventurers + equipment in the rowboat, that might bring up the total weight to a couple of hundred kg.

Each decanter produces 30 gallons per round. Together, that's 60 gallons per round or 10 gallons (45kg, or 45,000 cubic centimetres) per second. (A fire hose is about 2 gallons per second.)

The water that comes out of a decanter is one foot wide. Assuming that's a diameter, the cross-sectional area of water from both combined is about 1,400cm2. The total volume coming out per second is 45,000cm3. That gives a speed of water from the geyser of .32m/s. Weirdly, this is less than a mile per hour, more of an ooze than a geyser.

...OK, this doesn't make much sense. Fire-hose nozzles are normally about an inch wide, not a foot wide. Perhaps the water from the decanter is one foot wide but very thin?

I can't answer this question, but my instinct is this should propel the boat forwards about the same speed as rowing it would.

I think the 1ft wide means it is 1ft wide at the end of the 20ft long stream, not coming out of the decanter.

Decanter of Endless Water wrote:

If the stopper is removed from this ordinary-looking flask and a command word spoken, an amount of fresh or salt water pours out. Separate command words determine the type of water as well as the volume and velocity.
"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.
The geyser effect exerts considerable pressure, requiring the holder to make a DC 12 Strength check to avoid being knocked down each round the effect is maintained. In addition, the powerful force of the geyser deals 1d4 points of damage per round to a creature that is subjected to it. The geyser can only affect one target per round, but the user can direct the beam of water without needing to make an attack role to strike the target since the geyser's constant flow allows for ample opportunity to aim. Creatures with the fire subtype take 2d4 points of damage per round from the geyser rather than 1d4. The command word must be spoken to stop it.

Agreed. A firehose's stream definitively does not stay one inch wide for very long.

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I assume the geyser is going straight up. Then the power exerted by one decanter is
power = mass of water * acceleration * height reached / time
= (30 gallons of water / round) * g * 20'
= (113.55 kg / 10 sec) * 9.8 m/s^2 * 6.1 m
= 1130.5 watts

You have two decanters, so they're exerting power of 2261 watts.

This is equal to the force on the rowboat times its velocity. The force is countered by the force of friction,
F_friction = (coefficient of friction mu) * F_normal
= mu * (mass of boat & its load) * g = mu * M * g

Therefore the velocity of the rowboat is
v = (power of decanters) / (mu * M * g)

I will use MD's estimate of ~200 kg for M. That leaves us with
v = 1.15/mu in meters/second
= 2.6/mu miles per hour

I have no real idea what mu should be, but probably not very large. I'll call it .1; then the boat goes at 26 mph. Which is respectable for a bottle-driven vehicle.

OK - since the math peeps are here -- I have often pondered if a DoEW was left overturned and running as a fountain in a desert (a Scruffy desert like Arizona, not the Sahara), how long would it take to actually make a lake? Some place with a cycle of temperatures from winter night lows of 35F, days of 55F and high summer nights of 85F and days of 101F. Is there a point where evaporation ends a constant water supply from pooling?

OH - and assume its at the low point of a great valley, so the water would not have an egress until the lake was at least a hundred feet deep.

Almost based an adventure on this, but never completed it because the very premise may be ridiculous.

You could always just craft flexible nozzles that fit on the end of the decanters to increase the water pressure. The engineering involved is trivial.

A big lake like Lake Superior is 12,100 cubic kilometres of water, but let's say we're going for a smaller one.

A single cubic kilometer of water (220,000,000,000 gallons) would take 44,000,000,000 seconds to produce with 5 gallons per second, about 1,500 years. This requires none to be lost through evaporation or drainage.

Clearly, at some point the loss of water will exceed the water produced.

Basically this is the same as "how big a lake would you get if you had a body of water being fed by a small stream but no rain"? The answer is: a pretty small one.

Exactly how big is more a question of geography than numbers.

Or even more evil... FLOOD THE WORLD!

2bz2p wrote:

OK - since the math peeps are here -- I have often pondered if a DoEW was left overturned and running as a fountain in a desert (a Scruffy desert like Arizona, not the Sahara), how long would it take to actually make a lake? Some place with a cycle of temperatures from winter night lows of 35F, days of 55F and high summer nights of 85F and days of 101F. Is there a point where evaporation ends a constant water supply from pooling?

OH - and assume its at the low point of a great valley, so the water would not have an egress until the lake was at least a hundred feet deep.

Almost based an adventure on this, but never completed it because the very premise may be ridiculous.

Depends on the humidity of the air above the lake and wind speeds, not to mention how many "winter" days vs "summer days"

https://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html

Note the calculator at the bottom.

The Decanter is producing approximately 11.3 kg/s of water. Assume the soil below is saturated and will not absorb further water.
assume dry air above x = 0
Assume Winter nights, winter days, summer nights and summer days are all in equal proportion.
Xs (35F) = .005
Xs (55F) = .009
Xs (85F) = .027
Xs (101F) = ?.04? (very approximately)

Xs(average) is assumed to be 0.02

Assume the average wind speed is 5 m/s (12 mph)

Note that the accumulated assumptions make this calculation very iffy.

this lake will average around 18000 square meters in surface area - it will vary a lot between summer and winter ( Xs(summer) ~ .033 , Xs(Winter) ~ .007, a factor of 4...)

The height of the geyser is 20ft. The boat goes 20ft. 40ft seems reasonable for a second.

Not as fast as a jetski in all likelihood, however by the look of it, once you get good, by shutting off one of the decanters momentarily you can get the rowboat to do a barrel roll...

...this lake will average around 18000 square meters in surface area - it will vary a lot between summer and winter ( Xs(summer) ~ .033 , Xs(Winter) ~ .007, a factor of 4...)

So the premise of a sustained small - or even fairly large - lake in the middle of a desert existing because of an open decanter is not too far fetched. THANKS!

Fuzzy-Wuzzy wrote:

I assume the geyser is going straight up. Then the power exerted by one decanter is

power = mass of water * acceleration * height reached / time
= (30 gallons of water / round) * g * 20'
= (113.55 kg / 10 sec) * 9.8 m/s^2 * 6.1 m
= 1130.5 watts

You have two decanters, so they're exerting power of 2261 watts.

This is equal to the force on the rowboat times its velocity. The force is countered by the force of friction,
F_friction = (coefficient of friction mu) * F_normal
= mu * (mass of boat & its load) * g = mu * M * g

Therefore the velocity of the rowboat is
v = (power of decanters) / (mu * M * g)

I will use MD's estimate of ~200 kg for M. That leaves us with
v = 1.15/mu in meters/second
= 2.6/mu miles per hour

I have no real idea what mu should be, but probably not very large. I'll call it .1; then the boat goes at 26 mph. Which is respectable for a bottle-driven vehicle.

If you calculate force, you are generating an acceleration, not a velocity. F = ma.

The boat's final speed will occur when the net forces are zero. Thus, you also would need the equation for drag, and the equation for drag, not friction, since the boat is moving through water and air, not sliding along the ground. What will happen is the boat will accelerate as long as the force from the decanters exceeds the drag force. Fortunately, the drag force increases in direct proportion to flow velocity squared, meaning as the boat speeds up, the drag force rises exponentially.

Anyway, the drag force is equal to 1/2 density * velocity^2 * reference area * drag coefficient.

I don't care enough to solve the equations. :) The boat moves as fast as the GM says it does and I'd reward the player for their crazy little idea.

the geyser has a DC 12 strength check recoil, so I would say it produces thrust similar to the effects of a 12 strength check.

The Doc CC wrote:
Fuzzy-Wuzzy wrote:

I assume the geyser is going straight up. Then the power exerted by one decanter is

power = mass of water * acceleration * height reached / time
= (30 gallons of water / round) * g * 20'
= (113.55 kg / 10 sec) * 9.8 m/s^2 * 6.1 m
= 1130.5 watts

You have two decanters, so they're exerting power of 2261 watts.

This is equal to the force on the rowboat times its velocity. The force is countered by the force of friction,
F_friction = (coefficient of friction mu) * F_normal
= mu * (mass of boat & its load) * g = mu * M * g

Therefore the velocity of the rowboat is
v = (power of decanters) / (mu * M * g)

I will use MD's estimate of ~200 kg for M. That leaves us with
v = 1.15/mu in meters/second
= 2.6/mu miles per hour

I have no real idea what mu should be, but probably not very large. I'll call it .1; then the boat goes at 26 mph. Which is respectable for a bottle-driven vehicle.

If you calculate force, you are generating an acceleration, not a velocity. F = ma.

The boat's final speed will occur when the net forces are zero.

I am well aware of both these facts, as well as power = Fv, which is what I was using with the latter by setting F = P/v equal to F_f.

The Doc CC wrote:

Thus, you also would need the equation for drag, and the equation for drag, not friction, since the boat is moving through water and air, not sliding along the ground. What will happen is the boat will accelerate as long as the force from the decanters exceeds the drag force. Fortunately, the drag force increases in direct proportion to flow velocity squared, meaning as the boat speeds up, the drag force rises exponentially.

Anyway, the drag force is equal to 1/2 density * velocity^2 * reference area * drag coefficient.

Ah! I never took a course that covered hydrodynamics, so of this I was unaware. Thank you! However, I stand by my equations should the player try sending the boat down a road this way. ;-)

The Doc CC wrote:
I don't care enough to solve the equations. :) The boat moves as fast as the GM says it does and I'd reward the player for their crazy little idea.

I certainly wouldn't pause the game to solve them! And I'd reward them, yes. But the OP seemed interested in the actual answer, so.

Considering that the decanter only has enough power to knock the wielder over, not move them, I would say that it wouldn't work to move a boat. Think about it - it can potentially knock over a Colossal creature, and a Fine sized creature that holds it won't be moved (just potentially knocked over). No creature hit by the stream is moved or even knocked down.

It is magic, with magically created force that can injure but not move. Doesn't have to make sense. If it had the power to move something, a Fine-sized creature should be able to ride one like a rocket. Yet at best it just knocks the holder off their feet.

Math nerd here. Don't bother using real-world physics. We have what we need here:

Decanter of Endless Water wrote:
The geyser effect exerts considerable pressure, requiring the holder to make a DC 12 Strength check to avoid being knocked down each round the effect is maintained.

For an average character to make the DC 12 check, they need a base +2, or 14 Strength. That is 175# max load.

Lifting and Dragging wrote:
A character can generally push or drag along the ground as much as five times his maximum load. Favorable conditions can double these numbers, and bad circumstances can reduce them by half or more.

The most force he can exert laterally is 5*175# = 875#

Any more, and he cannot overcome it.

Likewise, he can hold against that same amount, and any more beyond will push/pull him.

So the max push of a Decanter is geyser mode is 875# at 20'.

/cevah