The RW Physics of the Decantur of Endless Water

[Treebranch]

Alright, I read what Coredump has to say, and even though I didn't check his math, it seems right given what his setup is.

The idea of the water stopping at 20 feet is just stupid...it would continue on if there were significant force. Given HOW Core set his situation up...he's dead on, the decanter has no place in the real world.

Now here's what I did! This is probably more of what you wanted. I made a few assumptions that help boost the potency of this item, though in retrospect, they make the strength check a joke...a human could NOT hold this thing and use it as a weapon.

ASSUMPTIONS:

The stream is 1' wide...yeah whatever, that's not gonna happen, they put that there for the game and I'm going to choose to ignore it. However, notice it doesn't say diameter, it could be a thin blast, like a setting on some garden hoses. But it doesn't matter, I'm not going to use this information.

The stream ends at 20 feet. er...what? Okay, so here's how I handled this. I just said that's if it's shooting straight UP. Again, D&D abstracts are BAD, so this is the best way I could think of to say "okay, here's WHY they say 20 feet." Keep in mind that if you point it any direction other than up, the decanter (with my math) will go a hell of a lot further.

So, here's the math:

To get the initial velocity, we use a simple equation:

20 = 0 + v(6) - .5*32.2*6^2 (which is position = vt and .5at^2, a = gravity in this case)

So the beauty is...the water is leaving the opening at 100 feet per second! (This is rounded, but not much.)

So now all we have to do is figure out the weight of 1 second of water. 30 gallons in 6 seconds, so 5 gallons in 1 second. Each gallon is 8.345404 pounds, so...

5*8.345404 = 41.72702 pounds

Okay, here's where an assumption has to be made again, and basically has to exist because it's a *magic* item. We have to assume that the water is initially at rest. So then it goes from 0 to 100 feet per second in 1 second. So its acceleration (the moment it leaves the decanter) is 100 ft/s^2.

Now, F = ma.

41.72702 * 100 = 4172 foot-pounds of force. Now let me put this in terms you can more easily relate to.

Don't know how much that is? Using conservation of momentum, where m1v1 = m2v2, we can see that something weighing 5 times as much as the 5 gallons of water would travel 1/5th as fast, or 20 feet per second.

So, 5 times the 5 gallons of water is 208.6 pounds. So basically, picture a 208.6 pound person running at you at 20 feet per second. Hrm, how fast is this?

20 / 1.4667 = 13.6 mph.

So basically a 210 pound person is running at you fast enough so that they could do a 4 1/2 minute mile. So tell me, are you able to stop them?

I'll work on the boat calculations in a bit - I'm just putting this up so you can see how poorly D&D models real life.

 

[JiffyPopTart]

Keep it capped up....

30 gallons per round=300 gallons per minute=18,000 gallons per hour=432,000 gallons per day.

Assuming any evil super-villian worth his salt could amass a collection of 100 Decanters of Endless Water to flood the world, he could produce 432 million gallons of water each day.

Average daily consumption of water in New York City: 1,240 million gallons. The poor super-villian is rationing each citizen to 1/3 his/her normal water use.

Volume of the 5 Great Lakes in North America: 6,000,000,000,000,000 gallons.
Time taken to refill the Great Lakes with his Super Water Deathtrap: 38,000 years.

Maybe he needs to look into opening some Gates into the Elemental Plane of Water?

DS

 

[babomb]

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...

I don't think that's right. .45 N seems WAY too small to be moving 5 gallons of water per second at a rate of about 40'/s. It's several orders of magnitude smaller than my answer, anyway.

I set mine up the same way, where it hits the ground at the 20 foot mark, with the decanter held 1 m off the ground, and got a similar horizontal velocity (approx. 13.5 m/s).

The width of the nozzle doesn't matter because we know the flow rate.

Now, since water's flowing out at a rate of 18.927 L/s (5 gal/s) with a velocity 13.494 m/s, that gives a change in momentum (impulse) delta-p of 255.4 kg-m/s in one second. (Edit: Just for clarity, the density of water is 1 kg/L, so the mass of the water that flows from the decanter in 1 s is 18.927 kg. delta-p= m*v, so the impulse is 255.4 kg-m/s. In retrospect, I could have simply multiplyed 255.4 kg/s by 13.494 m/s and skipped the following step, but the two cases are functionally identical.)

F = delta-p/delta-t (Newton's second law)

F = (255.4 kg-m/s)/1 s = 255.4 N

for an acceleration of 0.2 m/s^2 on a 1200 kg boat.

So how fast does the boat go at terminal velocity? I don't know. I'll have to calculate the drag later.

 

[Treebranch]

Hey guys - try it my way and see what ya get. Reason that the "geyser" is 20' long when shooting straight up. Because it's certainly usable straight up, maybe to attack aerial things. This will save you the headache of figuring out "well, it takes x seconds to fall y feet, where y is the height the decanter is most likely held at," so it must have this velocity.

The item says 20 feet...it'll always be 20, whether a giant holds it or a human does. Easiest way to compensate for this is just to calculate it as going straight up.

 

[Coredump]

Please take this in the context of 'peer review'. I am only commenting to see if I missed something, or to possibly help in case you missed something.

The idea of the water stopping at 20 feet is just stupid...it would continue on if there were significant force. Given HOW Core set his situation up...he's dead on, the decanter has no place in the real world.

Well, I according to my assumptions, it will work just fine in the real world. (assuming you are okay with magic decanters in the first place.) It is a decanter with a 1.75" opening, and thus will shoot water that will land 20' away, and deliver 30 gallons per round. The math works (except for the 1' wide part) Now, I am not as sure about the DC12 str check or the 1d4 damage... but those are results, the decanter is okay as is.

The stream ends at 20 feet. er...what? Okay, so here's how I handled this. I just said that's if it's shooting straight UP. Again, D&D abstracts are BAD, so this is the best way I could think of to say "okay, here's WHY they say 20 feet." Keep in mind that if you point it any direction other than up, the decanter (with my math) will go a hell of a lot further.

Actually, defining the stream as straight up, while in many ways makes more sense, will lead to a *less* powerful stream, and it will not make it 20 when aimed parallel to the ground. (assuming it isn't very high up)

To get the initial velocity, we use a simple equation:

20 = 0 + v(6) - .5*32.2*6^2 (which is position = vt and .5at^2, a = gravity in this case)

So the beauty is...the water is leaving the opening at 100 feet per second! (This is rounded, but not much.)

This was confusing. While I could tell there was an error, it took awhile to figure out what it was.
First, lets look at this generally. You are saying that water is travelling 100 feet per second, yet it will take 6 seconds to reach 20 feet.
The mistake you made was assuming the time was a given to reach the end of the stream. Only the distance was known. So first you need to use X and A to find T, then you can find V.

Secondly was why the numbers you used seemed to work. What happened is that water goes up, and then it comes down. Assuming it has a velocity of 100'/s, it gets to 20' in about .21 seconds, keeps travelling up until it gets to a maximum of 155',then starts falling back to earth. After about 6 seconds total, it again passes by the 20' mark, and at 6.2 seconds, hits the earth again.
Your calculations were dealing with the 'return' trip.

To reach a maximum height of 20', requires only an initial velocity of 18'/s

Okay, here's where an assumption has to be made again, and basically has to exist because it's a *magic* item. We have to assume that the water is initially at rest. So then it goes from 0 to 100 feet per second in 1 second. So its acceleration (the moment it leaves the decanter) is 100 ft/s^2.

Well, it would have to have an *initial* velocity of 100 (or 18) fps. So the acceleration would be almost instantaneous. But it would not be a constant acceleration, since it doesn't keep accelerating.
Using 100'/s^2 would mean it would never stop.... since it is accelerating much faster than gravity is slowing it down.
OTOH, 100'/s^2 is too slow, since it would take a full second to get to top speed.

Now, F = ma.

41.72702 * 100 = 4172 foot-pounds of force. Now let me put this in terms you can more easily relate to.

Except it can't be accelerating at 100'/s^2 as a constant... To work at all, there is an 'instantaneous' acceleration, but no accel after that.

Don't know how much that is? Using conservation of momentum, where m1v1 = m2v2, we can see that something weighing 5 times as much as the 5 gallons of water would travel 1/5th as fast, or 20 feet per second.

A stream of water is not the same as a solid object. And taking 'one second' worth of water is an arbitrary amount. Why not 6 seconds, or .1 seconds?

edit: formatting

.

 

[Coredump]

I don't think that's right. .45 N seems WAY too small to be moving 5 gallons of water per second at a rate of about 40'/s. It's several orders of magnitude smaller than my answer, anyway.

It seemed small to me too. I will go over the numbers again later to see if I missed something.

The width of the nozzle doesn't matter because we know the flow rate.

Now, since water's flowing out at a rate of 18.927 L/s (5 gal/s) with a velocity 13.494 m/s, that gives a change in momentum (impulse) delta-p of 255.4 kg/s.

I like it. And it seems much more elegant than my solution. Not sure if it being a fluid changes this at all. But I will look at it more later.

F = delta-p/delta-t (Newton's second law)

F = (255.4 kg/s)/1 s = 255.4 N

Yet realize. This means that the 100kg person holding this is accelerating at 2.55 m/s^2. That is 0-60mph in about 10 seconds.

More later... this is fun... in a super geeky kinda way.

 

[Treebranch]

Lol, have to admit that you're right on that one, Coredump. I dunno what I was thinking.

But when you say a stream of water isn't the same as a solid object...ehh, it has a center of mass, and conservation of energy works on it too. BTW, my comment stating it has no place in the real world meant with the 1' stream...it doesn't work

And yeah...afternoon of drinking combined with not having physics in well...years...yeah. Dunno why I didn't just *think* about that initial equation, hehe.

I'll look into it, but really, you should use it shooting straight up as your base case, otherwise you're skewing it. (since the 20' is omnidirectional) And yeah, after giving it about 3 seconds of sober thought, you're right, it won't go 20' if aimed horizontally. But still, thanks for the wake-up, hehe.

Quick aside...with an initial velocity of 18 ft/sec...it goes up 5 feet. Maybe you used 9.8 for gravity rather than 32.2?

However I should mention that there's nothing wrong with my acceleration assumption, and notice that I did state "as it leaves the decanter." I did not say this acceleration was constant. Think of it as a potato gun. (Then think of that gun shooting LOTS of potatos like a machine gun...then you see why it's fair to treat a "stream" as a solid in regards to momentum, etc. Because really, on a very tiny level, water is solid.

And "1 second" of water is hardly arbitrary - it puts everything on a "1 second" scale. Don't say you can't do that either - you obviously DO know what you're talking about, so you know that if you shot a 5 gallon "chunk" of water every second for 6 seconds, you'd shoot just as much water and deliver the same amount of force as a constant stream - think about rocket fuel calculations. You figure out how much momentum is leaving the rocket, and then how fast the rocket (ie it's momentum) has to go to compensate. Force is just the change in momentum - the same thing applies here. So to figure out the force at a given second, you need the mass at a given second, which is 5 gallons of water.

 

[Henry]

Someone commented on an evil villain flooding the world with this item - I'd be more worried about what other effects it could have on the weather patterns, on the electrolyte balance and concentrations of elements in the Sea water, and what effect these would have long-term on wildlife and the environment. If some enterprising magic-user opened a business of these things to eliminate world clean drinking water woes, it could have the same effect as using thousands of captured and melted comets for drinking water.

 

[Treebranch]

Okay, I'll admit that my first analysis had a rather fundamental...and rather stupid flaw in it. Here's a corrected version.

I'll still use straight up as the base case calculation, since that removes guesswork of "oh, a person holds it this high...blah blah".

And btw, conservation of energy is a universal law...it applies to fluids equally, hehe.

1/2 mv^2 = mgh.

So you get v = sqrt(2gh), where h is 20 feet, and g is 32.2 ft/s^2

Or...v = 35.888 ft/s initial velocity.

I still say that m1v1 = m2v2 works. And it's a fact, conservation of momentum holds for all things. Now allow me to clarify why acceleration = velocity in this case.

In one second, 5 gallons of water changes velocity from 0 ft/s to 35.888 ft/s. Therefore, the acceleration (while leaving the decanter) is 35.888 ft/s.

So, consider the decanter as exerting 35.888 * 41.727 foot pounds of force on the water. The water, by action/reaction must likewise exert this force on the decanter, and the decanter again my action/reaction must exert this force on the holder.

So the actual value is (ROUNDED) 1500 foot-pounds of force.

My initial calculations were VERY flawed, though my method is correct. The only question to really answer is the true velocity at the opening, as 20' upwards will indeed not be 20' horizontally unless held higher than the average adventurer would hold it. Again, a problem created by the abstraction of D&D.

Again, for those who don't believe me, I take this quote from a physics website:

"A rocket has an exhaust velocity of 3000 m/s and burns fuel at 100 kg/s. How much thrust will it generate?

The force which the rocket exerts on the exhaust is the rate of change of the exhaust's momentum.
In one second, 100 kg of fuel changes its velocity by 3000 m/s as it blows out the back of the rocket. Its momentum changes by 100 kg x 3000 m/s or 300,000 kg m/s. Its rate of change is then 300,000 kg m/s and the force on the exhaust is 300,000 N. By the law of action and reaction, the force on the rocket is also 300,000 N.
"

If the decanter threw out enough water, it'd work just like a rocket. Ever use one of those water rockets as a kid?

 

[tomBitonti]

Here are my rough calculations:

The physics of an jug of endless water;

The following works out the momentum transfer
from a jug of endless water assuming an orifice
1 1/2 inches in diameter and flow rates of one,
five, or thirty gallons per round.

The following use pounds as mass.

I use momentum transfer, assuming that the momentum of the
water expelled by the jug must be balanced by a gain of
momentum in the jug (or the holder of the jug). I also
assume that the water expelled from the jug need not pay
for the effective momentum gain because it is expelled in
the jug's frame of reference.

One pint weigh 1 pound (Remember, "A pint a pound the world around.")
One gallon has 8 pints, weighing 8 pounds.

One round has 6 seconds.

So:
One gallon / round gives
8 pounds / 6 seconds == 1 1/3 pounds / second
Five gallons / round gives
40 pounds / 6 seconds == 6 2/3 pounds / second
Thirty gallons / round gives
240 pounds / 6 seconds == 40 pounds / second

Each pound of water carries momentum according to its velocity out of
the jug.

Working from:

1 lb == 454 gm
1 cu in == 2.54 ^ 3 cc

So, with 16.39 cc / cu in

Water Mass Water Volume
1 lb 27.7 cu in
1.33 lbs 36.8 cu in
6.67 lbs 185 cu in
40.0 lbs 1110 cu in

(Alternatively, water weighs about 62 pounds / cu ft,
which gives 27.9 cu in for one pound.)

A circle 1.5 inches across has 1.77 sq in.

For 27.7 cu in of water to flow through 1.77 sq in in one second
the water must flow at 15.6 in / sec, or 0.89 mi / hour.

For 36.8 cu in of water to flow through 1.77 sq in in one second
the water must flow at 20.8 in / sec, or 2.4 mi / hour.

For 185 cu in of water to flow through 1.77 sq in in one second
the water must flow at 105 in / sec, or 6.0 mi / hour.

For 1110 cu in of water to flow through 1.77 sq in in one second
the water must flow at 627 in / sec, or 36 mi / hour.

Momentum transfer is at the rates:

One gallon / round:
1 1/3 pounds / second * 0.89 mi / hour ==
1.2 pound miles / hour

200 lbs @ 0.000275 g
100 lbs @ 0.00055 g
50 lbs @ 0.0011 g
5 lbs @ 0.011 g

Five gallons / round:
6 2/3 pounds / second * 2.4 mi / hour
16 pound miles / hour

200 lbs @ 0.0037 g
100 lbs @ 0.0073 g
50 lbs @ 0.015 g
5 lbs @ 0.15 g

Thirty gallons / round:
40 pounds / second * 36 mi / hour
1440 pound miles / hour

200 lbs @ 0.33 g
100 lbs @ 0.66 g
50 lbs @ 1.32 g
5 lbs @ 13.2 g

These values vary inversely to the square of the radius, that is,
doubling the radius divides the values by four, while halving the
radius multiplies the values by four.

For example, a 5 pound jug with a 1/2 inch orifice jetting 30 pounds
of water per round would be subject to 13.2 * 9 or about 120 g's of
acceleration.

Thx,

Tom Bitonti