The RW Physics of the Decantur of Endless Water

[Coredump]

Bingo

Oh those cursed decimal places....

I thought .45N was way too low also. And now I know why...

When I determined pressure, I used the density for water as 1.0
except I was using N and meters as units. Water is 1,000 kg per m^3. Thus my numbers were off by a factor of 1000. Thus it would be 72000 Pa, and 450N of force.

Except I also made a typo with the area on the next section. I used .045; but that is the diameter. The radius would have been .0225

Thus P=F/A is 72000=F/.0015 or F = 115N

 

[Coredump]

And btw, conservation of energy is a universal law...it applies to fluids equally, hehe.
And it's a fact, conservation of momentum holds for all things.

For your uses, this is pretty much true. But both of these statements have qualifiers. They are not always true for all things.

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.

Yeah, I figured out the mistake I made... when I re-did it, I got about the same as you did.
But.... note for later. This calculation assumes 35.888 is the *initial* velocity. Not the velocity after a full second of acceleration.

I still say that m1v1 = m2v2 works.

This is, in essence, the exact same method that babomb used. I can't say that it doesn't work, I just have conflicting instincts. Sometimes it seems better than my method, and sometimes worse.

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.

Sorry, but the velocity is from the beginning. Thus acceleration is, in effect, instantaneous. But for the method you are applying (conservation of momentum, ala the rocket example) the acceleration isn't an issue anyway.

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.

Using english units isn't quite as elegant as metric. And your method here is not the same as the example below. But using the example below, you get 207N as an answer. (Part of the problem is using lb as mass and as force. Plus you are trying to combine acceleration with conservation of momentum.)

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

Cool. What website was it. I am trying to brush up on this stuff some more... and I have exhausted my old textbooks.


You and Babomb used (essentially) the same logic, and came to the same conclusion. The difference in final force stems from how you measured the initial velocity. (horizontal or vertical)

Not sure how to rectify the differences between what you two got and what I got.... Have to look into this some more.

edit: formatting

 

[BSF]

*laugh*
This is a fun converstaion. Alas, I understand little regarding the physics end of it. Still, the message I think I am reading is that the decanter of endless water is terribly ineffective if you try to model it off real-world phsyics. Is that correct?

 

[Treebranch]

I'm not combining acceleration with conservation of momentum at all really. I'm using them together, but I'm not inventing a new law or anything.

Here's a better way to explain it I think.

We agree that in 1 second, 5 gallons is expelled from the jug, correct? Okay, good. And we agree that m*v is equal to momentum. Now, realize during 1 second, the momentum changes from 0 to 1500. Force is the change in momentum (dp/dt I do believe). dt is established - 1 second. dp must then be 1500, which must equal m*a.

There ya go, a better explanation without using dv/dt. All you have to do is note the change in momentum (0 to 1500 in one second - which must equal force).

No foul play there. And as I mentioned, the only thing that really is arguable is initial velocity.

EDIT: And the website for that is http://www.people.vcu.edu/~rgowdy/mod/018/xmp4.htm#4. I did a google search to make sure I wasn't crazy.

EDIT Again: Changed explanation from dv/dt to dp/dt. Much easier to see that force is simply the change in momentum.

 

[Majere]

• “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.

1) Assume water begins Stationary before "entering" the "pump"

Denisty of water (lbs/gallon) = p
Mass of water pushed back each round : 1p,5p,30p

Velocity of water, (Assume a 3"x3" opening):
Volume of 1 gallon of water in cu. feet = V
Therefore Velocity, 16v,80v,480v

Momentum of water:
16pV, 400pV, 14400pV
This is equal to the momentum given to the raft.
The raft has a mass (in lbs)= M

Therefore the change in velocity of the raft
16pV/M
400pv/M
14400pV/M

The raft would feel a deceleating for to to viscous drag, and would accelerate until a "terminal velocity".
To work this out would be lending to some wild approximations on the drag of the raft, the time over which the water is accelerate and such.

And I dont have time to find p or V

Majere

 

[D+1]

I think you all might be going about this incorrectly. The solution is not in physics calculation (especially since you seem to agree that the physics of the game stats is bogus anyway) but in a bit of real-world experimentation. The backpressure is not defined by velocity, size etc. of the water stream - it's defined as a DC12 str check to avoid being knocked down. I think that is waht you need to solve for - how much real-world physical pressure is represented by a DC12 str check to avoid falling down.

DC12 is fairly easy - you and I, without noteworthy strength or skill will make it about 50% of the time when not actually BRACED for it. Get a friend and a nice, soft mat to fall onto. Then you need something to measure the force you're going to push him with. Ideally, you'd have a machine to simply set the force to and push you with that force. When you get to a point where it knocks you on your butt half the time - there's your answer.

Or am I actually missing something? I'd like to have something like hard numbers for this too because I've long had this wierd notion to have an "air raft" of sorts that is lifted and propelled by a series of decanters. I have wondered just how practical it would be in a D&D world based on the "known" quantity of the gyser backpressure.

In essence my design would be a disc (size indeterminate) with decanters on the underside and controls that adjust the angle of the decanters much as the cyclic controls of a helicopter control the pitch of its blades. Straight down provides simple lift, and slight angles to any given direction translates some of the lift into forward momentum into the opposite direction. I've just always thought it would be cool to see such an apparatus sloshing across a desert in a D&D campaign but haven't ever gone so far as to calculate specifically how many decanters would be needed for how much lift:mass ratio and without simply inventing new magic items. That's too easy. The operation of it should follow normal D&D rules for activation of magic items. The trick is simply figuring the real-world amount of force being generated that can then be more practically applied to other ideas.

 

[Scion]

how about this, it means almost nothing for the real world, but might work for d&d

str 12 human, light load 43lbs. We'll just round up to 50 to make it easy.

We will say that this sort of craft has an acceleration of 10ft/round. For every 50lbs beyond the first you reduce this acceleration. Max speed is whatever you want to set it to (lets say 50ft/round, poor manuverability).

So, at 50lbs worth of creatures/gear (discounting the size of the boat, this is d&d, it has no bearing ) it accelerates at 10ft/round (reaching max speed in 5 rounds)

100lbs = 5ft/round (max speed in 10 rounds)
150lbs = 3.3ft/round (max speed in 15 rounds)
200lbs = 2.5ft/round (max speed in 20 rounds)
etc.

While not terribly physics-like it will work for d&d standards. You could even say that a poorly built craft counts as 50lbs of gear, normal craft 0lbs, and a masterwork craft counts as -50lbs (or cuts the total weight in half). Each extra decanter will double the acceleration rate, but probably not change the max speed.

Easy

 

[apsuman]

I can not believe that I am actually posting here but...

since it is MAGIC would it be okay to say that the decantur opens a hole to the elemental plane of water and water comes out with a force for those in front of the geyser but produces no foce for those behind (or holding) the decantur? In a sence the water pushing out of the hole "pushes back" (that whole equal an opposite force thing) into the plane of water, bypassing the person holding the decantur.

 

[Coredump]

Apsuman, of course it being magic changes everything... I got involved because it gave me an excuse to get my physics books out of the basement and do some problem solving.

But even from the beginning I felt that the approach D+1 is using is the best. There is 'some' backpressure, since you must roll a DC12 str check to hold onto it. But since there is no check afterwards, it could just be an initial check, and no force after that.

.;

 

[reanjr]

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,

You need to know more than the force of the decanter. This problem is done using somewhat complicated calculus. One must first know the surface dynamics of the watercraft in the water, as the faster the craft goes the more force is applied by the water. It's a terminal velocity problem, basically. I'd suggest using the Str 14 idea.