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*Pathfinder & Starfinder
The RW Physics of the Decantur of Endless Water
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<blockquote data-quote="Coredump" data-source="post: 1638941" data-attributes="member: 6939"><p>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.</p><p></p><p></p><p>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.</p><p></p><p></p><p>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)</p><p></p><p></p><p>This was confusing. While I could tell there was an error, it took awhile to figure out what it was.</p><p>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.</p><p>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.</p><p></p><p>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.</p><p>Your calculations were dealing with the 'return' trip.</p><p></p><p>To reach a maximum height of 20', requires only an initial velocity of 18'/s</p><p></p><p></p><p>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.</p><p>Using 100'/s^2 would mean it would never stop.... since it is accelerating much faster than gravity is slowing it down.</p><p>OTOH, 100'/s^2 is too slow, since it would take a full second to get to top speed.</p><p></p><p></p><p>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.</p><p></p><p></p><p>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?</p><p></p><p>edit: formatting</p><p></p><p>.</p></blockquote><p></p>
[QUOTE="Coredump, post: 1638941, member: 6939"] 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. 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. 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) 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 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. 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. 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 . [/QUOTE]
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