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The RW Physics of the Decantur of Endless Water
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<blockquote data-quote="Treebranch" data-source="post: 1639087" data-attributes="member: 21049"><p>Okay, I'll admit that my first analysis had a rather fundamental...and rather stupid flaw in it. Here's a corrected version.</p><p></p><p>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".</p><p></p><p>And btw, conservation of energy is a universal law...it applies to fluids equally, hehe.</p><p></p><p>1/2 mv^2 = mgh.</p><p></p><p>So you get v = sqrt(2gh), where h is 20 feet, and g is 32.2 ft/s^2</p><p></p><p>Or...v = 35.888 ft/s initial velocity.</p><p></p><p>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.</p><p></p><p>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.</p><p></p><p>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.</p><p></p><p><strong>So the actual value is (ROUNDED) 1500 foot-pounds of force.</strong></p><p></p><p>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.</p><p></p><p>Again, for those who don't believe me, I take this quote from a physics website:</p><p></p><p><em>"A rocket has an exhaust velocity of 3000 m/s and burns fuel at 100 kg/s. How much thrust will it generate?</em></p><p><em></em></p><p><em>The force which the rocket exerts on the exhaust is the rate of change of the exhaust's momentum.</em></p><p><em>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.</em></p><p><em>"</em></p><p></p><p>If the decanter threw out enough water, it'd work just like a rocket. Ever use one of those water rockets as a kid? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=";)" title="Wink ;)" data-smilie="2"data-shortname=";)" /></p></blockquote><p></p>
[QUOTE="Treebranch, post: 1639087, member: 21049"] 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. [B]So the actual value is (ROUNDED) 1500 foot-pounds of force.[/B] 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: [I]"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. "[/I] If the decanter threw out enough water, it'd work just like a rocket. Ever use one of those water rockets as a kid? ;) [/QUOTE]
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