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Natural Science Forum / Physics / General Physics / October 2004jumping from a bridge
when I jump off a bridge into water, how deep will I dive? > when I jump off a bridge into water, how deep will I dive? If it's quite shallow water, not very far at all. John Lowry Flight Physics > when I jump off a bridge into water, how deep will I dive? "Just make sure you do it right the first time 'Cause nothin's worse than a suicide chump" http://www.lyricsdomain.com/6/frank_zappa/suicide_chump.html Dirk Vdm > when I jump off a bridge into water, how deep will I dive? if I answer this how long till you get it? In sci.physics, Rolphe Fehlmann <fehlmann39@bluewin.ch> wrote on Sat, 16 Oct 2004 14:25:11 +0200 <41711327$1_1@news.bluewin.ch>: > when I jump off a bridge into water, how deep will I dive? An interesting question, but one needs more data. [1] Height of bridge off the water. [2] Density of the air (retrievable). [3] Density of the water (constant/retrievable). [4] General form-factor of the jumper (ball? rectangle? rag doll? torpedo tube? glider??) This is mostly for drag computations during the airborne part of the descent. One could contemplate a cannonball descent as well (ouch) versus a graceful diving-in headfirst into a sufficiently deep pool or river. [5] Weight and volume of the jumper. I can make some assumptions here but it does make a difference; women are more bouyant on average, for example, than men. If the bridge is sufficiently high one has to contend with the shock of hitting the water as well; this could break one's neck or back, effectively eliminating the "I" and having only the corpse come back up.  Signature#191, ewill3@earthlink.net It's still legal to go .sigless. > In sci.physics, Rolphe Fehlmann > <fehlmann39@bluewin.ch> [quoted text clipped - 22 lines] > one's neck or back, effectively eliminating the "I" and > having only the corpse come back up. Another variable: how penetratable is the mud on the bottom? John Lowry Flight Physics In sci.physics, John T Lowry <jlowry100@earthlink.net> wrote on Sun, 17 Oct 2004 17:18:39 GMT <PPxcd.7894$gy1.5057@newsread1.news.pas.earthlink.net>: >> In sci.physics, Rolphe Fehlmann >> <fehlmann39@bluewin.ch> [quoted text clipped - 24 lines] > > Another variable: how penetratable is the mud on the bottom? If one introduces mud, one must also introduce the depth of the waterbody. But yeah, that's a good one. :-) > John Lowry > Flight Physics Signature#191, ewill3@earthlink.net It's still legal to go .sigless. > > [quoted text clipped - 23 lines] >> but it does make a difference; women are more bouyant on average, >> for example, than men. <snip> Doug Durian has done some nice work with impact craters in granular media, for example: Low-Speed Impact Craters in Loose Granular Media <http://prola.aps.org/searchabstract/PRL/v90/i19/e194301?qid=2713afdab2d0b5cd&qse q=7&show=10> J. S. Uehara, M. A. Ambroso, R. P. Ojha, and D. J. Durian Phys. Rev. Lett. 90, 194301 (2003) I wonder how the analysis would carry over to a liquid..... SignatureAndrew Resnick, Ph.D. Department of Physiology and Biophysics CWRU School of Medicine tanspose 'op' for mail I suggest the following approach: There are the following forces that need to be conidered: DOWNWARD FORCES - Gravitation: Fg=m*g [80*9.81] UPWARD FORCES IN AIR - air resistance: Fa=cw/2*r*A*V^2 [cw value of a human being is: 0.78; [A=1256]] UPWARD FORCES IN WATER - static buoyancy: Fa=r*g*V [g=9.81; V=1 (for a first approch)] - dynamic resistance in water: Fa=cw/2*r*A*V^2 [A=1256] Now all those forces need to be converted into energy and an equation needs to be formualted. Anyone can help? Nicolas > In sci.physics, Rolphe Fehlmann > <fehlmann39@bluewin.ch> [quoted text clipped - 22 lines] > one's neck or back, effectively eliminating the "I" and > having only the corpse come back up. > I suggest the following approach: > [quoted text clipped - 42 lines] > > one's neck or back, effectively eliminating the "I" and > > having only the corpse come back up. Your assumptions are unstated. I think you assume vertical entry and vertical alignment of body. I don't, so I would get different results. To do this correctly you need to state them all. The direction of the dive may not be down. If the body is arched you may move in a circle and come up without going as deep, of course with some increase in forces. And bouancy depends, among other things, on whether you've breathed in or out. >> I suggest the following approach: >> [quoted text clipped - 54 lines] > And bouancy depends, among other things, on whether you've breathed in > or out. IMHO those are all details. Let's just simply the problem and try to solve it when the body is straight. Can anybody help on that and continue what I've started above? If you jumped from this bridge ,,the screws will suck you under the ship and the sharks behind us will eat whats left. besides its 100 feet . Unless you hit the poopdeck its 50 feet. we tip the tower down to get under golden gate. The tower is 75 feet over the bridge. Up there it looks like you could jump and hit the water ,,real hard. Deepwater 2 Global star > If you jumped from this bridge ,,the screws will suck you under the ship > and the sharks behind us will eat whats left. [quoted text clipped - 4 lines] > Up there it looks like you could jump and hit the water ,,real hard. > Deepwater 2 Global star Guys, you all talk too much. You try to be clever Mr. Feynman II, but it seems you guys can't even solve a simple problem of classical physics. > > If you jumped from this bridge ,,the screws will suck you under the ship > > and the sharks behind us will eat whats left. [quoted text clipped - 7 lines] > Guys, you all talk too much. You try to be clever Mr. Feynman II, but it > seems you guys can't even solve a simple problem of classical physics. Careful when generalizing, lest ye tread on toes. BTW, stop responding to Frazir. What do you do when you find out an energy creature feeds on energy? Thats right... There was no question to respond to. If you stand in front of a truck , how hard will it hit you is not a queston to respond to. YOU WOUNT penetrate the water at all because the bay is frozen solid ... and you dint jump ,,I tossed you off : )) That comment's a sure fire way to prevent one from pulling out the calculus... It's not simple "classical physics" at all. Without seeing your original post, the nature of the surface of the water is one of the principle confounding factors, as is the unpredictable, semi-elastic nature of the collision between the body & the water. You could do the math using a rigid body, but it'd be wrong. > > If you jumped from this bridge ,,the screws will suck you under the ship > > and the sharks behind us will eat whats left. [quoted text clipped - 7 lines] > Guys, you all talk too much. You try to be clever Mr. Feynman II, but it > seems you guys can't even solve a simple problem of classical physics. > I suggest the following approach: > [quoted text clipped - 6 lines] > - air resistance: Fa=cw/2*r*A*V^2 [cw value of a human being is: 0.78; > [A=1256]] This problem requires the solution to the differential equation dv/dt = g - uv^2, where u = cw/2*r*A/m Which then becomes dv/(g - uv^2) = dt Let a = (g/u)^(1/2) Integrating we get: (1/2a)(ln|v+a| - ln|v - a| = t + C Simplifying we get: (v + a)/(v - a) = Ce^(2at) Or: v = -a(1 + Ce^(2at))/(1 - Ce^(2at) If at t = 0, v = 0, then C = -1 Simplifying, we get v = -a( 1 - e^(2at) )/(1 + e^(2at) Unfortunately, (assuming my math is right, which I don't by any means guarentee) this result is only useful if we know how long it took him to hit the water. If you want to find his position as a function of velocity, the differential equation is much more complicated. > UPWARD FORCES IN WATER > - static buoyancy: Fa=r*g*V [g=9.81; V=1 (for a first approch)] > - dynamic resistance in water: Fa=cw/2*r*A*V^2 [A=1256] Solving this problem will require the complicated solution mentioned above. Actually, this one is a bit more complicated yet, since it has an added displacement term. I don't have the time today to work through that solution, maybe later. A. > Nicolas > [quoted text clipped - 24 lines] > > one's neck or back, effectively eliminating the "I" and > > having only the corpse come back up. > > I suggest the following approach: > > [quoted text clipped - 9 lines] > This problem requires the solution to the differential equation dv/dt > = g - uv^2, where u = cw/2*r*A/m Neat. That's the same form as the equation describing a mass accelerating from rest sliding down an inclined roller track: v' = B - Av^2 Look for article by Mel Lep with subject containing "Dynamics on a slide" for solution. I wrote: > ... as the object falls, it is spinning up > to rollers to match its linear velocity. So the condition for [quoted text clipped - 7 lines] > environment. ... here is an example where all the energy is > injected into macroscopic modes of freedom ... So the arguments for an opposing force dependent on v^2 must be rather general. |
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