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Natural Science Forum / Physics / General Physics / July 2008Water wave question..
If a 1 centimeter sphere that floats in water half way normally just sitting there and has a mass of 1gram, were dropped from 2 centimeters above the water... What would the wave length from the first peak to the second peak be, and what would the height of that peak also be? (anyone know an applet on the web that might do this stuff maybe?) If you want to make fun of me and my crazy ideas please do post an answer to the above question. :)  SignatureJames M Driscoll Jr Spaceman On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > If a 1 centimeter sphere that floats in water > half way normally just sitting there and has a mass of 1gram, > were dropped from 2 centimeters above the water... > What would the wave length from the first peak to the second peak be, > and what would the height of that peak also be? You almost have enough information, but you also need the depth of the water. Do you know why? > (anyone know an applet on the web that might do this stuff maybe?) > [quoted text clipped - 4 lines] > James M Driscoll Jr > Spaceman > On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 6 lines] > You almost have enough information, but you also need the depth of the > water. Do you know why? Oops yup gotta have depth. Lets say 6 centimeters then. On Jul 3, 9:10 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 9 lines] > Oops > yup gotta have depth. Do you know why? > Lets say 6 centimeters then. > On Jul 3, 9:10 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 13 lines] > > Do you know why? Yes, but that is not the issue here so you can ignore what I know since the answer has nothing to do with what I know about the situation except the facts I have given already. The measurements. Do you have an answer or are you going to just twist away for ever as usual? On Jul 3, 9:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 9:10 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 15 lines] > > Yes, BS, but never mind.... > but that is not the issue here so you can ignore > what I know since the answer has nothing to do with what I know > about the situation except the facts I have given already. > The measurements. > Do you have an answer or are you going to just twist away > for ever as usual? I'm not twisting, just taking my time. Are you in a hurry? Is this a problem you have to solve soon otherwise you'll have to give someone's money back? So, the way to solve this is to find the frequency of the oscillation of the ball in the water and the speed of wave propagation in the water. And then the distance between the ripples will be the speed of propagation, divided by the frequency. Do you agree? Do you know how to do it yet? If you do, what do you think is the next step? PD > On Jul 3, 9:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 19 lines] > > BS, but never mind.... Only BS in your head because you always have something to make up about my knowledge limits. >> but that is not the issue here so you can ignore >> what I know since the answer has nothing to do with what I know [quoted text clipped - 12 lines] > propagation, divided by the frequency. Do you agree? Do you know how > to do it yet? If you do, what do you think is the next step? Yes that is what I am looking for and.. I am only looking for the first wave set created (wave 1 and 2) calculated measurements. I am trying to find someone to do this idependantly of my "experiment" If I tell you anything about what I figured, it would not be independant at all. You choose YOUR next step. Independant experiment being setup here. Use your independance and only ask me the basics of the objects being used. You have all the basic now don't you? SignatureJames M Driscoll Jr Spaceman On Jul 3, 10:21 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 9:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 22 lines] > Only BS in your head because you always have something > to make up about my knowledge limits. No, it's BS because first you underspecified the problem, and then you overspecified the problem (so that the sphere could not have all the properties you gave it). This demonstrated for any semiliterate reader what your knowledge limits are. So, I'm probing a little further to see if you really do have some idea how to find the answer to the problem, or whether you are just tossing it out to see who gives you the answer. > >> but that is not the issue here so you can ignore > >> what I know since the answer has nothing to do with what I know [quoted text clipped - 28 lines] > James M Driscoll Jr > Spaceman > No, it's BS because first you underspecified the problem, and then you > overspecified the problem (so that the sphere could not have all the > properties you gave it). This demonstrated for any semiliterate reader > what your knowledge limits are. So now you are stating a hollow wood ball that has a mass of 1 gram and 1 centimeter diameter will not float. LOL How many times do you want me to dance your twist before you actually just answer the actual question with the facts stated? > So, I'm probing a little further to see if you really do have some > idea how to find the answer to the problem, or whether you are just > tossing it out to see who gives you the answer. I posted it to see others answer, so far not one "real" answer has been given. Proof of a typicle physicist twisting bullshit about experimentation. So, Use what is given and answer with what is given. Even if your answer is simply the ball will not float and will make "??? size waveheight" and ""??? size wavelength. Sheesh WTF is wrong that you have to fight and not answer at all? You get an F also. LOL SignatureJames M Driscoll Jr Spaceman On Jul 3, 1:54 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > No, it's BS because first you underspecified the problem, and then you > > overspecified the problem (so that the sphere could not have all the [quoted text clipped - 3 lines] > So now you are stating a hollow wood ball that has a mass of > 1 gram and 1 centimeter diameter will not float. That's correct. I invite you to make a hollow wooden ball of diameter 1 cm and mass 1 gram and drop it in water. That will be the true test, won't it? If you find that this is hard to make, then you can take a hollow wooden ball of 10 cm and mass 1 kg (which has the same density as the little ball) and it still will not float. > LOL > How many times do you want me to dance your twist before [quoted text clipped - 5 lines] > > I posted it to see others answer, OK, so you have no idea how to find the answer... > so far not one "real" answer has been given. > Proof of a typicle physicist twisting bullshit about experimentation. [quoted text clipped - 5 lines] > Sheesh > WTF is wrong that you have to fight and not answer at all? I'm not fighting, I'm taking my time and inviting you to participate in the thinking. You just seem to be in baby-bird mode, whining when people don't give you what you want when you want it in the manner you want it. > You get an F also. From you? Oooooh. > LOL Right. LOL. > -- > James M Driscoll Jr > Spaceman > On Jul 3, 1:54 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 9 lines] > 1 cm and mass 1 gram and drop it in water. That will be the true test, > won't it? Maybe, but of course that is still just a complete twist away from you answering the question at all. > OK, so you have no idea how to find the answer... I am waiting to see if my answers are close. So far you have given none. So I have nothing to say yet. Independant answers from mine is what I am looking for as I stated already. > I'm not fighting, I'm taking my time and inviting you to participate > in the thinking. > You just seem to be in baby-bird mode, whining when people don't give > you what you want when you want it in the manner you want it. And yet still no answer. That is "fighting" giving an answer yet again. but you know what. You can forget this one and see if I have enough facts in the updated version that removes the floating problem completely. SignatureJames M Driscoll Jr Spaceman On Jul 3, 2:18 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 1:54 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 16 lines] > > I am waiting to see if my answers are close. I'm not asking for your answers. I'm asking for one step you use to generate your answer. If you generated your answer with a dartboard, let me know. You've got absolutely NO poker face, Spaceman. I'd be happy to play cards with you for money some day. > So far you have given none. > So I have nothing to say yet. [quoted text clipped - 16 lines] > James M Driscoll Jr > Spaceman > I'm not asking for your answers. I'm asking for one step you use to > generate your answer. If you generated your answer with a dartboard, > let me know. > > You've got absolutely NO poker face, Spaceman. > I'd be happy to play cards with you for money some day. That you would be wrong about. In fact I am not allowed to play poker because I am always laughing either way and they don't let me play anymore. :) and I will admit PD, I goofed on this whole thing so chalk it up as my bad. :) SignatureJames M Driscoll Jr Spaceman On Jul 3, 10:21 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 9:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 47 lines] > at all. > You choose YOUR next step. OK, how about we take turns? I'll take a step, and then you take a step. That's called collaboration. I have an idea how to get the whole way there, so when you take a step, I'll let you know whether that's the direction I had in mind, and you can do the same for me. I've already done a step, but I'll take another one just in the spirit of good will. The frequency of a floating object that bobs up and down in the water is given by Hooke's law, where the restoring force is the imbalance between the buoyancy force and the weight. In equilbrium, the buoyancy force equals the weight: d_obj * g * V_obj = d_liq * g * V_sbm where d_obj = density of the object d_liq = density of liquid V_obj = volume of object V_sbm = volume submerged Since you want the sphere to be halfway submerged in equilibrium, for small oscillations, we can take the belly band of the sphere to be roughly a cylinder, which should be good enough for the calculation we need to do. OK, so far? So what's the next step? > Independant experiment being setup here. > Use your independance and only ask me [quoted text clipped - 4 lines] > James M Driscoll Jr > Spaceman > On Jul 3, 10:21 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 75 lines] > > OK, so far? So what's the next step? For you to use all that and give an actual answer instead of a twist around the final answer of the question given. The only "step" wanted was the final answer. It looks like I will have to make the question 10 pages long so you will grasp it to be able to answer it. LOL SignatureJames M Driscoll Jr Spaceman On Jul 3, 1:57 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 10:21 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 81 lines] > It looks like I will have to make the question 10 pages > long so you will grasp it to be able to answer it. So if I told you the answer was "6", how would you know how that answer was arrived at? How would you have any basis for knowing whether it's right? I know you haven't done the experimental test, because you don't have a 1 cm ball with a mass of 1 gram that floats halfway in the water. > LOL > > -- > James M Driscoll Jr > Spaceman > On Jul 3, 1:57 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 87 lines] > answer was arrived at? How would you have any basis for knowing > whether it's right? If you told me the answer was 6 I would know you do not have a complete answer. you have no units and are missing the other half of the question. and yet again your no answer is proof of the fighting of giving an answer at all. > I know you haven't done the experimental test, because you don't have > a 1 cm ball with a mass of 1 gram that floats halfway in the water. Even if that were the fact about the ball not floating (but it is not) you are still just not answering at all. That is the typicle relativist way, I should have remembered. my bad.. No answer needed now PD, new question "fixed" to not include the need for the floating at all. Will you answer that one, or just twist all over the place again? LOL SignatureJames M Driscoll Jr Spaceman On Jul 3, 2:21 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 1:57 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 90 lines] > If you told me the answer was 6 I would know you do not have a > complete answer. Ah, quite so. So if I told you my answer was 6 cm and 0.4 cm, how would you know how that answer was arrived at or whether it is at all close to correct? > you have no units and are missing the other half of the > question. [quoted text clipped - 17 lines] > James M Driscoll Jr > Spaceman > Ah, quite so. So if I told you my answer was 6 cm and 0.4 cm, how > would you know how that answer was arrived at or whether it is at all > close to correct? I would not, but it would have been added to the data. anyways. It's all water under the broken bridge now. I goofed bad, see my other posts. We can close this chapter and call it Spacemans greatest f.ck up yet. :) SignatureJames M Driscoll Jr Spaceman >> I know you haven't done the experimental test, because you don't have >> a 1 cm ball with a mass of 1 gram that floats halfway in the water. > > Even if that were the fact about the ball not floating (but it is not) You said it was floating. >>> I know you haven't done the experimental test, because you don't >>> have a 1 cm ball with a mass of 1 gram that floats halfway in the [quoted text clipped - 4 lines] > > You said it was floating. The ball was not a centimeter actually. Ruler operator malfunction. It looks like it is more like 1.3 centimeters actually. but I don't really care about it now. I will be breaking out the better measurement devices from now on. This is the end of this silly sinking/floating mess up. :) SignatureJames M Driscoll Jr Spaceman Spaceman Water being none compressible creates this action Ball hits the water and it comes up to the sides of the ball and pushes away. First ring of water the height of how high it reached up on the side of the ball. Lets say 9 inches. Next wave ring comes up 3 inches and its obeying the inverse square law. Hmmm I'm not sure on that. I am sure its an up down action and if you put a paper boat in the water it would not move to the lake shore. When these circles of waves hit the sandy bottom close to the shore they will start to ripple from the friction.and like all energy they will end up as heat. Waves with their up down action are tricky,and give an optical illusion of motion. Most ocean waves are just up and down until they get close to shore where a surfer can get on top of them to ride the wave to shore. Best to keep in mind not all water waves are the same type and some do travel fast as 300mph,and these are caused by a great push of an Earth quake under non-compressible water Bert. On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > If a 1 centimeter sphere Is that a diameter or a radius? If a diameter, the specific gravity of the sphere is 3/pi, or 95.5%, and then only 4.5% of the volume will stick out of the water, like an iceberg. If a radius, then the specific gravity is four times smaller than that, and now only 1/4 of the sphere will be below the surface, with most of it sticking up above like an empty styrofoam cooler on a lake. Notice that neither of these is "half way" in the water. > that floats in water > half way normally just sitting there and has a mass of 1gram, [quoted text clipped - 9 lines] > James M Driscoll Jr > Spaceman > On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: >> If a 1 centimeter sphere > > Is that a diameter or a radius? Yes of course, It should be diameter That is a default when some says such things, isn't it? anyways.. The diameter = 1 centimeter > If a diameter, the specific gravity of > the sphere is 3/pi, or 95.5%, and then only 4.5% of the volume will [quoted text clipped - 3 lines] > above like an empty styrofoam cooler on a lake. > Notice that neither of these is "half way" in the water. If it floats at half way would that not indicate the density allows such? On Jul 3, 9:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 2, 4:53 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 17 lines] > If it floats at half way would that not indicate the density > allows such? But you've specified the density already. You gave the dimension of the sphere, and so you know the volume, and you specified the mass. So you already have the density, and you can't suppose it to be an additional variable. > On Jul 3, 9:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 24 lines] > you already have the density, and you can't suppose it to be an > additional variable If I stated it will float half way, how did you come out with it will have 4.5% or 75% instead of the half way like I stated. The sphere floats half way. The water is 6 centimeters deep 1 centimeter is the diameter of the sphere. so What else is needed? What more do you need to know? Do you need to know the spere is hollow to allow the 50% floating? I would think you could pick that fact up without that much worry. so. There you go then. anything else? SignatureJames M Driscoll Jr Spaceman On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 9:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 27 lines] > If I stated it will float half way, how did you come out with > it will have 4.5% or 75% instead of the half way like I stated. Because you specified the volume of the sphere by telling me how big it is, and you specified the mass of the sphere. The density of the sphere is ALWAYS the mass divided by the volume. That's why I asked what the 1 cm referred to, so I knew how big the ball is. It's like the area of a wall, James. You can't tell me you've got a flat wall that is 12' wide and 8' tall that has an area of 42 square feet. But while we're at it, I see that I made an arithmetic error. If the 1 cm is a diameter, then the specific gravity of the ball is 6/pi, or 1.9. It is therefore heavier than water and will then sink to the bottom of the water and not bounce at all. > The sphere floats half way. > The water is 6 centimeters deep [quoted text clipped - 12 lines] > James M Driscoll Jr > Spaceman > On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 33 lines] > it is, and you specified the mass of the sphere. The density of the > sphere is ALWAYS the mass divided by the volume. So hollow stuff always screws you up completely? If you want, you can forget it all. But if you really want. The hollow sphere floats half way. the size and mass and how it floats are all measured facts. You just missed the fact it must be hollow I guess. :) > That's why I asked what the 1 cm referred to, so I knew how big the > ball is. > > It's like the area of a wall, James. You can't tell me you've got a > flat wall that is 12' wide and 8' tall that has an area of 42 square > feet. Bumpy walls can do such easy. You are stuck thinking too much 2D instead of thinking of other dimensions of the object, such as it being hollow. :) > But while we're at it, I see that I made an arithmetic error. If the 1 > cm is a diameter, then the specific gravity of the ball is 6/pi, or > 1.9. It is therefore heavier than water and will then sink to the > bottom of the water and not bounce at all. Hollow ball, you lose. LOL How much are you going to twist away from the given facts of the ball? If it floats half way, it would have to be hollow. any real scientist would have figured that out. :) SignatureJames M Driscoll Jr Spaceman >> On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> >> wrote: [quoted text clipped - 40 lines] > the size and mass and how it floats are all measured > facts. Impossible. The sphere's being hollow does not alter the net density -- mass divided by volume. You specified both, thus fixing the density. > You just missed the fact it must be hollow I guess. Hollow won't help. Overall mass of water displaced versus the overall mass of the object tells you whether it floats or sinks. Your sphere must sink. If you are claiming that experimentally you have found that your 1 gram sphere of diameter 1 centimeter floated in water, then I say that you are either mistaken in at least one of the parameters or are lying. Your choice. > :) > [quoted text clipped - 8 lines] > You are stuck thinking too much 2D instead of thinking > of other dimensions of the object, such as it being hollow. Bumps would *increase* the surface area! A flat wall of 12' x 8' would have a surface area of 96 square feet. A bumpy one would have *more* than 96 square feet. Are we to believe, from this dialog, that you don't know how to calculate the area of a rectangle? >> But while we're at it, I see that I made an arithmetic error. If the >> 1 cm is a diameter, then the specific gravity of the ball is 6/pi, or >> 1.9. It is therefore heavier than water and will then sink to the >> bottom of the water and not bounce at all. > > Hollow ball, you lose. Being hollow makes no difference if mass and volume are specified. > LOL > How much are you going to twist away from the given > facts of the ball? > If it floats half way, it would have to be hollow. > any real scientist would have figured that out. If you had a clue you would know that what you are saying is nonsense! > Impossible. The sphere's being hollow does not alter the > net density -- mass divided by volume. You specified > both, thus fixing the density. Are you saying hollow balls are impossible? >> You just missed the fact it must be hollow I guess. > [quoted text clipped - 5 lines] > water, then I say that you are either mistaken in at > least one of the parameters or are lying. Your choice. So you are saying hollow balls can not exist. LOL > Bumps would *increase* the surface area! A flat wall of > 12' x 8' would have a surface area of 96 square feet. > A bumpy one would have *more* than 96 square feet. No sh.t, that is what I stated. but we can drp this sh.t because it is totally irrelevant to the water wave question. > Being hollow makes no difference if mass and volume are > specified. Wrong. 100% wrong. You get an F yet again. You like to ignore bouyancy facts huh? They don't fit into your "solid only" rules about density. LOL > If you had a clue you would know that what you are saying is > nonsense! You are the one that is proving to be clueless here Greg. Hollow balls are burning you real bad. LOL SignatureJames M Driscoll Jr Spaceman >> Impossible. The sphere's being hollow does not alter the >> net density -- mass divided by volume. You specified >> both, thus fixing the density. > > Are you saying hollow balls are impossible? Something wrong with your reading comprehension skills? No, I'm not saying that hollow balls are impossible. I'm saying it makes no difference to the net density of the ball if its volume and mass are already specified. Are you saying that being hollow changes the weight of the 1 gram ball? Are you saying that being hollow changes the volume of the 1 cm diameter ball? Which is it? >>> You just missed the fact it must be hollow I guess. >> [quoted text clipped - 15 lines] > No sh.t, > that is what I stated. Here's the exchange: PD wrote: It's like the area of a wall, James. You can't tell me you've got a flat wall that is 12' wide and 8' tall that has an area of 42 square feet. James wrote: Bumpy walls can do such easy. So you implied that making the wall surface bumpy (despite PD having specified a flat wall) would *decrease* the surface area from 96 square feet to 42 square feet. So what you stated was exactly the opposite of what you just said you stated. You are therefore either mistaken twice, or a liar. Your choice. > but we can drp this sh.t because it is totally irrelevant > to the water wave question. Not if it shows that you can't even work out surface areas, and you insist that fixed displacement volume and mass don't determine the net density of an object! >> Being hollow makes no difference if mass and volume are >> specified. [quoted text clipped - 5 lines] > They don't fit into your "solid only" rules about > density. James, you'd be better to learn something about a topic before holding forth on it. It would avoid your making such an obvious fool of yourself. Unless, that is, that is your goal, in which case you seem to have hit upon a very efficient operating mode. >> If you had a clue you would know that what you are saying is >> nonsense! > > You are the one that is proving to be clueless here Greg. > Hollow balls are burning you real bad. Now there's a sentence I never expected to hear. >>> Impossible. The sphere's being hollow does not alter the >>> net density -- mass divided by volume. You specified [quoted text clipped - 10 lines] > the 1 gram ball? Are you saying that being hollow changes > the volume of the 1 cm diameter ball? Which is it? I am saying the hollowness facts can change the bouyancy. Why do you twist and change my statements all the time? You really need to do that huh? That is sad as usual. <Snipped rest of the usual relativist twisting of statements made.> On Jul 3, 12:13 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > >>> Impossible. The sphere's being hollow does not alter the > >>> net density -- mass divided by volume. You specified [quoted text clipped - 17 lines] > > <Snipped rest of the usual relativist twisting of statements made.> Nope, Spaceman, it is you that does not get it. You can make a solid ball out of wood that is 1 cm in diameter and 1 gm in mass. You can make a hollow ball out of steel that is 1 cm in diameter and 1 gm in mass. The two balls will have identical densities. The density is given for an object by mass/volume, regardless how the mass is distributed inside the object. This is why a boat that is made out of steel can still float on water, because it has the overall density of styrofoam (because the mass is distributed into a hollow shell). If it did not have the density of styrofoam, it would not float. PD > On Jul 3, 12:13 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 23 lines] > You can make a solid ball out of wood that is 1 cm in diameter and 1 > gm in mass. You can? WOW won't that float? LOL > You can make a hollow ball out of steel that is 1 cm in diameter and 1 > gm in mass. > The two balls will have identical densities. but different bouyancies. DUH! > The density is given for an object by mass/volume, regardless how the > mass is distributed inside the object. So wood won't float when it is the same mass as hollow steel. LOL > This is why a boat that is made out of steel can still float on water, > because it has the overall density of styrofoam (because the mass is > distributed into a hollow shell). If it did not have the density of > styrofoam, it would not float. Hollow ball that floats half way sheesh! My god you just refuse to use what is given as the experiment huh? The ball is wood and hollow and has 14.7 psi air inside (same as outside pressure) Sheesh man! (o psi difference but it still is "compressed" air once taken to outer space...) LOL I love twisting to your twists.. but.. really. Why don't you just answer the freakin question instead of pulling your twisting bullshit. Are you afraid of your answers to the given question as usual? SignatureJames M Driscoll Jr Spaceman On Jul 3, 1:49 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 12:13 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 26 lines] > You can? > WOW won't that float? Nope. > LOL > [quoted text clipped - 3 lines] > > but different bouyancies. Nope. Identical buoyancies. > DUH! > > > The density is given for an object by mass/volume, regardless how the > > mass is distributed inside the object. > > So wood won't float when it is the same mass as hollow steel. And the same dimensions. Right. You can try it to be sure. > LOL > [quoted text clipped - 5 lines] > Hollow ball that floats half way sheesh! > My god you just refuse to use what is given as the experiment huh? But it's not an experiment. It's something you made up in your head. You can't make it in a real experiment. And you haven't tried, to find out. > The ball is wood and hollow and has 14.7 psi air inside (same as outside > pressure) Makes no difference. > Sheesh man! (o psi difference but it still is "compressed" air once taken > to outer space...) [quoted text clipped - 3 lines] > Why don't you just answer the freakin question instead of > pulling your twisting bullshit. Well, because you gotta make the problem stated so that it is not ambiguous, first of all. Then you gotta make the problem so it isn't self-contradictory, second of all. > Are you afraid of your answers to the given question as usual? > > -- > James M Driscoll Jr > Spaceman > On Jul 3, 1:49 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 30 lines] > > Nope. So you made one and it sank? I don't believe you. But why should I .. you won't answer that it would even if it does float. :) SignatureJames M Driscoll Jr Spaceman On Jul 3, 2:22 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 1:49 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 32 lines] > > So you made one and it sank? I've made enough things with dimensions and masses set, so that I know what will sink and what will float, yes. It's a really basic lab taught in the first week of many physics courses, and I've done scads of those. You haven't, it appears. > I don't believe you. I'm not asking you to. I'm asking you to do it yourself and find out for yourself whether you are full of sh.t or not. > But why should I .. > you won't answer that it would even if it does float. [quoted text clipped - 3 lines] > James M Driscoll Jr > Spaceman On Jul 3, 12:13 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > >>> Impossible. The sphere's being hollow does not alter the > >>> net density -- mass divided by volume. You specified [quoted text clipped - 12 lines] > > I am saying the hollowness facts can change the bouyancy. Only if it changes the mass or the dimensions too. Changing the hollowness ALONE without changing the mass or the dimensions has absolutely no effect on the buoyancy. > Why do you twist and change my statements all the time? > You really need to do that huh? > That is sad as usual. > > <Snipped rest of the usual relativist twisting of statements made.> >> On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> >> wrote: [quoted text clipped - 41 lines] > facts. > You just missed the fact it must be hollow I guess. Do you realise what a fool this makes you seem ? You claim to have a ball with a volume of 0.5236 cm^3 and a mass of 1g, and you expect it to float in a liquid of density 1.0 g/cm^3 Does Archimedes mean nothing to you? Did she die in vain? >>> On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> >>> wrote: [quoted text clipped - 47 lines] > > Does Archimedes mean nothing to you? Did she die in vain? she huh? :) I think he would hate that more than my screwup. :) SignatureJames M Driscoll Jr Spaceman > "Spaceman" <space...@yourclockmalfunctioned.duh> wrote in message > Does Archimedes mean nothing to you? Did she die in vain?- Hide quoted text - ROFL! Archimedes was female? I'd suggest that you had better research your sources. Harry C, p.s., OG, you usually post correct information, which I respect. I try to do the same. Now I also respect the fact that many female authors had to publish under male names during largely the 18th and 19th centuries, but to make a post that support Achimedes being a female, that is simply a bit of a stretch! For me as a male animal, it is equivalent to suggesting that Jesus was female. That too could be possible, but where in either case is their evidence? Could lead to an interesting discussion. On Jul 3, 8:27 pm, "OG" <o...@gwynnefamily.org.uk> wrote: > "Spaceman" <space...@yourclockmalfunctioned.duh> wrote in message > Does Archimedes mean nothing to you? Did she die in vain?- Hide quoted > text - ROFL! Archimedes was female? I'd suggest that you had better research your sources. Harry C, Wooosh! You've clearly never heard of Tony Hancock! :-) On Jul 3, 9:40 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 3, 9:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 33 lines] > so > What else is needed? If it floats halfway and the 1 cm is the diameter, then it can't have a mass of 1 gram. > What more do you need to know? > Do you need to know the spere is hollow to allow the 50% floating? [quoted text clipped - 6 lines] > James M Driscoll Jr > Spaceman > If it floats halfway and the 1 cm is the diameter, then it can't have > a mass of 1 gram. I am a broken record now. Hollow ball, hollow ball.hollow ball. sheesh! :) >> If it floats halfway and the 1 cm is the diameter, then it can't have >> a mass of 1 gram. > > I am a broken record now. > > Hollow ball, hollow ball.hollow ball. Being hollow doesn't change the mass or volume that you specified. The sphere will still displace the same amount (volume & mass) of water corresponding to the sphere's volume. If the mass is likewise fixed at 1 gram, then its density is fixed, too. >>> If it floats halfway and the 1 cm is the diameter, then it can't >>> have a mass of 1 gram. [quoted text clipped - 8 lines] > to the sphere's volume. If the mass is likewise > fixed at 1 gram, then its density is fixed, too. The sphere will displace the same amount from the volume of water, the density is not only determined by the "volume and size". Hollow balls prove you wrong about your density thoughts. You really love to get F's as grades huh? LOL SignatureJames M Driscoll Jr Spaceman >>>> If it floats halfway and the 1 cm is the diameter, then it can't >>>> have a mass of 1 gram. [quoted text clipped - 12 lines] > volume of water, the density is not only determined > by the "volume and size". Right. It's determined by the mass and the volume (density is specified in units of mass per unit volume). You fixed the mass and the volume, so you also fixed the density by implication. > Hollow balls prove you wrong about your density thoughts. Apparently they don't. They do, however, nicely elucidate your complete incompetence in matters technical and scientific. >> But you've specified the density already. You gave the dimension of >> the sphere, and so you know the volume, and you specified the mass. [quoted text clipped - 14 lines] > I would think you could pick that fact up without that much worry. > so. Why do you complicate the scenario so? What's the point? Just what is it that you want to get out of the exercise? Superficially it appeared that you were looking for the wavelength and height of the wavetrain. Now you're out fishing for the sphere anatomy? >>> But you've specified the density already. You gave the dimension of >>> the sphere, and so you know the volume, and you specified the mass. [quoted text clipped - 17 lines] > Why do you complicate the scenario so? What's the > point? What complications, I gave the info, and the complications are coming from not figuring the facts about the info. Sheesh. > Just what is it that you want to get out of > the exercise? Superficially it appeared that you > were looking for the wavelength and height of the > wavetrain. Now you're out fishing for the sphere > anatomy? No, I gave the anatomy of the ball basically, It floats half way. It is being ignored to twist away from the question to begin with. Would it have to be hollow or not with such dimensions? simply factual answer... YES. Hollow ball! dang I have to fix my record. :) In sci.physics, Spaceman <spaceman@yourclockmalfunctioned.duh> wrote on Wed, 2 Jul 2008 17:53:32 -0400 <V6mdnUpF3uW6ZPbVnZ2dnUVZ_tzinZ2d@comcast.com>: > If a 1 centimeter sphere that floats in water > half way normally just sitting there and has a mass of 1gram, [quoted text clipped - 6 lines] > please do post an answer to the above question. >:) It might be worth noting here that a 1 cm diameter sphere would have a volume of 523.6 cubic millimeters, and therefore such a sphere as you describe would sink (the water displaced would weigh 523.6 milligrams, not quite enough to keep the sphere afloat). The energy released from the drop, from drop point to surface, is approximately (0.001) (9.81) (0.025) [the 0.025 is because of center of mass] = 245.25 milliJoules. There is additional energy as the item continues to drop through the water but that energy will probably not manifest as a wave; I'm frankly not sure what it will do. There are a number of issues such as surface tension and motility/viscosity that factor in. As for making fun of your "crazy ideas", there will be other opportunities. :-P This is not a bad question, actually, though it's a bit incomplete. Signature#191, ewill3@earthlink.net Windows Vista. It'll Fix Everything(tm). ** Posted from http://www.teranews.com ** > In sci.physics, Spaceman > <spaceman@yourclockmalfunctioned.duh> [quoted text clipped - 31 lines] > other opportunities. :-P This is not a bad question, > actually, though it's a bit incomplete. Perhaps the question could be revised to make it consistent and perhaps more tractable. Something like: A 1 cm diameter sphere of a uniform density such that it will float half in, half out of water is poised above a large body of water of uniform depth of 6 cm. Initially the sphere is at rest and just touching the water's surface at its lower most point. The apparatus is located in a uniform gravitational field of 1g acceleration. Air resistance effects can be ignored. If the sphere is allowed to drop freely from its initial position, what is: a) The equation of motion of the sphere over time b) The wavelength of the ensuing wavetrain c) The amplitude of the wavetrain over time Lowering the initial height of the sphere is meant to prevent complete submersion of the sphere and the ensuing turbulance that would result (with too much initial potential energy the sphere might also pop right back out of the water at least once, which would over complicate the problem with surface tension issues, turbulance issues, different environments, etc.). There's still a lot to consider including the drag of the sphere in motion, varying surface area in contact with the water, water displacement rates, etc. Not a trivial excercise! > Perhaps the question could be revised to make it > consistent and perhaps more tractable. Something like: There we go, the old lets change the experiment bullshit. The typicle relativist method of diversion to not use the "actual experiment" and find the experiment to fit what we think instead of what we are trying to figure out. LOL > A 1 cm diameter sphere of a uniform density such that it > will float half in, half out of water is poised above a [quoted text clipped - 3 lines] > located in a uniform gravitational field of 1g > acceleration. Air resistance effects can be ignored. You have altered the experiment, You get an F in science class today. <snipped rest of altered experiment thoughts since they are not compatible with the experiment proposed.> Nice diversion Greg. To bad that would get you an F in science class. :) SignatureJames M Driscoll Jr Spaceman >> Perhaps the question could be revised to make it >> consistent and perhaps more tractable. Something like: [quoted text clipped - 16 lines] > > You have altered the experiment, Only because the idiot setting the test is proposing something that is physically impossible. You claim to have a ball of diameter 1cm that has a mass of 1g that floats 50% submerged in water. Either the diameter isn't 1cm or the mass isn't 1g or the liquid isn't water > In sci.physics, Spaceman > <spaceman@yourclockmalfunctioned.duh> [quoted text clipped - 17 lines] > (the water displaced would weigh 523.6 milligrams, not > quite enough to keep the sphere afloat). hollow sphere. the half floating tip should have been a clue mr cd salesman. :P > The energy released from the drop, from drop point to > surface, is approximately (0.001) (9.81) (0.025) [the [quoted text clipped - 5 lines] > There are a number of issues such as surface tension and > motility/viscosity that factor in. Basic old fashion still Water, nuff said > As for making fun of your "crazy ideas", there will be > other opportunities. :-P This is not a bad question, > actually, though it's a bit incomplete. No, It is complete now. The answer is what is not complete yet. Still waiting for the wavelength of first wave set created and wave height of the second wave from the drop point. :) SignatureJames M Driscoll Jr Spaceman On Jul 3, 8:24 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > In sci.physics, Spaceman > > <space...@yourclockmalfunctioned.duh> [quoted text clipped - 48 lines] > James M Driscoll Jr > Spaceman Not yet complete, and data is NOT consistent. (1) If diamater = 1 cm and it floats 1/2 way, you can work out the mass; as others have pointed out, it will not be 1 gram. (2) Similarly if radius = 1 cm. (2) If mass = 1 gram and it floats 1/2 way, you can work out the radius and the diameter; neither will be 1 cm. So, take your choice: either (1), (2) or (3), In the real world we inhabit you just cannot have both radius or diameter = 1 cm, and mass = 1 gram if it is going to float 1/2 way. OK, so if you agree to make a choice, then (as others have suggested) you need to know the water's depth, although if it is "sufficiently deep" an exact value probably won't matter. One more thing you need to know that you have not specified: you say you drop it from a height of 2 cm. Is the center at a height of 2 cm, or is the bottom of the sphere at a height of 2 cm? (One could also quibble that if the water is shallow one needs to know the shape of the container and whether or not the bottom is horizontal or has a slope, etc. But never mind that.) Now, rather than responding that all this is avoiding your question or twisting your words, or whatever, why not just admit that your original problem specification was incomplete and somewhat erroneous? Is it so terrible to admit you are human and can make mistakes? R.G. Vickson > On Jul 3, 8:24 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 55 lines] > and it floats 1/2 way, you can work out the mass; as others have > pointed out, it will not be 1 gram. So a 1centimeter diameter hollow ball that has a mass of 1 gram is impossible? :) <snipped rest of non hollow ball to fit the half floating reality diversions> > Now, rather than responding that all this is avoiding your question or > twisting your words, or whatever, why not just admit that your > original problem specification was incomplete and somewhat erroneous? > Is it so terrible to admit you are human and can make mistakes? No mistake, Sorry that hollow balls act that way and ignore the math you are using. :) <mean comment time> I think the real problem is hollow heads are trying to answer a the question that is based upon a physical reality that can and does occur in reality. </mean comment time> Repeat after me, A hollow ball can float half way with the dimensions and mass given. If you don't think so, You better re-think your thinking completely. :) SignatureJames M Driscoll Jr Spaceman -- James M Driscoll Jr Spaceman >> Not yet complete, and data is NOT consistent. (1) If diamater = 1 cm >> and it floats 1/2 way, you can work out the mass; as others have >> pointed out, it will not be 1 gram. > > So a 1centimeter diameter hollow ball that has a mass of 1 gram is > impossible? Sure it's possible. But it won't float because its density will be about 1.9 gm/cm^3, which is greater than that of water at 1.0 gm/cm^3. > :) > <snipped rest of non hollow ball to fit the half floating reality [quoted text clipped - 8 lines] > Sorry that hollow balls act that way and ignore the math > you are using. James doesn't understand density or bouyancy. > <mean comment time> > I think the real problem is hollow heads are trying to answer [quoted text clipped - 5 lines] > A hollow ball can float half way with the dimensions and mass > given. Sure it can... In a pool of mercury. > If you don't think so, > You better re-think your thinking completely. Riiight. >>> Not yet complete, and data is NOT consistent. (1) If diamater = 1 cm >>> and it floats 1/2 way, you can work out the mass; as others have [quoted text clipped - 6 lines] > will be about 1.9 gm/cm^3, which is greater than that of > water at 1.0 gm/cm^3. So Greg, If I now also had to tell you that the air is compressed inside the sphere, would you think different about it "floating" or would you still say such is impossible just because you can not figure out the rest of the actual properties from the info I gave? :) According to your thoughts compressed air could not help things float better because the diameter and mass are all that matter. :) As I stated, you will find all sorts of reasons you do not need to answer and so far you have gotten F's all the way and I have had to give you more simple facts you should have been deriving from the facts given. You need to think,. instead of just use your textbook part of your brain. It seems that is impossible for you and you need your hand held through the entire experiment. :) > James doesn't understand density or bouyancy. James is playing upon your ignorance of bouyancy. :) > Sure it can... In a pool of mercury. Or in water if the air pressure inside were great enough. Again, you get an F in science. SignatureJames M Driscoll Jr Spaceman >>>> Not yet complete, and data is NOT consistent. (1) If diamater = 1 >>>> cm and it floats 1/2 way, you can work out the mass; as others [quoted text clipped - 13 lines] > such is impossible just because you can not figure out the rest > of the actual properties from the info I gave? You specified the mass of the sphere and its volume. Those are the only relevant parameters to determining its density. Composition doesn't matter, internal geometry doesn't matter. Are you saying that *adding* mass in the form of air to the hollow interior will make the sphere larger or lighter? > :) > According to your thoughts compressed air could not help > things float better because the diameter and mass are all that > matter. That's right. Do are you saying that you think otherwise? Things like balloons, air mattresses, and rubber rafts use compressed air to fill out and support their geomentric structures, giving them volume without adding a lot of weight that would be otherwise be required if the same were attempted by rigid structure (struts and braces) alone. > As I stated, you will find all sorts of reasons you do not need > to answer and so far you have gotten F's all the way and I have [quoted text clipped - 9 lines] > > James is playing upon your ignorance of bouyancy. James *still* doesn't understand bouyancy. >> Sure it can... In a pool of mercury. > > Or in water if the air pressure inside were great enough. > Again, you get an F in science. Adding mass to a fixed volume container will not make it lighter. > You specified the mass of the sphere and its volume. > Those are the only relevant parameters to determining [quoted text clipped - 4 lines] > to the hollow interior will make the sphere larger > or lighter? It can make it more bouyant. I am not adding mass, I merely don't have all the mass as one material and one of the mass types involved is more bouyant than the other. Apparently that is something you like to ignore so you can just not answer the original question and give your typicle twisting away from the actual experiment. LOL >> :) >> According to your thoughts compressed air could not help >> things float better because the diameter and mass are all that >> matter. > > That's right. Do are you saying that you think otherwise? I know otherwise. The more air pressure inside the more it will float. from the bouyancy differences of air and water. > Things like balloons, air mattresses, and rubber rafts > use compressed air to fill out and support their geomentric > structures, giving them volume without adding a lot of > weight that would be otherwise be required if the same were > attempted by rigid structure (struts and braces) alone. So you can not grasp adding bouyancy from air pressure that does not stretch the container or only stretches the container to the size given. LOL > James *still* doesn't understand bouyancy. James know how to make things float that you apparently could never understand! LOL > Adding mass to a fixed volume container will not make it > lighter. I never said it would make it lighter. The mass I gave includes the pressurized air. I said it can make it more bouyant. You are very ignorant of that fact huh? SignatureJames M Driscoll Jr Spaceman >> You specified the mass of the sphere and its volume. >> Those are the only relevant parameters to determining [quoted text clipped - 9 lines] > as one material and one of the mass types involved is more bouyant > than the other. Look up bouyancy. It depends *only* on the weight and the volume of fluid displaced. > Apparently that is something you like to ignore so you > can just not answer the original question and give your > typicle twisting away from the actual experiment. Look up boutancy. You'll discover that it is you who doesn't have a grasp of the matter, and that your scenario is therefore impossible (yet you say that it is a real experiment that you have done!). >>> :) >>> According to your thoughts compressed air could not help [quoted text clipped - 6 lines] > The more air pressure inside the more it will float. > from the bouyancy differences of air and water. Hey James, full scuba tanks are much heavier than empty ones, yet they are full of all that bouyant air. What's up with that? >> Things like balloons, air mattresses, and rubber rafts >> use compressed air to fill out and support their geomentric [quoted text clipped - 5 lines] > air pressure that does not stretch the container > or only stretches the container to the size given. A 1cm diameter sphere is still a 1cm diameter sphere no matter what the filling is. If you also specify the overall mass, then its density and bouyancy are fixed. Period. >> James *still* doesn't understand bouyancy. > [quoted text clipped - 9 lines] > I said it can make it more bouyant. > You are very ignorant of that fact huh? See? James has no understanding of bouyancy. A 1cm diameter sphere is still a 1cm diameter sphere no matter what the filling is. If you also specify the overall mass, then its density and bouyancy are fixed. Period. James doesn't understand Archimedes' Principle, a 2200 year old concept. >>> You specified the mass of the sphere and its volume. >>> Those are the only relevant parameters to determining [quoted text clipped - 12 lines] > Look up bouyancy. It depends *only* on the weight > and the volume of fluid displaced. And I gave you the fact. It is floating half way. apparently you still want to just ingore that fact so you never have to answer the question at all. >> Apparently that is something you like to ignore so you >> can just not answer the original question and give your [quoted text clipped - 4 lines] > scenario is therefore impossible (yet you say that it > is a real experiment that you have done!). Look up bouyancy of compressed air in cylinders and then think spheres instead. It is you that is still twisting away from giving an answer at all. > Hey James, full scuba tanks are much heavier than > empty ones, yet they are full of all that bouyant > air. What's up with that? They are heavier Greg, but alas, they are more bouyant. Or do you think the air does not matter at all? > A 1cm diameter sphere is still a 1cm diameter sphere > no matter what the filling is. If you also specify the > overall mass, then its density and bouyancy are fixed. > Period. You really do think compressed air does not make something float better? > A 1cm diameter sphere is still a 1cm diameter sphere > no matter what the filling is. If you also specify the > overall mass, then its density and bouyancy are fixed. > Period. Parrot time.. LOL > James doesn't understand Archimedes' Principle, a 2200 > year old concept. Greg wants to ignore parts of the principle when it comes to compressed air. :) SignatureJames M Driscoll Jr Spaceman In sci.physics, Spaceman <spaceman@yourclockmalfunctioned.duh> wrote on Thu, 3 Jul 2008 13:22:39 -0400 <t5GdnQlkC760lvDVnZ2dnUVZ_oLinZ2d@comcast.com>: >> You specified the mass of the sphere and its volume. >> Those are the only relevant parameters to determining [quoted text clipped - 6 lines] > > It can make it more bouyant. And how would you do that without changing the experiment? You specified 1 cm diameter and 1 gram mass. If you meant 1 cm *radius*, you might have a chance, as the sphere would then displace more water (as it would be larger -- 8 times larger, as far as volume is concerned, in fact). > I am not adding mass, I merely don't have all the mass > as one material and one of the mass types involved is more bouyant > than the other. Makes no difference. > Apparently that is something you like to ignore so you > can just not answer the original question and give your [quoted text clipped - 10 lines] > I know otherwise. > The more air pressure inside the more it will float. Makes no difference unless it changes the sphere size. Bouyancy = mass of displaced water (volume * 1 kg/liter) - mass of item. Put in as much air as you want; it will only make the item heavier. > from the bouyancy differences of air and water. > [quoted text clipped - 8 lines] > or only stretches the container to the size given. > LOL Apparently you cannot grasp the notion that bouyancy doesn't care about the air pressure, but only about the volume of the item (which determines the water displaced) and the mass of the item. >> James *still* doesn't understand bouyancy. > [quoted text clipped - 8 lines] > The mass I gave includes the pressurized air. > I said it can make it more bouyant. So adding air to a fixed volume container will make it lighter, even though adding mass to a fixed volume container will not? How can that be so? Obviously I'm missing something very important here in Tire Salesman Physics(tm)...?? > You are very ignorant of that fact huh? Signature#191, ewill3@earthlink.net /dev/signature: No such file or directory ** Posted from http://www.teranews.com ** > In sci.physics, Spaceman > <spaceman@yourclockmalfunctioned.duh> [quoted text clipped - 14 lines] > And how would you do that without changing the experiment? > You specified 1 cm diameter and 1 gram mass. And I specified it floats half way. No change in experiment on my side, only in facts on the sides that wish to not answer at all. > If you meant 1 cm *radius*, you might have a chance, as > the sphere would then displace more water (as it would > be larger -- 8 times larger, as far as volume is concerned, > in fact). So pressurized air can not make something more bouyant? Why not? > Put in as much air as you want; it will only make the item > heavier. So you have facts that compressed air does not make higher bouyancy than non compressed air would? Could you point me to such facts? > Apparently you cannot grasp the notion that bouyancy > doesn't care about the air pressure, but only about > the volume of the item (which determines the water > displaced) and the mass of the item. And you have what proof of that? Where is your proof that compressed air will not make something more bouyant? > So adding air to a fixed volume container will make it lighter, > even though adding mass to a fixed volume container will not? Higher bouyancy and making something lighter are two different things. > How can that be so? Obviously I'm missing something > very important here in Tire Salesman Physics(tm)...?? It must be that Cd salesman do not know anything about compressed air that tire salesmen so know about. :) Do you think a tire that sinks with 2 psi in it will not float when 40 psi is put in it or even 100 psi? Do you really think that adding "bouyant mass" does not make something more bouyant? :) SignatureJames M Driscoll Jr Spaceman >> You are very ignorant of that fact huh? >> In sci.physics, Spaceman >> <spaceman@yourclockmalfunctioned.duh> [quoted text clipped - 19 lines] > only in facts on the sides that wish to not answer > at all. Your "facts" are inconsistent. If the problem is badly formulated, you cannot expect reasonable answers. >> If you meant 1 cm *radius*, you might have a chance, as >> the sphere would then displace more water (as it would [quoted text clipped - 3 lines] > So pressurized air can not make something more bouyant? > Why not? Because air has mass, hence weight. Adding weight to the same sized object will not make it more bouyant. >> Put in as much air as you want; it will only make the item >> heavier. > > So you have facts that compressed air does > not make higher bouyancy than non compressed air would? > Could you point me to such facts? It's a high school experiment in first year chemistry class to "weigh" air and various other gases. Weigh an empty balloon, then fill it with a sample of gas an re-weigh (note that a balloon filled with air weighs *more* than an empty balloon). Some gases have a lower molar mass than air (like hydrogen or helium), so they displace more weight of air than the mass of the gas itself, making the resulting balloon plus gas weigh less than the air displaced -- hence the upward bouyant force is greater than the weight and the balloon wants to rise. >> Apparently you cannot grasp the notion that bouyancy >> doesn't care about the air pressure, but only about [quoted text clipped - 4 lines] > Where is your proof that compressed air will not > make something more bouyant? Scuba tanks. They displace the same volume of water whether full or empty and so produce the same buoyancy force. They are heavier when full, though, so the net upward force is *less* for full tanks, *more* for empty tanks. >> So adding air to a fixed volume container will make it lighter, >> even though adding mass to a fixed volume container will not? > > Higher bouyancy and making something lighter are two different things. Not if the volume is fixed. The buoyancy force depends *only* on the weight of the medium displaced, which is strictly determined by the volume of the object immersed in that medium. Add the buoyancy force to the weight of the object (being mindful of direction of the vectors) to obtain the net force on the object. >> How can that be so? Obviously I'm missing something >> very important here in Tire Salesman Physics(tm)...?? > > It must be that Cd salesman do not know anything about compressed > air that tire salesmen so know about. More like hot air, given the nonsense you're promulgating. > Do you think a tire that sinks with 2 psi in it > will not float when 40 psi is put in it or even 100 psi? > Do you really think that adding "bouyant mass" does not > make something more bouyant? There is no such thing as "buoyant mass". There is buoyancy, which is a force due to displacement of a medium, and mass which results in a gravitational force. Buoyant force and gravitational force are often directed oppositely, although you can set up situation where this is not so (such as balloons free-floating in a car that is accelerating or decelerating). > Your "facts" are inconsistent. If the problem is > badly formulated, you cannot expect reasonable > answers. Greg, you are correct I hate to say. :) I goofed bad. I actualy missed the blank spot between the 0 and the end of the ruler.. (stupid ruler) Next time.. no rulers.. and onyl micrometers and more accurate measurement completely. So.. I say I am sorry for all the crap I babbled and.. we can all forget this and tell Dirk to add it to his site for my goof ups. :) SignatureJames M Driscoll Jr Spaceman >> Your "facts" are inconsistent. If the problem is >> badly formulated, you cannot expect reasonable [quoted text clipped - 12 lines] > we can all forget this and tell Dirk to add it > to his site for my goof ups. James, my opinion of you has just risen by several points. Admitting an error takes more courage and character than clinging to a wrong idea out of stubborn principle. >>> Your "facts" are inconsistent. If the problem is >>> badly formulated, you cannot expect reasonable [quoted text clipped - 17 lines] > character than clinging to a wrong idea out of > stubborn principle. Thanks Greg, Now If I could only get you to understand the 2 clock problem when they both have the same tick rate per second but different times on thier faces as being a clock malfunction and nothing more than such. But, I know that will end up coming to you in "time" some day. And I do hope you get such way before you have any final moments on Earth or anything even close. :) SignatureJames M Driscoll Jr Spaceman >>>> Not yet complete, and data is NOT consistent. (1) If diamater = 1 cm >>>> and it floats 1/2 way, you can work out the mass; as others have [quoted text clipped - 36 lines] > Or in water if the air pressure inside were great enough. > Again, you get an F in science. My god. Do you realy believe that? How do you imagine air pressure alters bouyancy? --John Park |
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