Talos

John, Well, this tells you what a volumetric weight is. http://www.online-ca...tric/weight.php And, as I thought it might, it has nothing to do with what Talos is on about. Obviously you have no idea what a volumetric weight is, as evidence this commercial definition for the purpose of costing. Arc, Sure, and when a rock sinks the water rising past it pushes the rock down. The air moving around a car is pushing it forward. An elevator works like the balloon. You know as well as I do that this is claptrap. This might help—if you can get your heads round it. http://www.newton.dep.anl.gov/askasci/gen99/gen99903.htm

StringJunky, By your logic then: a brick and a feather free-falling in a vacuum would fall at different rates because the Earth is pulling more on the brick than the feather, which is not correct. The gases with a higher molecular weight than helium push the helium atoms out of the way and assume a lower position. Your comment is the most sensible so far. However, free falling in a vacuum is not applicable in an atmosphere. The volumetric weight of a feather is more than an equivalent volumetric weight of air. This means air will not descend and lift the feather up like it would a balloon. When we inflate a balloon with helium its volumetric weight decreases. When a point is reached beyond which the balloon’s volumetric weight is less than an equivalent volumetric weight of air, the air will descend and lift the balloon up.

Studiot, Buy yourself a helium balloon and some joss sticks. In a closed room light your joss stick/s and allow the room to fill with smoke. When the smoke laden air has become static, release your balloon from floor level and watch the smoke laden air descend around it due to gravity*, while simultaneously moving underneath the balloon and rising as it lifts the balloon up in an equal and opposite reaction. * the helium has weight because it is in a gravitation filed, but its weight is less than the air. Hence gravity has a stronger pull on the air than it does on the helium. John, I’ve really got you hot under the collar haven’t I. This gives me great pleasure.

John, He isn't going to let a mere fact get in the way of his idea. I predict the he will email a famous physicist and ask a pointless question. It’s not my idea. And famous physicists' don't answer pointless questions. I suspect it is you that has asked pointless questions and got no answers. You, studiot and Strange win the prize for claptrap. And you are bringing this forum to the point of ridicule. Mark my words, you will all be proved wrong.

Studiot, you say: Let us suppose that the inflated balloon is resting on the ground, so it is as low as the lowest level of air. Just where exactly will the alleged descending air descend to? It gets under the balloon thereby pushing the balloon up which increases the balloons potential energy. The increase in the balloon’s potential energy comes from the air’s loss of potential energy as it descends. Recall one of my earlier posts. Answer from Andrew Hanson, CPhys Outreach Manager National Physical Laboratory Hampton Rd | Teddington | Middlesex | UK | TW11 0LW THE DISPLACED AIR COMING DOWN LOSES POTENTIAL ENERGY AND THE BALLOON GAINS IT. How many times must I repeat this before you cotton on?

Strange Right. So when a balloon reaches the ceiling and stays there, it is held in place by a continuous downwash of air. Got it. Thanks. That makes so much sense. It makes no sense at all. When the balloon reaches the ceiling and stays there, it doesn’t require any more lift. It only required lift by means of a downwash of air in order to get it there. Studiot So if the helium balloon rises 1000 times this are you saying there will be a downwash of air 14 kilometres below the balloon? No. The downwash only occurs locally where the balloon is in space, and it continues locally around the balloon all the way up to your 1000 times 100 metres.

Studiot, So are you suggesting that 15 years ago balloons rose for one physics reason, but today they rise for an entirely different one? NO. I'm suggesting that there's more to it than buoyancy. Strange, "Perhaps you can ask them if airplanes fly due to buoyancy, as you claim." I never claimed that. I'm claiming that the same physics apply to a balloon as they do to an airplane, and it isn't buoyancy. It's to do with equal and opposite reactions. Lift for an airplane or a balloon is an equal and opposite reaction to a downwash of air.

John, you say “perhaps you can stop bothering these people.” I can assure you that they are not bothered, as you put it, but they are more than happy to respond. Your ideas are 15 years behind the times. I challenge you and, anyone else, to contact the physics departments of any university in the UK or America and, put your argument to the head of department. Of course you won’t, because I have challenged you and you will be proved wrong—but others might. These links will get you all the universities in the UK and USA. http://en.wikipedia.org/wiki/List_of_universities_in_the_United_Kingdom#Universities_alphabetically http://www.utexas.edu/world/univ/state/

John, I was addressing Strange’s accusation that my argument was some sort of failed argument from authority. I don’t need any authority because I know what I’m talking about. This was for Strange’s benefit. You say, “You really don't understand do you?” Oh yes I do. It is you and your cronies that doesn’t understand. And anyhow, “why should I believe you when higher intellects agree with me.

On 04 Dec 2014, at 11:21, Talos Perdix <talosperdix@tiscali.co.uk> wrote: Dear Professor Richard Ellis, Does a helium balloon gain potential energy when it goes up? Yes, any object that rises in a gravitational field gains potential energy. Jonathan Richard Ellis (CBE FRS) Jonathan Richard Ellis is a British theoretical physicist who is currently Clerk Maxwell Professor of Theoretical Physics at King's College London.) Strange You say: “Ah, thank you. I guessed it was some sort of failed argument from authority but Google failed me.” I say: “How much authority do you want?”

All of you. Are you disagreeing with this independant explanation? "AS THE BALOON RISES, THE DISPLACED AIR COMING DOWN LOSES POTENTIAL ENERGY AND THE BALOON GAINS IT."

Studiot. You are forgetting, or deliberately ignoring the fact that there are 3 terms on each side of the inequality—which leads me to ask “what part or parts of the inequality: m(air)gh > m(helium)gh is factually incorrect?” _____________ Studiot, you say: A body continues in its state of rest or motion unless acted on by an external force. I agree. There is no mention of potential energy, or any requirement that any potential energy possessed by this body be used, or that the body must possess any particular potential energy. I agree again, but I do ask, what do you think the force is comprised of. A body can still suffer change of motion with no change of PE and Newton's First Law still applies. I agree again. Equally a body can suffer change of PE, with no change of motion. I disagree. If a bodies potential energy changes it must either go up or down. There is eventual change of PE in this case, but at the outset of motion, no PE is exchanged. I think you mean—before the onset of motion, in which case I agree again. That is PE only changes as a result of work being done by the external force. I agree yet again—but I ask again, what is the external force comprised of? John, you say. So, your own evidence shows that the density of the air affects the weight. (and that's why they might seek to measure it in a vacuum) Did you not realise you were arguing against yourself? I am not arguing against myself. I have said that the weight of air is due to gravity and this affects the density—not the other way round. As regards the generator question that you raised, let me say that this is right up my street. Let me also say, that you have been proven wrong in asserting a helium balloon loses potential energy when it goes up in the air. When you can see your error for yourself and explain it to me, then and only then, will I deal with the generator issue.

Studiot. Potential energy……….m(air)gh > m(helium)gh You say “It is factually incorrect because at ground zero h = 0 so the inequality you state is factually incorrect.” I say: Then why does the balloon rise?

John and co. Here's an answer from Andrew Hanson, CPhys of the National Physical Laboratory that you quoted. Outreach Manager National Physical Laboratory Hampton Rd | Teddington | Middlesex | UK | TW11 0LW My question was: I can measure the weight of a deflated balloon. I can also weigh a bottle of compressed helium. If I now inflate the balloon from the compressed helium and then weigh the bottle again does the difference in weight equal the weight of helium in the inflated balloon? YES. THOUGH WEIGHING THEM WILL BE TRICKY AS FOR THE BOTTLE, THE WEIGHT CHANGE WILL BE VERY SMALL COMPARED WITH THE WEIGHT OF THE BOTTLE AND FOR THE BALOON, THE AIR ABOUT IT IS DENSER, SO BOUYANCY WILL HAVE TO BE TAKEN INTO ACOUNT. AT NPL WE WOULD WEIGH THIS IN A VACUUM. Is it true that the inflated helium balloon therefore has weight? YES, BUT SO DOES THE AIR ABOUT IT WHICH BEING DENSER WILL RESULT IN BOUYANCY SO THE BALOON WILL FLOAT UPWARDS. If so, will the balloon gain potential energy when it rises up into the earth’s atmosphere? IT DEPENDS ON WHAT YOU COUNT AS YOUR SYSTEM. I THINK YOU HAVE TO INCLUDE THE EARTH'S ATMOSPHERE. AS THE BALOON RISES, THE DISPLACED AIR COMING DOWN LOSES POTENTIAL ENERGY AND THE BALOON GAINS IT. Nothing in this answer contradicts anything I've said.

John, you say “for a helium balloon, measured in air, the weight is negative. So the more height you have the more negative the energy is.” I say, weight has nothing to do with where it is measured. Gravity gives a mass its weight, and it can never be negative. If I have a 1 kg weight and a 2 kg weight on either side of a balance scale the 2 kg weight will descend and the 1 kg weight will rise. I could argue, like you, that the 1 kg weight is negative, but that doesn’t alter the fact that it gains potential energy when it rises.