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It doesn't.

I am a little puzzled by your response above. Of course, your statement in itself is correct, but I was referring to the pressures acting horizontally from the vertical column of air, as discussed in the treatment given by studiot. Two or more such adjacent columns would, presumably, have equivalent horizontal pressures, which would cancel out, so making no overall horizontal pressure within the body of the gas. I think you may not have understood what I was saying. Very good. Thank you for contribution. Thank you studiot for your recent contribution.

Thank you for your contribution.

Agreed; a change in momentum is, in fact, an impulse, which is force*time. Thank you. The horizontal pressure problem has gone away. I realized a few hours ago that there are no such pressures in the atmosphere, of course, provided there is no wind. Thank you everyone for your help.

You must not think of your cue as the force of gravity. The cue imparts momentum to the snooker ball, whereas gravity causes the ball to have weight. No, this is confusing and irrelevant. What we should be considering is the possible effects on molecules travelling downwards under gravity of the "sideways" travelling molecules in the gas. Sure, the "sideways" molecules exert equal pressures sideways, but there will be no overall sideways movement unless the whole gas body is in movement, eg as in a wind. The total sideways momentum of the total molecules is conserved no matter how many collisions occur. The initial condition of the downwards molecules has no sideways component to take into account. So the total effect on the "sideways" pressure is unchanged by the collisions, including the initial "downwards" molecules. Again, NO! If you miss- cue and strike the ball off-centre, or your ball hits a second ball off-centre, then naturally the two balls will travel apart sideways, although continuing to some extent forwards, and the various resulting motions will have momentum conserved in any direction you choose to consider. And I am sure you are genuinely trying to help me. But, as I've said, some of your posts are confusing and I cannot understand them. Basically, I think we are in agreement about the "sideways" pressures, but my problem is to understand how these pressures are, presumably, equal to the pressure acting downwards as discussed in "studiot's" excellent post. So, perhaps someone else can post the missing explanation.

Why? The density of the air decreases with altitude because of gravity. This is very well explained in the excellent post by studiot The collisions transfer the weight of the air. Agreed. As above. Please refer to the post by studiot You seem to have missed my post about studiot's excellent treatment dealing with the effects of the distributed mass of the air molecules. I have gratefully acknowledged this explanation. Any other problems you may have with my earlier posts on this matter would, I hope, be considered before you post again about my earlier ideas. Nevertheless, I should still be pleased to receive suggestions to explain what perhaps I might call the "sideways pressure" issue. I never made any such suggestion. Please give a reference to any of my posts which gave you this idea, because this needs putting right! Does it not depend which way you point the cue? Sorry, but this is confusing. I think we are discussing the effects of collisions with the molecules proceeding vertically downwards under gravity. On your assumption, there will be no net CHANGE in momentum as a result of the collisions with the"sideways" molecules, so the "sideways" pressures will be the same as before the collisions. So I'm not happy yet. I still should like an explanation for the overall "sideways" pressure, which presumably in practice would be equal to the downwards pressure, even though this is caused by the weight of the air above.

Of course. I didn't even think it needed saying. I'm not even sure why you would need to ask. Thank you very much for your response. I needed to know your position on this because it is fundamental to the discussion, and as I mentioned, your previous posts had sadly confused me about your thinking on this matter. No! This violates the Law of Conservation of Momentum! It is true that the collisions can make the snooker balls bounce off and go in all directions, but the total momentum must be conserved. This means that although some may go right, then some will also go left, and when the various angles and components of momentum are taken into account, the overall change is zero. The momentum of each ball is mass*velocity, but since velocity is a vector, so is momentum. If you define the positive direction to be to the right, then the direction to the left is negative, and these signs must be applied in the calculations. So I think we still need an explanation for the "sideways" pressure. No!! Please refer to my post of just a few minutes ago. There is more to it than that. Multiple collisions consist of separate collisions, and momentum conservation applies to each and all. So there is no overall change in momentum. Please refer to my very recent response to "Strange", where I have gone into it in rather more detail.

Conservation of momentum only tells you for any single collision, there will be equal momenta in the sideways directions (for both x and y). As Strange has implied, you have to consider collisions at all angles. This will quickly make the motion stochastic. Certainly momentum is conserved for one collision, but so it is conserved for all the collision under consideration. So the total momentum after the collisions is equal to the total momentum before the collisions. But it must be remembered that we are dealing with vectors, so we need to use the vector components in any chosen direction, which necessarily takes the angles into account.

Why would you conclude that? In fact, it is (as noted in another thread) conservation of momentum that defines the angles that the two balls will move: https://mechasco.wordpress.com/2014/05/23/billiard-balls-and-the-90-degree-rule/ I came to that conclusion because of my reading of your earlier comment, which I quote below, together with my post which occasioned your reply. I have put your comment in bold, hopefully to avoid confusion. If there were any truth to that, then games such as snooker or billiards would be impossible. But snooker and billiards ARE possible. Therefore, this supports conservation of momentum. I'm sorry that things got rather confused. However, now in view of your latest reference to "mechaso", I am lead to believe that you DO, in fact, accept the Law of Conservation of Momentum. This is fundamental to our discussion, so please are you willing to accept this Law? I await your reply.

I have been thinking more about your reply to my post. As in the quote above, you seem to be implying that because snooker and billiards ARE possible, then that means that Newton's Law of Conservation of Momentum is not true. So, I repeat my question. Do you not accept that Newton's Law of Conservation of Momentum is true?

Do you not accept Newton's Law of Conservation of Momentum? Given a "perfect" set of snooker balls and table, this law would certainly apply. But you have to take into account ALL the balls involved in the overall collision process.

Thank you for your post. Yes, the mathematical analysis provided by studiot shows clearly the effect of gravity on the air molecules and the consequent air pressure, which I understand. But this is necessarily in the vertically downward direction because of the gravitational force vector. As for sideways motion, surely Newton's Law of Conservation of Momentum applies here. I suppose there is another factor I must have missed, and I should be pleased to learn about it.

Thank you again, studiot, for what is clearly a very informative post. It will take me some time to study it properly. And no, I didn't come across the book by C J Smith; no doubt before my time. Regards. I thought that my choice was appropriate for the topic, but maybe I was wrong.

Thank you for your friendly and helpful response. I fully understand the reason for your questions above. Here is my feedback. I am a (very) retired physicist. I received my degree in 1959; and perhaps this partly explains some of my posts. I have no experience with fluid statics or fluid dynamics. I am only slightly aware of the random walks ideas.

Of course, you are correct, it does apply. Sorry, I must have missed these questions. My answers are below. 1) Gravitational force of attraction of the Earth 2)Mass*Gravitational Force 3)When a molecule of air impacts upon the Earth's surface, it is "reflected" off again. There is a change of momentum in this collision, which applies an impulse to the Earth's surface, that is (force*time). When summed for all the impacting molecules over a specified area, we get the total force on that area, and hence the pressure. OK, ?