I never changed myh analogy.
Your claim:
"I explained the problem with your analogy and how pressure worked,"
My response:
"Nope, you changed the analogy."
Is it REALLY so hard to read properly? YOU started talking about MY analogy, not yours. NOONE said you changed your analogy. This is some shit reading right there.
You changed it into something inexplicable involving no kind of closed system and outside forces.
When did I change your analogy? Yeah, never. Lying only hurts your credibility.
I don't think you understand any of what you're talking about.
I do understand perfectly what I am talking about. You however don't seem to understand it, since you make contradictory claims and pretend the contradiction doesn't exist.
Take a can of compressed air. take a tube of toilet roll. Are the forces acting on each the same?
Maybe, maybe not. It depends on various factors. Something you should already know.
I'm not talking about spatial dimensions?
Yes. You are saying space is a fluid. Space has 2 definitions - The spatial dimensions or a large place that has nothing. Since a fluid is something, the second definition of space (a place where there is nothing) can't be true, because there is something if there is a fluid. Which means you have to be talking about the spatial dimensions. Unless you are making up your own definitions.
You got different answers from a highly specific case.
Ahh, so you do not know what "generalized" means. HINT: It's the opposite of specific.
You do not address the possibility of, for example, 8n.
See generalizing. Generalizing addresses every possibility within the given interval. For my generalization, it works for every real number.
You did not use a generalized equation at all, you focused on one case.
*sigh*
Your box has sides w(width)b(breadth)l(length). According to you, first I have to calculate the volume of the box to get "how many molecules are headed for each side/end" so wbl. Then, the difference in pressure between two sets of surfaces is the distance ratio. Let's say sides 1(area=bl), 2(area=wl) and 3(area=wb). 1 have distance w, 2 has b and 3 has l. Between 1 and 2 the ratio is w/b, between 2 and 3 the ratio is b/l and between 3 and 1 the ratio is l/w. Answer for 1&2 = wbl/w/b=b^2*l, answer for 2&3 = wbl/b/l=wl^2, answer for 3&1 = wbl/l/w=w^2*b. However, these are still volumes, hence they are not ratio differences or pressure. So you told me to divide by their surface areas to remove the units. For 1&2 => 1 = b^2*l/bl=b, 2 = b^2*l/wl=b^2/w. For 2&3 => 2 = wl^2/wl=l, 3 = wl^2/wb = l^2/b. For 3&1 => 3 = w^2*b/wb = w, 1 = w^2*b/bl = w^2/l. All of these are distances however. If we want to properly remove the units, we have to divide by these lengths by something. Since we have to remove all units, we have to divide these distances by themselves. Left is a 1. As you can see, we are not getting anywhere. This is all according to your math. And it is stupid, as none of this is pressure.