Sorry for the late reply, especially since I'm not the one dealing with a broken hand. I've been a bit of a vegetable though.
I'm starting to agree with you on the recoil thing. Even though water comes out of SuperCAP's nozzle in one direction, the flow goes in another direction through the backpack. Changing that momentum, through the turns in the tubing, creates a recoil force.
However, I still believe the amount of area determines how much force acts on the stream. pressure * area = force.
You can't say that more area in the chamber means more force acting on the stream. Quote:
| Originally Posted by Ben
And why do you care about "force at the nozzle?" What are you trying to prove with this? It seems irrelevant to me.
|
I only brought up the stream tests and all to show that the force depends on the area of the nozzle - and to show that that force is thus not dependent on force in the chamber. In other words, force in the chamber doesn't matter. Pressure is what ultimately determines force "at the nozzle," regardless of chamber size.
Jo mixed up "flow" and "displacement." Flow is volume per time, displacement is just the volume. The meaning is still clear though.
Quote:
| Originally Posted by Ben
This disproves my theory how? I don't see what the problem is. There's still only one part that doesn't have the other parts cancelled out. Watch a spherical CPS bladder discharge. The end moves in.
|
I don't have a spherical PC gun, but looking at balloons, they reduce in all diameters as the volume is depleted.
I think my old concern with your theory still stands. The rubber on the sides of a cylindrical bladder still exerts force, even if it does no work. As you fill the bladder, more rubber is stretched, and the force is increased.
But as you said here, there's only one part that doesn't have the force cancel out. It's just that bladder expansion and work is different for spherical and cylindrical PC guns.
Quote:
| Originally Posted by Ben
I've already posted several links that state that dV (a cross section of the volume) is what is used in the Navier-Stokes equations. Therefore, the mass matters less than you may think. Even so, as I explained previously, more often than not the force is many times higher than the weight of the water. So, the mass of the water is a non-issue. It essentially is negligible. You're arguing without thinking! You seem to have missed this point somewhere.
|
I'm not talking about force created from gravity and the mass of water. I'm talking about when the force is applied - more mass means less acceleration.
Quote:
| Originally Posted by Ben
Again, tell me more things I know. The difference here is that the piston is actually reapplying the force. That's how Pascal's principle works.
And I refuse to argue you further on this issue. A force is a PUSH, PULL, OR LIFT. Where's the push, pull, or lift? The piston is necessary to create the push.
Also, if you really wanted to do a more accurate test with "conservation of momentum," fire water guns with a force gauge to measure the recoil force. Seems to be an easier test.
|
A piston does little more than divide. What about regular air pressure guns? No piston. But you'll say that it's air pushing on water. But what about water pushing on water? As you said before (I'll search for it and quote if necessary), it's force, but not useful force. However, it is useful force if it's pushing on water in the nozzle. Force pushing on water in the chamber is useless.
Do my tests really need to be more accurate? I got an unnoticeable force applied by the small stream - I thought I saw the needle move, but I think it might have been a hallucination. I got maybe 2/3 of a pound of force from the large stream. You may say "imprecise" all you want, but the results are there. Sure I used a scale meant for hundreds of pounds, but there was clearly a difference in force used to accelerate the streams.
I haven't been able to do the tests since I've been sick.
Quote:
| Originally Posted by Ben
I am getting really annoyed that you fail to read my explanations to this issue you keep hammering. As far as I can tell, you have not read that once.
I speculated previously (in my head, not online, that's still speculation) that mass does matter to a certain extent. And dV indicates that. dV is a small cross section of the total volume. A larger volume will have a larger small cross section. I even posted links about this to prove it... please read them.
|
I'll take another look. But once again, it would be a lot easier if you would quote the pages you link to. Much of the stuff goes over my head, so I probably miss a lot of it.
Quote:
| Originally Posted by Ben
Again, you've missed several key points I made in my posts. If you continue to do this, I will simply not respond any longer aside from a request to read my posts. I spend half my posts telling you to read my previous posts.
To mention another thing you seem to have avoided (perhaps intentionally), to achieve the same flow with a wider pipe, less velocity is necessary. Therefore, less acceleration is necessary. Therefore, a wider pipe should have comparable flow with less acceleration. Not hard to understand.
|
I believe I got to it later - I mixed and matched the quotes. Perhaps not such a good idea after all.
Quote:
| Originally Posted by Ben
The total volume of water is what matters. Each individual molecule certainly is controlled by the pressure. Each individual molecule gets closer to each other and the forces of electric charge are actually causing the force. What matters overall (and all the physics I have pointed out show this) is the force. Why do you continue to ignore Navier-Stokes and everything I have posted that says flow is determined by force? And if you do not quote and answer this in your next post, I will simply ask you to read my posts again.
|
Why do you keep on saying "Navier-Stokes" without explaining how it relates to anything? I told you, even if it goes over my head, feel free to quote a single formula. Jo or iSoaker will likely understand it. I get the feeling it doesn't actually relate force and flow, because you haven't shown that it does. When you tell me to read the links, I get nothing out of it.
Quote:
| Originally Posted by Ben
I know an applied force creates a pressure. However, the entire fluid most definitely is not under a pressure. Thereforce, there is no differential. That's what I have been saying all along.
|
How can you say it's not under a pressure? I just said that you apply force over an area in a piston pumper. Is there something wrong with my logic? Because you just said the opposite.
The pressure is obviously (nearly) constant throughout.
Quote:
| Originally Posted by Ben
Desperate? I consider your continued and intentional ignoring the Navier-Stokes equations desperate. Yet, I do not call you out for it, until now.
And I did not say they were wrong. They never once said that pressure is the deciding factor in flow. They said more pressure is benefitial. I say that too. But they carefully put a qualifier on there. There obviously are other factors. Why do you interpret what they wrote differently?
|
I call your continued vague references to the Navier-Stoke equations desperate. Good job on discovering them, but it's time to show that they actually prove your argument. Is it really so hard to find an equation that says "Q = F * something"? (Or something along those lines.) Yes, it is.
But back to what I called desperate - the fact that you suddenly changed your mind about the authoritativeness of the patent. As soon as I used it to prove my argument, not yours, you claimed that it was not such a great source after all.
Once again, I'm getting really annoyed by your brushing off of their claim of a pressure-range relationship. Look back at the first post in this thread. I said that pressure and range have a positive correlation, and my qualifier was the Cv. They're just proving my argument.
Quote:
| Originally Posted by Ben
The 1250 was rounded to the nearest 50. I like square numbers. And actually, (2^2)*pi*100 equals 1256.54 pounds of force. I don't know what you're doing to get that wrong.
Umm, you're the one who's ignoring net force. I said net force before you ever did. I don't know what you're saying here.
Certainly, the water applies force back. But, it does not apply 1250 pounds of force back. I would doubt that the water and pipes apply more than 50% of the force back. The entire force model is a simplification based upon what we know. The force definitely is reduced due to the water, but it is not completely reduced.
|
Sorry, I used the physical pipe sizes, not their designations...I believe it was 4.027" in diameter or something, according to McMaster.
[edit]Actually, it was 4.026 according to McMaster and the Engineering Toolbox.[/edit]
Because the pressures are equalized, there's 100 PSI on each side of the piston. 1250 PSI are being applied in each direction. Thus, the net force is 0. There's a bit more net force when you open the ball valve.
Disregarding the friction, I'm positive the water would apply (2^2)*pi*100 - (0.25^2)*pi*100 = 3.9375*pi*100 pounds of force back. The net force would be (0.25^2)*pi*100 pounds.
When you fire SuperCannon II, do you really think there's half a ton of force on your arms? Does it feel like there's a polar bear sitting on top of you? I didn't think so. But an interesting test (besides building a smaller cannon) would be to measure the recoil force of SuperCannon II.
Quote:
| Originally Posted by Ben
Cool. Then why does the SS 300's perform well with a Cv that stinks and low pressure? High force is the only explanation. Why you continue to play this down isn't up to debate either: you are intentionally avoiding it.
And proportional does not mean that pressure is the only value in there. Generally certainly accounts for ability to flow, along with turbulence, as well as the force applied.
And why doesn't anyone use the Cv equations to calculate flow created? To reiterate something I posted before, nothing you have posted says that the equation is used to calculate flow. And I have taken a good look. It's useful to see how much flow is lost through valves. That is in fact why it was made. I'll go onto your beloved Physics Forums and ask about this myself if you want me to.
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Does its Cv really stink? If the narrow part is the same as the nozzle aperture as you've suggested, then its Cv isn't much worse than it would be with just the nozzle, and it would have less turbulence. Plus, I'll once again mention that they said the ball valve would help by increasing flow.
There's no point arguing with you on the "generally" issue. They neither mention force positively nor negatively, which I feel means it's insignificant, but there's nothing I can do to change your interpretation. All I'll say is that they mention force nowhere in the article (within context), since numbers of mentions seems to count.
Regarding your last paragraph, I've seen the discussion. I'll get to it at the end of this post. It appears they're siding with the pressure heads (nice term, by the way
).
Quote:
| Originally Posted by Ben
He doesn't find the exact amount of flow because it involves math well beyond the student who asked. That is why there was no small equation directly showing the relationship between force and flow. It becomes complicated. And I think you knew this. That's why there are an entire set of equations for different situations. They're called the Navier-Stokes equations and they're used explicitely for flow created by force.
|
Well then, I think we can drop that water bottle scenario.
Quote:
| Originally Posted by Ben
This REALLY ANGERS ME. Again, I will not respond again if you ignore certain parts I post. I EXPLICITELY posted the part that says force creates flow. As usual, you seem ignorant to it.
Go to http://www.maths.qmul.ac.uk/~hve/MAS209/
Read Chapter 6 for the best part, but most any chapter will do. The introduction says " We now turn back to the study of fluid motions as determined by some equations of motion. We mean by this in particular (in the traditional Newtonian sense) an equation which relates the acceleration of the motion of a fluid to the forces that are generating the motion."
I could pull up many many more pages on this if you want it. But I hope that I don't have to. |
I still am interested in what variation of N-S relates flow to force. All they've said in what you've quoted is that acceleration and force are related. That much is obvious even through Newton's second law.
How can you write off Wikipedia's comment on the N-S equations while using a similar comment in the PDF to prove your own point?
Quote:
| Originally Posted by Ben
I too say range and pressure is related. And the thought that the patent was implying Cv is up to debate. The fact that the water gun involved was called a low pressure water gun and had wide chambers with small pipes, yet still performed excellent, indicates to me that they meant something other than ability to flow.
|
They certainly mentioned the ball valve for flow, you make it sound like the tubing was an extension of the nozzle, and the pressure-range correlation is what I mentioned in my first post. If you think the water gun performance model needs to be much more complicated than the Cv equation, then feel free to create one.
You've also repeatedly said before that pressure does not matter. Recently you've agreed that it at least helps determine the range.
Quote:
| Originally Posted by Ben
Good. That's what the Cv still is supposed to work on. Nozzles and valves. Not entire water guns.
|
I already explained how a water gun is like a valve. There's a pressure differential, an outlet, and an inlet of sorts.
Looking at the debate in the Physics Forums, they've said they don't know what you're talking about with "flow lost."
Quote:
| Originally Posted by Ben
I too say pressure is related to flow. But it is not the only factor and Cv does not account for everything else. That and I've gone through those threads and none of them use Cv to calculate flow created. Those threads don't help your argument.
And please pull up a calculator that uses the Cv equation to calculate flow created by a pressure. I've looked at them and I haven't found one yet.
And why do you ignore my .gov and .edu links? I don't ignore yours. I post what's wrong with them. Please read every post I posted between your last one and the one you posted before there. And by read, I mean read, not skim.
|
Once again, the Cv equation works. Look back at this thread to see all my examples. The Physics Forums thread you started also appears to agree with the Cv equation. What I'll agree with is the fact that it's definitely simplified. However, as you can see from the first post in this thread, all I said was that a constant nozzle/gun/Cv leaves a pressure-range correlation. Recently you posted a link to an explanation of basic factors in Cv - and larger tubing obviously means higher Cv.
I don't know how those calculators calculate the flow. However, they definitely use an inputted Cv for something, and it definitely isn't "flow lost" because that apparently doesn't exist. Also, their equations did not relate force to flow.
I've already dealt with your .gov and .edu links. The .gov talks about an irrelevant problem in which they don't even calculate flow, not to mention it wasn't necessarily written by a NASA guy. The .edu(s) only seem to talk about the N-S equations that you constantly refer to, but a force-flow equation is notably absent.
Quote:
| Originally Posted by Ben
Navier-Stokes. I will say no more. If you can not go on to any page describing those equations and see that they explicitely state the correlation, I can not make you. Just because the equations are more advanced than you are does not mean that they are wrong. Fluid dynamics is a very complicated field and requires complicated equations. Your ignorance does not mean the equations are wrong.
Also, Wikipedia said the following only once:
...
And I agree. Proportionality does not say that one thing is determined solely by another. And the fact that this was said only once does not help your argument.
|
I've noticed you've said no more. Just saying the name doesn't mean a thing. You have yet to type out an equation for us to see.
I've seen them talk about a correlation - a force-acceleration one that means nothing.
And not even once does anybody mention a flow-force correlation. That really doesn't help your argument. And I will again refer to my first post.
Quote:
| Originally Posted by Ben
You use 57 feet and 52.5 feet. Though, I would prefer more tests and I intend to do more. I am completely sure that the LR system will shoot further.
|
Those two figures use bias towards your arguments. How do we know 57 was not to the spray? How do we know it wasn't just a better nozzle with no spray but more puddle range? Why don't we compare it to 60 feet?
Well, I'll be looking forward to your results. 'Tis a shame you broke your hand though.
Quote:
| Originally Posted by Ben
Again, you have read me poorly. It's not narrowist near the nozzle. The internal diameter is about the same as the nozzle. In fact, I think it is the same. I bet that was done purposely to make a more well formed stream.
It also doesn't have high flow. You know it's flow. It's about 5X. The same flow as the CPS 1000 with a smaller internal diameter.
Anyway, the real Cv value that matters is the nozzle. That reduces flow more than anything, yet you imply that the ID of the pipes in the main factor in Cv. Please read my previous post on this. I will not waste my time to find it for you. You haven't even stated anything about this, so I can only assume that you have not read it.
|
If the ID is the same, then couldn't the entire tubing past the barrel be considered the nozzle? I don't see how that would reduce the Cv - having a long stretch of aperture-sized tubing does little in terms of turbulence and Cv.
I don't understand your second paragraph.
Quote:
| Originally Posted by Ben
An example: a nozzle is a valve. So yes you can change the velocity.
|
Let me quote what you were responding to:
Quote:
| Originally Posted by SilentGuy
Then why does it not use delta velocity (or acceleration, if you will)? Because the flow is constant throughout the valve. You can’t have 50 m/s at the inlet and 40 m/s at the outlet because 10 m/s were lost. (Disregarding changes in diameter.)
|
Note the last sentence - disregarding changes in velocity.
Let me rephrase that - you can't have a change in flow, or "lost flow" as you like to call it. Flow is constant throughout the system. Cv can't be used to calculate "lost flow" because "lost flow" isn't even possible.
Quote:
| Originally Posted by Ben
Using a simplification not intended for that purpose gets the wrong answer.
|
???
All I said was that a water gun is a valve. Just because they don't call it a "water gun valve," and just because Cv is usually used to rate valves, doesn't mean you can write off the equation in this situation. The Physics Forums people think Cv is a fine method.
Quote:
| Originally Posted by Ben
I could use much of the same argument for you, but I'll use a different one. Using an equation meant to calculate flow from force is just like using any other equation. Using an equation to calculate flow lost backwards to calculate flow created from a pressure is using an equation for something it was not meant to be used for.
So why is Navier-Stokes wrong, but this equation that I can't find an example of being used in this manner correct? The Cv equation is a simplification. There are many equations for different situations that can not be used in others, and this is one of them.
|
Why do you keep on going on about "flow lost?" It doesn't exist. The Physics Forums people have said so, the Cv equation doesn't say a thing about delta flow, and flow is constant throughout a system.
N-S isn't wrong. It just doesn't relate force and flow in the first place. It does relate pressure and flow, according to Wikipedia. In fact, I should be using N-S for my own argument.
Quote:
| Originally Posted by Ben
I think I may be missing the point. What was the test you did trying to show? I don't understand how it disproves anything I'm saying.
As for how everyone accepted your results, that would be a lie. In fact, I don't think anyone aside from myself and joanna commented on them. Complete acceptance? I don't think so.
|
My test just showed that the force used to accelerate a stream depends on nozzle area and is independent of chamber area. That's it. It doesn't matter how much force you have in a chamber of whatever size - it just boils down to the pressure over the area of the nozzle. And if the nozzle area is fixed, then velocity depends on the pressure.
I was under the impression iSoaker had commented on the results, but he might not have seen them. At any rate, he hasn't rejected them.
And I still don't see (1) how or (2) why you disagree with the results. One stream applied unnoticeable force, and the other applied much more. I'll do a test with the scale vertical when I get better, but keep in mind what you said - the force is much larger than the weight of the water, so you can ignore the weight of the water.
Quote:
| Originally Posted by Ben
The Cv is very different and this equation will not account for it. I will measure it if you want. In fact, I plan to this weekend.
And when it comes down to it, the Cv that matters is the NOZZLE, not the piping. That is one major reason why your model is wrong. Again, actually read some of my posts to find where I argue this.
|
We'll see you're results. We've beaten the Cv horse to death, and the Physics Forums people agree with me, but you keep on talking about a nonexistent "flow lost."
SuperCAP and SuperCannon II have the same pressures and forces. One has a worse nozzle and a smaller Cv. My model uses the Cv to explain the worse range, although outside of the model, I'll agree the nozzle also makes a difference. Do you have a force model at this point? If so, is its explanation of the situation terribly different?
Please go back and read my first post. I explicitly said the main factor in determining Cv is the nozzle. Interior tubing also matters, especially if there's turbulence there. But before I ever even considered the rest I mentioned the Cv of the nozzle.
Quote:
| Originally Posted by Ben
Why wouldn't the flow stop nearly instantly with the air? Explain that to me. I want to see your reasoning.
|
System A: With the first ball valve open.
System B: With the first ball valve closed.
Both systems are ideal.
Open the second ball valves. Initially, both systems have identical pressure gradients, so they have identical flow. After a minuscule amount of time, a certain amount of water has been displaced (volume/time = flow). System A, which had lots of air compressed into a small space, still has pressure because the air is still very much compressed. System B, on the other hand, has no pressure because barely any water had been compressed and all the excess water has already been displaced. System B stops the flow because there's no pressure gradient. System A continues to have flow because there's still a pressure gradient because the pressure is maintained.
Quote:
| Originally Posted by Ben
Well, last time I checked the square root still kept everything relatively proportional. 100 PSI is still 58% than 40 PSI more with the square root. And higher velocity does not always mean more range. That is an error you continually make. Especially in water guns, higher velocity can create more drag, which can destroy streams. This problem is more prounounced on the situations that create it because small streams are easier to break up, and they also move faster.
|
The higher velocity = higher drag simply proves my point. You suggested that by my argument, "a water gun with notably higher pressure will peform more than slightly better than one with the same force but lower pressure." My model only correlates the pressure and the velocity. The velocity will by 58% greater for SuperCannon II at 100 PSI, but the range will only be 10% or whatever you had because of the range-velocity correlation. I agree with you that higher velocity does not mean that much higher range.
Quote:
| Originally Posted by Ben
Well, good luck justifying why my results will go in the other way after I build.
|
Once again, we'll see about that. No need to exchange rhetoric until then.
Quote:
| Originally Posted by Ben
Navier-Stokes. I say no more.
|
Well, that's saying a lot...so far it hasn't seemed like N-S says much about a force-flow correlation. I've already addressed this issue. It just says F=ma. And Wikipedia says N-S relates pressure and force.
Quote:
| Originally Posted by Ben
The total area matters, not the force per unit area.
And Cv is reduced mainly by the nozzle. Why you have continually ignored this is up to debate. Please read the part of my posts about this.
|
You yourself said it was force per unit volume. That implies force per unit area if it's volume per distance.
Why don't you read my first post? Also, reread what you had quoted: "In practice, the wider chambers will only have marginally more flow because the total Cv is barely larger." The total Cv is barely larger because the Cv is mainly reduced by the nozzle - as we've both stated.
Quote:
| Originally Posted by Ben
I don't know that the "diagram" is. Could you explain?
Your stream tests also do not show that force is reduced per unit area. Less water output means less impact. I don't see how that proves anything.
I will not explain again also why your interpretation of Pascal's principle is wrong again. Please read any post I have made.
While it took you days to post your horribly incomplete reply to many posts I made, this reply took me only a bit more than an hour. Keep them coming.
|
Why do I get the feeling nobody reads my post?
No matter, I'll repost the picture:
According to your theory, all 1250 pounds of force would be used regardless of the size of the nozzle. And thus that regardless of water output, the impact is the same.
Quote:
| Originally Posted by Jo
It's perfectly possible to hold a hose with no nozzle on it and have no recoil from it. (assuming the hose isn't kinked or bent.)
Try it. Then put a nozzle on, and the recoil should be quite noticable.
My point about recoil is that the most major vector in any water gun shot is the backwards recoil, not the vertical one. Even in SuperCAP, you've mentioned nearly being knocked over backwards by the rear recoil. You've never noted that you almost became airborne because of the vertical recoil however - oddly enough.
I'm going on about nozzle force, because I'm explaining my full viewpoint. You say flow doesn't happen without force. I'm agreeing, but not with your source of that force (Hey, that rhymes!)
|
Are you sure about that hose thing? Sounds right from my experience...but it doesn't seem like it could be right...interesting.
If there's vertical recoil as Ben says (after your post), I'm wondering if it's not felt as much because the backpack also weighs a lot. Interesting too.
Quote:
| Originally Posted by Ben
It's the direction of flow that matters. Any direction it is moving in will make an equal but opposite force. That's why both have recoil. The nozzle increases the velocity however, so there is the most transfer of momentum there. If you don't like this explanation, I'm sure the new member who posted about water rockets will, so ask them.
|
There should be recoil both up and back. The reaction from the flow moving down the backpack and then turning through the pipes requires a bunch of force vectors.
Quote:
| Originally Posted by Ben
Also, I've grown quite tired of this argument. The more I look into it, the more I become convinced that the entire situation is too complicated and requires closer study. I asked at PhysicsForums and I didn't really get a straight answer. On one hand, they said the Cv equation could be used, but on the other hand they said Bernoull's equation could be used to find the velocity of the the flow. And before you pressure-heads get carried away, the velocity has to be multiplied by the area to get the total flow. I worked through the equation and got that unrestricted flow equals sqrt(2*A*F). I would post the derivation, but it would take me a while to type up and this already has taken me forever. I brought up some inconsistencies between pressure and performance and one person said "at that size you have to consider the size of the tubing and the size of the reservoir and the calculations get a lot more complicated."
Interestingly enough, I calculated the Cv valve of Supercannon II with the fire hose nozzle. The reported value for flow was 20% more than what actually occurs. That's somewhat close I suppose, but I think it suggests at the very least that there are other major factors in play. I suppose fire trucks are more powerful. Too bad.
I am convinced that Navier-Stokes takes more factors into consideration and it is what we're looking for. With that said, I'm planning some rather expensive tests right now to find whatever correlation there is between anything. I'm open to anything, but I'm seeing too much conflicting information. I'll accept being wrong if that is the case, but now I'm thinking we're all wrong.
|
I've taken a look at their argument, and it seems to be pointing in Jo's and my favor.
Here's the link for those who haven't found it (yet).
Quoting a few people:
Quote:
| Originally Posted by russ_waters
In f=ma, Force determines acceleration if a given mass is constant. But for fluid systems, mass is itself a rate and varies according to pressure and cross sectional area. Using pressure instead of force takes away that complication and allows you to directly find velocity.
|
Quote:
| Originally Posted by FredGarvin
What your opponents are saying is correct. The Cv is a measure of the flow capacity of a valve, orifice, etc...The definition of Cv is the flow of water in GPM created by a 1 psig delta p across the valve. It has nothing to do with "flow lost" (I'm not really sure what that means).
|
Now, I'll agree with you that we need to learn much more about these physics. All I can say is that the new tests, with a bunch of constants but the right variables, should give us some real results. I still think a 10% difference in range is a decent margin of error due to the slight differences in Cv for a 2" chamber and a 4" chamber (or whatever you're going to use).
That 20% difference between the SuperCannon II Cv and the nozzle Cv does help clear up some stuff. The nozzle (or whatever part is the smallest) clearly does make a big difference, although the rest is somewhat important. I wonder how it would be with that nozzle on SuperCAP (although it would require a 1.5" threaded connection).
Quote:
| Originally Posted by Ben
I would suggest reading the part I quoted twice already. I have pointed that part out enough and that is one major reason I tell you to read my posts. It explicitly states that force determines flow. If you do not quote what I quoted, I will assume that you have not read it. As for why I can't give you an equation, well, if you don't know partial differential equations, there's not much you can do with N-S. I've said they were complicated...
|
It regurgitates F=ma. I might not be able to understand any of the math (and looking through the links, I don't), but I'm sure Jo will.
Quote:
| Originally Posted by Ben
Read my post please. I directly addressed this.
|
Hadn't gotten to your post.
Thanks for writing all that out with one hand. Hope this post isn't way too much, feel free the redundant parts or do whatever.
04-27-2007, 10:42 PM
