Super Soaker Central

On friday afternoon, I began to think about the physics os water weaponry, as usual. Anyway, on thing that came into my head was, was that is CPS actually constant?
STOP READING HERE IF YOU CAN ANSWER MY QUESTION.

In my mind, the CPS bladder works in much the same way as a spring. According to Hook's Law, The distance that a spring is stretched is proportional to the force applied. If this is the case then CPS would not be constant, but Linear.

Please correct me if I am horribly wrong.

Storm

CPS isn't constant; you can even get more constant pressure with air pressure water guns. But apparently Hooke's Law isn't a very good model, since rubber is a viscous fluid that tries to return to its original shape in weird ways. Ask Ben.

Thanks Silence, keep posting to see if we can get Bens attention. ( apoligies for spamming)

Storm

Elastic pressure isn't truly constant no. However, it's not linear either. While you're right in that the force increases as the bladder expands, the surface area also increases.

Now since

pressure = force / area

Then as the force increases, the area it acts on increases with it, producing constant pressure. Or at least it would (I think) if rubber's elastic behaviour was linear. As has been said, it's not, and I think that non-linearity is the cause of deviation from constant pressure.

Disclaimer: I haven't done the maths.

But as the bladder expands, more force is applied. When a user pulls the trigger the pressure will decrease in a predictable fashion, that is the pressure will not ?fluxxuate?. That is what I meant by linear. The pressure drop would be represented on a graph with a line.

Storm

It will not be linear. Rubber is a non-Hookean material so it can not be approximated by Hooke's law. The only way anything resembling a line in pressure vs. time would happen is if the tubing was used like rubber bands to push a piston.

I don't know if you've read the Latex rubber tubing page on the website but I've explained this there:

How it works

Latex tubing achieves relatively constant pressure though a simple mechanism. The tubing itself expands, first in diameter and then in length. It DOES NOT expand completely in length and then in diameter as some people believe before using it.

By expanding, the force applied on the water increases. However, at the same time the inner surface area of tubing is increasing. The force and surface area increase in nearly linear proportion. Therefore, by the basic equation pressure = force / area, if the force and area are increasing in approximately linear proportion, the pressure is approximately constant.

Following the basic equation pressure = force / area, if we assume that force is related to the wall thickness and we know that the area is related to the ID, then two relationships become clear. The thicker the tube, the higher the operating pressure. The smaller the internal diameter, the higher the operating pressure.

What cantab said is my best understanding of it. I'll explain what he said in a different way. Think of the tubing as a bunch of rubber bands. When expanded, each rubber band applies the same amount of force as the others. If we approximate the shape of a bladder with a cylinder, then as more water is pumped in the length of the cylinder increases. As the length increases, so does the "number of rubber bands". The total force applied increases. However, the surface area the force is applied to increases as well (the surface area of a cylinder), so the pressure (pressure = force / area) does not increase or increase much.

For the sake of illustration, let's say that force = constant*length and surface area = pi*diameter*length.

...therefore pressure = constant*length / pi*diameter*length... which simplifies to pressure = constant / pi*diameter, which is independent of the length.

Let me reiterate that what I did above is only an approximation meant to illustrate that the pressure build up is small because the surface area increases as the total applied force increases.

The real way you'd figure this out with math would be through a surface integral, but I know little about the relationship between force and distance in rubber so I've never figured that out. It would be rough at best anyway approximating with a cylinder, which is what I'd have to do unless I figure out a series of equations to describe the shape of the bladder accurately.

As Silence said too, in real life rubber CPS isn't truly constant... I've figured out that the pressure drops by about 25% over the duration of the shot. You can get effective 0% drop with a regulator and less drop than 25% with a LPD air pressure water gun. If the pressure's high enough, the fact that you get diminishing returns from higher pressures will make the performance even more constant to the point where you won't notice the drop easily, as was the case in my test LPD water gun.

Ah, I understand now. Thankyou for all your responses.

Storm

If the Pressure was "Constant" there would not have to be any pumping after the pressure was built up once. in a literal point of View that is.

That's not correct for two reasons. Don't think of a pump as building up pressure. Think of it as transferring energy to the system which is stored as a pressure. And you'd eventually run out of water so you'd have to move some more into the pressurized system.

Darn I forgot that water guns shoot water.. :s lol



Magic still count then?