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The physics of force transmission in air versus water pumping

“Whereas a rock is not compressible, you are.
When you’re submerged, water pressure squeezes in on you and reduces your volume. This increases your density.

(Be careful when swimming—at shallow depths you may still be less dense than water and be buoyed to the surface without effort, but at greater depths you may be pressed to a density greater than water and you’ll have to swim to the
surface.)

http://www.gdn.edu/PT_Faculty/cmcke…_6thru8prob.htm

Originally Posted by marinera
I doubt the internal force of erection will sum with what you read on the gauge, since the gauge is measuring the negative pressure acting on the matter inside the cylinder - that would be, the force which is directed toward the walls of the cylinder.

Measuring by length. Not girth. Girth is compressed. Length is enhanced for the same reason and for the buoyancy - although this can depend by the depth of the cylinder. The total expansion, at any given pressure, will be lower in water, though. It would be like pressing with your hands on your whole shaft and a clamp at the base - you can have a bit more length, but the reduced girth will more than compensate this.

Let me ask one thing : it is easier for you to walk when submerged in water or in air? That’s it. You don’t want to transmit force, you want to subtract force.


Did you ever take physics? Its not possible to cause expansion in length AND compression in girth in the same fluid medium, as the force is applied uniformly. Buoyancy would generate mere ounces of force as opposed to lbs of force generated by the vacuum. Plus, this would only work in the direction opposite of gravity or “up”. I hold my tube downward, so buoyancy should shorten my length.

I do think the surface area in the force vector direction will give a force proportional to that surface area, but not absolutely sure. It would still mean that all surfaces would move toward expansion in a negative pressure environment, just that the force vector direction with the greater surface area would have greater net forces effecting it. This is why I believe girth tends to expand more in pumping than length. The “pull” for length would just be generated by the surface area perpendicular to the axial vector, which would be the glans primarily. Which may explain why blisters tends to form on the tip of the glands.


Last edited by sparkyx : 08-19-2011 at .

Originally Posted by marinera

You don’t want to transmit force, you want to subtract force.

Transmission includes both positive and negative. Its like a metal pole where you hold one end, and I hold the other. Whether I push or pull, the pole TRANSMITS the force. Water does that without loss of energy, air will absorb some of it. This is why when pumping up a bicycle tire, the pump gets hot. The heat occurs from air absorbing some of the energy as it is compressed, and it generates heat. Because air will absorb energy, less is TRANSMITTED to the penis, in both compression and expansion.


Last edited by sparkyx : 08-19-2011 at .

Originally Posted by sparkyx
Did you ever take physics? Its not possible to cause expansion in length AND compression in girth in the same fluid medium, as the force is applied uniformly. ….


Hey listen Sparkyx, did YOU ever take physics?

If you are submerged in water, the water pushes you up. Your body floats in water. But if you want to move to east or west, that is harder. This is due to density of water, it’s weight. Got it? Is that true or not? Now, apply that to your penis when submerged in water : up = length; east and west = girth.

Originally Posted by sparkyx
Did you ever take physics? Its not possible to cause expansion in length AND compression in girth in the same fluid medium, as the force is applied uniformly. ….


Water pressure is the result of the weight of all the water above pushing down on the water below. As you go deeper into a body of water, there is more water above, and therefore a greater weight pushing down.

http://www.ehow.com/about_6556057_w…ase-depth_.html

Lets say you have a box that is filled with water. One of the side of the box contains retangular hatch. How much torques does the water exert on the hatch?

Pressure depends on depth, so the force is not uniform. You well have to integrate over the surface, breaking it up into strips of constant force at constant depth.
http://www.physicsforums.com/showthread.php?t=74731

Originally Posted by dtwarren1942

The density of a material is defined as the mass per unit of volume. It’s a measure of how tightly the atoms of a material are packed. It has nothing to do with the hardness of the material. For example, water has a density of 1000 where as air has a density of 1.29. Accordingly, if you apply 1000/1.29 or 775.2 units of pressure/ force against a closed container of air, the density of the air at this pressure/force will equal the same density as water. In other words, air becomes just as incompressable as water when 775 units of pressure/force are applied against the air. (Pressure in this scenario is based on Pascals - newtons/area. The calculation is way above my pay grade)

Phew, I need a brake.

Interesting, but irrelevant; we are removing density by increasing vacuum/relative pressure. I don’t want to ADD 775 units of pressure to my cylinder when I’m pumping.

I think your calculation may be fundamentally flawed as pertains to units. Pascals are a truly tiny unit of pressure. Atmospheric pressure is 100,000 Pascals (100 kPa: see http://en.wikipedia.org/wiki/Ambient_pressure) , so an increase to 100,775 is really not much: not even one percent. The atmosphere varies by that all the time, and I can’t remember in my lifetime when air was as dense as water as a natural occurance.

I think your calculation may be fundamentally flawed in another way. The density of water and the density of air are 1000 and 1.29 respectively at the same pressure level. Your calculation of 1000/1.29 to achieve the same density is yielding a percentage increase in pressure, not an absolute value. 1.29 times 775.2 equals 1000; not 1.29 subjected to an increase in pressure of 775.2 over baseline thus yielding 1000.

So, I would have to subject my dick to 775.2 times atmospheric pressure for it to be surrounded by air which is as dense as water. Hmmm, I think I’ll pass.

Plus, I don’t know that compressibility and density are necessarily linearly correlated, and things of similar density may not have the same compressibility.


For Lampwick, becoming hung like a donkey was the result of a total commitment.

Originally Posted by sparkyx
….. Buoyancy would generate mere ounces of force as opposed to lbs of force generated by the vacuum. ..


Really? How muc water fits in your pump? Doing a quick, rounded calc, I think a 9”x2,5” diameter tube can contain roughly 2 lbs of water. This is the buoyancy force. Not so low.
(I’m not sure the buoyancy force is already included in what you read on the gauge, though. If so, even without vacuum, when you insert your penis inside the tube you should read a variation in the gauge.).

Originally Posted by sparkyx
….. Plus, this would only work in the direction opposite of gravity or “up”. I hold my tube downward, so buoyancy should shorten my length.
….


Given the shape of your penis and the fact that it is held by the weight of your body, it would penetrate as well in water and the compressive force on the sides would squeeze it as well, so the net effect hardly can be a shortening effect - than you add the force of the vacuum, and you see why your penis is lenghtened.

Out of curiosity : when you push your pump perpendicular to your body, how the hell you can read the length on the tube?

Originally Posted by sparkyx
….. This is why I believe girth tends to expand more in pumping than length. The “pull” for length would just be generated by the surface area perpendicular to the axial vector, which would be the glans primarily. Which may explain why blisters tends to form on the tip of the glands.


Blisters are caused mainly by the friction against walls. Since most of penises are slighly curved, expecially when extremely engorged, one side of the glans will have friction against the cylinder, causing the blister.

A nice effect, from a theorical point of view, of the higher density of water respect to air, is that it should mantain the penis more straight when over engorged. I’d check if this really happen, because it would suggest that water pumping is safer; but it also would suggest that water refrain expansion, compared to air, so at any given level of vacuum the penis would expand less.

Sparkyx, I think you married a thesis here. How much is not the point.

Let’s go back to the example of the brakes. If there is air, brakes don’t work fine. Why? Tell me please.

Originally Posted by marinera
Really? How muc water fits in your pump? Doing a quick, rounded calc, I think a 9”x2,5” diameter tube can contain roughly 2 lbs of water. This is the buoyancy force. Not so low.


You have endless energy! Amazing, really.

Buoyancy is the density DIFFERENTIAL, and your penis being mostly fluid density is probably very close to water. Remember if a human being exhales all his air, he will sink in salt water, which is more dense than the fresh water used in pumping. Toss your steak for dinner in the pool and let me know if it floats or sinks to the bottom.

The buoyant force of a floating object
F = (Vw)

F = force,
V = volume of fluid displaced in ft3
w = specific weight of sea water, 64lbs/ft3

But this assumes displacing water ( sea water in this case) is with air density as it appears you are. So for this example more accurately would be;

F=V (w1-w2)
with w2 being the specific wt of the penis which is probably very close to sea waters value. If the penis is denser than water, it will result in a negative force and would sink. Again, toss your steak into a pool and see what happens.


Last edited by sparkyx : 08-19-2011 at .

Originally Posted by marinera
Sparkyx, I think you married a thesis here. How much is not the point.

Let’s go back to the example of the brakes. If there is air, brakes don’t work fine. Why? Tell me please.

The air absorbs more energy (through compression) than fluid. This loss of TRANSMISSION, results in the braking power being too low to overcome the inertia of the vehicle. Force IS being transmitted, but not enough to stop the vehicle.

And yes I needed a year of Physics for my B.S degree.


Last edited by sparkyx : 08-19-2011 at .

Sparkyx, my friend, brakes won’t work with air inside because the fluid will expand more when pressed. Without air, fluid will not expand. So air allows expansion, fluid don’t allow expansion.

Now, $1,000 question : do you want your penis expands or not? If you don’t want it expands, put in fluid. If you want it expands, put in air.

Originally Posted by sparkyx

You have endless energy! Amazing, really.

….

Endless energy has nothing to do here. 1 decimeter cubic of water weighs about 2 lbs, right? Think simple.

Originally Posted by marinera
Sparkyx, my friend, brakes won’t work with air inside because the fluid will expand more when pressed. Without air, fluid will not expand. So air allows expansion, fluid don’t allow expansion.

Now, $1,000 question : do you want your penis expands or not? If you don’t want it expands, put in fluid. If you want it expands, put in air.

Not sure what you are saying here, but you are incorrect.

Force produced by pushing down on the brake peddle is transmitted very efficiently though the fluid medium without loss of energy.

When air gets in the line, the force then is used to compress the air and much of it is lost in that compression of the air, and doesn’t make it to the brake pads, resulting insufficient force to stop the car.

Listen, as always, its been fun, thanks.

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