Spear Speed at Depth

[Dangerous]

There was a thread about little fuzzy creatures and shaft speed at depth. The question was, is there any hydraulic reason for apparent shaft loss of speed at depth.

Just saw this and maybe there is a loss of zip at depth, due to lower temp.

http://www.freediver.net/freedivelis...faq_bands.html


Dang

[KJNDIVER]

There is no hydrodynamic reason for loss of shaft speed at depth with a banded gun. Now a pnuematic gun is a different story. The deeper the depth, the less power you will have due to compressibility of air. The only reason for a a difference in shaft under water from a banded gun would be the deterioration of the bands and having less power to propel the shaft through the water.

[yonfuki]

I don't know if this makes a difference, but American bands are hollow thus having a "compressable" air chamber inside the bands, whereas the Euro-style bands are solid. Now can any figure out the physics behind a hollow band at depth?

[Dangerous]

Absolutely,

The point in the discussion - see link, was that lower temperature decreases band performance (higher temperature, to a point, increases)

Quote:"How Temperature Affects Bands
Warming your bands will cause them to retract more. Unlike most every material in the world - elastomers in the stressed/stretched state, when heated, recoil. (Hold a rubber band up to your face when you stretch and relax it and notice when it gives off and sucks up heat). Your bands will have more elasticity, less soak off, and more recoil in warm vs. cold water. The equilibrium state for the molecular chains in a polymer is when they are all coiled up and bunched up and randomly oriented. When you stretch it, you decrease this disorder and align the chains. When heat is applied to the system, energy is available for the chains to go back to the way they want to be, and the rubber will recoil."

Dang

[one_lucky_shot]

I have a Rino Classic can that take these solid "Europe" bands? and if so what is better the solid or hollow?

[yonfuki]

Actually rubber becomes resistant to elasticity with decreased temperature, and therefore would actually want to return to it's coiled form even more. However, anytime you strech or bend anything it will create its own heat due to molecular friction and the more heat added to any given elastomer will cause it to slowly loose it's "memory" (try leaving a rubber band out on the deck of your boat and then try to strech it). Eventually, the elastomers become exhausted from the frictional heat and the molecular bonds begin to degrade, thus the band breaks. Sucks, doesn't it?

[KJNDIVER]

On the hollow and solid bands; if there is a difference in performance, it is not due to the hollow core that is "compressible". The band gets its power from from being stretched, and wont lose power to the 'air' trapped in the band. The spring constant of the band should stay the same, no matter the depth. I say should, because I have not spent hours researching this or anything and just using good sense. If the solid band has more power it is because there is more material to stretch and thus a greater spring constant, thus more power.

As for the water temp, I dont think it would make a whole lot of difference unless make one dive down in the keys in 80 degree water and you next dive up in the North Sea with 40 degree water. And I woould have to agree with Yonfuki, the colder water would make the material contract more, and the warmer water would make them easier to stretch.

[Dogmatrix]

Would the increased density of the "deeper" water also affect the band dynamics? ect?
just a thought... I am no engineer.

Rob

[Murat]

I don't know about elastic properties, but....


At deeper water, the pressure is much more. This means greater forces will act on spear. I foget its formula but i think it was F=Pressure*Surface Area. But this is not gonna effect the flight of the shaft, coz spear has symetircal shape. So the total forces acting on the shaft is zero. I am thinking about how cold effect the travel of spear but i can't remmember anything now.

[Murat]

Originally posted by Dogmatrix
Would the increased density of the "deeper" water also affect the band dynamics? ect?
just a thought... I am no engineer.

Rob

I am not chemist or physic engineer but i don't think water density is related with depth.

[Dogmatrix]

Sorry about that murat... I used the wrong word, was thinking about the effect that increased pressure might have on the rubbers contractile qualities at depth. not sure if there is any effect or not.

Wrote that post very late at night... just before bed
wasn't thinking very well

Rob

[KJNDIVER]

At deeper water, the pressure is much more. This means greater forces will act on spear. I foget its formula but i think it was F=Pressure*Surface Area. But this is not gonna effect the flight of the shaft, coz spear has symetircal shape. So the total forces acting on the shaft is zero. I am thinking about how cold effect the travel of spear but i can't remmember anything now.

You are correct with all those statements except why the total forces are zero. It is not because of symetrical shape, but because the object is fully submerged in the water and the greater forces present act everywhere on the shaft and are equal in magnitude and opposite in direction. If you have an object that is subjected to 100 lbs of force in this direction --->, and also 100lbs of force in the opposite direction <---, you have increased the pressure forces, but the object will not go anywhere because everything is balanced. So the added forces at depth counteract each other as they act in opposing directions. It is the added forces introduced by the bands in a specified direction that make the system unbalanced and propels the shaft forward.

I am not chemist or physic engineer but i don't think water density is related with depth.

This is not entirely accurate, but for all practical purposes this is true. Water density at 10,000 ft would be greater than water density at the surface.

But :
1. No way in hell are you goin to dive to that depth to be affected by the change.

2. The change would not be all that significant, relatively speaking it would next to nothing.

I am thinking about how cold effect the travel of spear but i can't remmember anything now.

The cold in no way what so ever will effect the travel of the spear thru the water.

I used the wrong word, was thinking about the effect that increased pressure might have on the rubbers contractile qualities at depth. not sure if there is any effect or not.

Depth and pressure do not affect BAND guns in any measureable amount. The band gets its power from being stretched and is not effected by outside pressure. The most important things that effect the power are the spring constant coefficient, and the distance being stretched. ( 1/2 * k * x squared) Where k is the spring contant and x is the distance stretched. The water temperature may affect the spring properties ( contractile quality) but if anything, it will make the band have more power due to the increased tendency to contract more. Heat causes things to expand, cold causes things to contract, basic physics. But that would only occur for an extended duration of exposure at low temps. The band is at ambient temperature while on the surface, surface water temp will not be all that different from air, relatively speaking. And the relatively short exposure times while at depth should not have enough of an effect to cause any changes unless you are diving in very cold water throughout the water column from surface to depth. i.e.: North Sea, or Great Lakes in Febuary.




Now, everything yall have stated does indeed effect pneumatic guns. These guns lose their power as the depth increases, but banded guns are not affected.

[Murat]

Originally posted by KJNDIVER
You are correct with all those statements except why the total forces are zero. It is not because of symetrical shape, but because the object is fully submerged in the water and the greater forces present act everywhere on the shaft and are equal in magnitude and opposite in direction. If you have an object that is subjected to 100 lbs of force in this direction --->, and also 100lbs of force in the opposite direction <---, you have increased the pressure forces, but the object will not go anywhere because everything is balanced. So the added forces at depth counteract each other as they act in opposing directions. It is the added forces introduced by the bands in a specified direction that make the system unbalanced and propels the shaft forward.



This is not entirely accurate, but for all practical purposes this is true. Water density at 10,000 ft would be greater than water density at the surface.

well,

Lets say we apply 100 newton force from both left and right to the POINT (notice that it has symetrical shape)
---->.<----- In this case total force is zero coz directions are opposite and object is SYMETRIC. But this rule does not hold for non-symetric objects. In non-symetric objects you may not find exact oppiste point to apply force and cancel the other force. Got it?. At leat this is what i know, of course i am not Einstein.

For water density: If we assume sea (water) structure is same everywhere. I mean it has homogeneous salt distribution or whatever. It should have the same density. Do you remmember basic explanations from chemistry? density= mass/volume, so 1 GLASS of water desity is same with 1 tank of water??? doesn't matter if glass has depth of 15cm but tank has 3 meter.

I can't remmember now why deeper whater has more pressure on objects but its not realeted with density as i remmember. Unfortunatelly i remove my physic book to somewhere else, couldn't find it.

[Dogmatrix]

OK so no change in density... water resists compression pretty well yeah? I guess salinity differences might affect this but not a factor where homogenous.

Murat perhaps deeper water might increase in pressure due to all the water on top of it.

KJNDIVER some good points there!

Rob

[KJNDIVER]

Lets say we apply 100 newton force from both left and right to the POINT (notice that it has symetrical shape)
---->.<----- In this case total force is zero coz directions are opposite and object is SYMETRIC. But this rule does not hold for non-symetric objects. In non-symetric objects you may not find exact oppiste point to apply force and cancel the other force.

This total force is zero, but it is not because of the symetry of the point or shaft. It is because the object is fully submerged and pressure forces act everywhere, equal and opposite in direction. In some instances, the shape of the object does come into play, but not here.

Now, if the tip were not symetrical, if it had some imperfection that favored one side or the other, the shaft would not fly straight or may be affected by speed. But this would have nothing to do with depth, as it would occur at 3 m just as it would at 30 m.

But this rule does not hold for non-symetric objects. In non-symetric objects you may not find exact oppiste point to apply force and cancel the other force. Got it?.

If this were true, it would create an unbalanced system. The sysem would no longer be static (not moving) but it would instead be dynamic (moving). So what you are saying is the water would create a force on its own to propel the shaft in a certain direction (the direction that is not counteracted). Does the shaft push in one direction or the other while you are underwater? Although it may not be what you meant to say,this is what you are saying.

At leat this is what i know, of course i am not Einstein. Do you remmember basic explanations from chemistry?

Well, I am not Einstein either, but I am a Mechanical and Industrial Engineering student that has currently completed 2 courses in Fluid Dynamics, which last time I checked, is a bit more advanced than basic chemistry.

density= mass/volume, so 1 GLASS of water desity is same with 1 tank of water??? doesn't matter if glass has depth of 15cm but tank has 3 meter.

I agree with you. At those differences it has no change in density. But if you reread my post, I mentioned difference of surface water, and 10,000 ft (3048 m). And even at those great differences, the density change is so insignificant, that it can be disregarded. Like I said, for ALL PRACTICAL PURPOSES. What part of that is hard to understand?




Got it?