Band rubber speed?

Jeff,
The speed is one factor, but in my opinion the main element is energy: the loaded bands have an energy determined by the force used to pull them, and by how far they were pulled. When released, "a good amount" of that energy is transferred to the shaft (and some is lost). The energy that is transferred to the shaft (most of it) will translate into speed, and is depended on the mass of the shaft: the heavier the shaft, the lower the speed so that the compound result would be the same energy.

So a heavier shaft will travel slower, but with the same energy as a lighter shaft.

As the shaft moves through water, it will loose a good part of that energy due to friction/collision with the water (so the shaft energy will be transferred to the water by moving or heating the water). The energy loss is dependent of 2 factors: the section of the shaft, and the SQUARE of the speed - so the speed is a more important factor in the energy loss than the section of the shaft. That explains why a heavier shaft will better retain the energy on the longer shots compared to a thinner shaft, so that at the impact with the target a heavier shaft will have a better chance to penetrate.

Band speed? Not so important in the range we're talking about (3 or 4 bands for most of our guns). I can say that adding a 4th 5/8 bands on my riffe did make a difference.

When would be the band speed become irrelevant? At the point where the mass of the shaft becomes irrelevant in the total energy... So if you would shoot a 1mm carbon shaft that doesn't flex and weighs about 1 gram, then band speed would be very important.

My 2 cents, looking for change..

Stefan

my first thought is that the longer band will contract at the same rate of a smaller band if they are stretched to proportionally the same difference...the limiting factor in the "band response time" will be a factor of how much it is stretched from it's original condition.

i think, that a 4" band stretched to 16" will "snap" faster, than a 5" band stretched to 15". since one is 1:4 and the other 1:3

the rate of return to the original shape would be something like the cross-section x the stretch factor

Dry fire the power band in a protected tube with out a shaft, record the snap of the sear and a Ti wish bone striking a end cap. Record the sound and look at the spikes on sound program and you get the speed .
Dan will ban me if I ever talked about the stupid unsafe Shite Ive done out of water.
i used to usea re-geared supper 8 for underwater tests when John Warren played around doing unsafe things. Most of the data I film and data I collected was lost in a fire in 1990 and at my age I have forgotten
most of the late night stuff. I have come to the conclusion the 4 of Mori's Mean Green are all the power
I will ever hunt with.

Cheers, Don

Quote from Tin Man

T
My thinking goes like this. Suppose I am pushing my kid in a wagon. No matter how big or strong I am, (or how big or small my kid) that wagon will never go faster than I can run. I can get three friends to help me, and we might get the wagon up to to speed sooner, but it will still never go faster than we can run. All the extra strength of my three friends is wasted, because the limit is how fast we can run.

This would relate to spearbands if you had a unit of measurement (15 feet for example) that you and your three friends had to get that wagon up to maximum speed. (your kid would be having a blast with this)
As long as the spearband can reach its max speed before it releases from the shaft, you're getting the most out of it's potential energy. ?? I think.

What if your running as fast as you can and give him a push?
Then he will be going faster than you can run. I'm off the original topic.

Hey Jeff can you post what I PMd you the other day. And although that video shows us that a rubber band contracts at one speed the chemical composition of each band will change its properties so that number cant be used for spearfishing bands.

Quote from Tin Man

How fast can a rubber speargun band contract, and what (if any) limitations would that place on gun design?

My thinking goes like this. Suppose I am pushing my kid in a wagon. No matter how big or strong I am, (or how big or small my kid) that wagon will never go faster than I can run. I can get three friends to help me, and we might get the wagon up to to speed sooner, but it will still never go faster than we can run. All the extra strength of my three friends is wasted, because the limit is how fast we can run.

So I was wondering. There must be some maximum speed at which a speargun band can contract. So at some point, adding more bands to a gun might not increase the shaft speed at all. Or, moving in the opposite direction, using a lighter shaft might not result in any additional speed.

Anyone ever seen info on how fast a rubber band can contract?

Display More

Short answer is speed of spear will never be faster or equal to band speed. The mass displaced of a band contracting alone is almost negligible compared to pushing a spear. F=ma, F is constant on a stretched band at a given distance. On a band alone m is very very small so a is almost infinte, compared to a with spear of mass m. Integrate over the distance of the band pull and velocity will be almost infinite with a band alone.

Now add some reality i.e. drag, friction, cavitation, etc and band speed will still be super fast.

On a rubber band F=kx, where k is a spring constant, x is the distance displaced, so a band exerts force as long as it is stretched, F>0 always until it reaches its unstretched length. On human running F=ma, you can't push your kid any faster because your top speed is constant hence a=0 then F=0.

The second your a=0 and velocity is max is like a band stopping at its length. Your analogy applies up to this point, you are like a band. Your velocity is limited by your strength and mass, a bands velocity is limited by how far it is stretched, mass, and its spring constant.

Distanced stretched and spring constant are "constant" on a spear gun. So again with a lighter mass you will get a higher velocity.

If shafts didn't whip I'm sure everybody would be using smaller diameter shafts.

Davie really cleared it up for me, thanks buddy.

I think the practical application for this kind of thought would be to design a device that used the bands from 350% stretch and released the spear at 50% as opposed to the current standard of 300% -0%.

this would show us the effect of the higher order band speed only without the dead band at the end of the shot...if the spear were the same length and the cross section with the same track, the spear from the test rig should exit the track sooner and with more energy imparted, not from the bands snapping faster , per se, but from using the more efficient part of their power release cycle.

Quote from LunkerBuster

I think the practical application for this kind of thought would be to design a device that used the bands from 350% stretch and released the spear at 50% as opposed to the current standard of 300% -0%.

Roller guns and pneumatics do something like this.

I think it's generally accepted that the lighter shaft will "bleed off" energy quicker from water resistance. What surprized me, however, was that the penetration difference between the 9/32 and the 5/16 shaft was still very large at close range. I was thinking of the target penetration as a crude measurement of relative "stopping power", sort of like the discussion by deer hunters about a .243 compared to a .44 caliber bullet. I'm just not sure if the density of the water just causes much more dissipation of energy, and hence shaft speed (manifested in less target penetration) than we thought, or if the band speed represents a limit we didn't consider before. :confused2:

Energy is conserved, if you put x amount of energy into loading a gun then x amount of energy will be expended shooting the gun. Doesn't matter what you shoot from a speargun it will have the same kinetic energy.

However if the same amount of Kinetic energy is put into 2 objects the heavier object will have more momentum.

There is a wall, it is when the force exerted by the bands equals the opposing forces such as drag, water resistance, and friction. This will cause the shaft not to accelerate. The opposing force is pretty small compared to the force of the bands, even near the end, so more bands will always equal higher speeds if the shaft can handle it.

Energy is only conserved in the absence of friction. In this scenario there is both friction from the spear on the track and the drag force of moving through the water. Because of this the energy stored in the band will never equal the kinetic energy of the spear.

Jeff the spring rate of the band is definitely not linear but that is a completely different problem.

I think some of the confusion here might be with the use of KE. In this scenario the conservation of mechanical energy doesnt apply because of the force of friction. Instead what is most applicable are the equations: sum(F)=ma. and p=mv. These are Newtons second law of motion and the equation for momentum (p). Also Force(drag)=1/2(p*(v^2)(C)(A) is critical. What is most important in this formula is the fact that the drag force is proportional to the velocity squared. In your test for penetration, using the same bands shooting different mass spears your acceleration will different. the lighter spear will, like you said have a higher velocity.

From the instant the bands no longer apply a force on the spear the only force acting on them will be the the force of drag. For the the cross sectional area for your two bands differs by about .05". so in your drag equation you have .5(v^2)(.5)(.49)=Fdrag(5/16). For the 9/32" shaft you have .5(v^2)(.5)(.44). These simplify to (.12)(v^2) and (.11)(v^2) respectively.

To summarize, the (9/32) shaft has a smaller mass, a higher drag force and a higher initial velocity.
the (5/16) shaft has more mass, less drag and a lower initial velocity.

Next comes newtons second law. sum(F)=ma=Fdrag
after the bands have released the shaft the only force on the spear is drag. just to simplify this without knowing the actual mass difference of the shafts: mass of(5/16)= 4. Mass of (9/32)=1. NOTE: this is an exaggeration to explain the difference. As a result the initial velocity of the smaller shaft is twice that of the heavier shaft. and experiences 4 times the drag.

Now according to Newtons second law sumF=ma the accel for (9/32)= (.11)(4)=.44 and for the (5/16) = (.12)(1)/4 =.03 creating a ratio of about 15:1. Because of this in very short distances the lighter shaft slows down significantly. This will change at longer ranges because the drag force will decrease as the velocity decrease. There will be a point some distance from the barrel that both will penetrate the same distance.

So even though this is a significant exaggeration the lighter shaft slows down a huge amount in the first few feet and because momentum = mv and the the shaft is lighter but not enough faster to compensate. Now because the heavier shaft has higher momentum it then hits the fish/target with more punch.

I really hope this makes sense and that I didnt loose anyone with the simplifications or the notation. Jeff this should explain your experiment. This also explains why heavier shafts can travel farther. Now I need to go do my actual physics homework.

Ok for the first part i meant cross sectional area of the spear and your right those numbers arent correct. they should be (.076 for the 5/16) and .062 for the 9/32. Sorry about that I wasnt very careful when I was transferring this from paper.

For the sumF=ma the 9/32 spear was assigned the mass of 1. so the .44 number is the force divided by 1. 4 is the the velocity (2) squared not the mass.

This Is pretty exaggerated and in practical application the difference in mass is not nearly what I described. I simply used those numbers to simplify the calculations. In hind sight probably a bad idea.

But throughout the entire shot the 9/32 spear looses power. If you were to graph the velocity vs time for both shafts in water you would see that the velocity of the two shafts quickly become very similar and at this point which isnt very long after the shot the lighter shaft is now at a disadvantage because is has less momentum. This is further compounded when striking the target. Impulse (J) = change in momentum/ time it takes. And because the target exerts about the same force on both both spears the lighter shaft slows down faster and penetrates less.

So while none of these factors seem that large individually they add up and result in penetration.

As for why would anyone use them if it seems like they have no advantage my thinking on this would be that even though they have less penetration it is usually enough to penetrate most fish that would be shot and there is less recoil making them a little more accurate for smaller guns. But when lots of penetration is really needed and recoil can be offset like in large tuna guns I would be very surprised to see a 9/32 in shaft used by anyone. Ideally you could make a 9/32 shaft have the same mass as a larger spear and then you would see then used more.

This is actually related to your other question of the rate of band contraction. More bands doesnt always equal more velocity and more velocity isnt always better. Shooting a heavier spear at the same velocity gives you a better punch than a really fast light spear. And even for objects that are very hydrodynamic water does have a significant effect and the only way to really "combat" this is with added mass.

This is so much more interesting than the projectile motion we do in physics.