Rubber Tube Contraction Velocity

How do you propel the projectile again? Is the pocket ejecting the projectile? How big is the pocket? Is it ambient temperature?

Target is what? Where?

Give up smoking before it is too late!

Energy input minus losses equals stored energy. 50# in both cases.

Energy Imparted into projectile minus losses equals distance traveled.

If both cases are equal.

Can you PM some inspiration my way?

Correct. GA

Actually, this would be correct if the tubing was weightless. There is more tubing to accelerate in the longer one.

Agreed. I was going to add the other energy loss of natural rubber not really being very elastic but the tubing mass difference will probably be the greater burden.

Calculate the potential energy from the work done on each tube.

For a given projectile mass, if the potential energy is the same in every case, so will be the projectile's 'muzzle' velocity.

Rixter is correct. Energy = Force x distance, but the force changes as the rubber tube contracts. If we assume force/distance is linear, the average force over the given distance is 25 lbs for both cases. The potential energy of the longer one is twice as large as the potential energy of the shorter one when fully stretched since the distance is twice as much and the average force is the same. (If it's not linear, you have to integrate f(x)dx.) Assuming all of the potential energy is converted to kinetic energy, the ball from the long one will have twice the kinetic energy (0.5 mv^2) when it is released than the short one. They have the same mass, so the velocity of the faster one is sqrt(2) x velocity of the slower one.

I can easily see the info in the chart,but I think it is referring to the same

bands,pulled to different lengths, not to 2 bands of different lengths at rest,with the

same pull weight at the point of release.

The info in the chart is common sense;if you pull the band further,the projectile will

go further,but that is not exactly the situation I am referring to.

The pull rate will decrease with decreasing length,so it does not maintain the same

thrust for the entire stroke.

Thanks for the good link though.

In another lifetime, I made several spear-guns for scuba fishing, using such tubes.

I recall the grade being called 'surgical tubing'.

Given that the projectile will immediately tend to succumb to gravity when it is released, it stands tio reason that the longer the contact and/ push of the tubing, the further the projectile will travel. I am not saying it will travel faster...only further.

Anyhow, it all gets distorted under water, so maybe I'm just blowing bubbles here!

I am looking at it like this:

The shorter tube expends all of it's energy(50 pounds) in 6 inches,the longer tube

expends all of it's energy(50 pounds) in 12 inches.

The force decreases with every inch of travel.

The shorter tube would impart a faster rate of acceleration to the projectile,thus

actually leaving the pocket of the tube sooner and arriving at the target sooner.

The maximum speed should be the same,but the shorter tube should get the

projectile to the target sooner,having accelerated the projectile at a faster rate.

Am I looking at this all wrong?

Or have I had too much "Georgia Moon"?

("Less than 30 days old" 100+proof.)

Sorry - Hooke's law won't help you with rubber. So, finish that glass, re-frame the question with steel springs, then look up Hooke's law...

Wrong. The stored energy is not 50 pounds; it is 50 pounds times the length. Double the length = double the energy.

Try a slingshot with 2 inches of tube length, and see what you get.

Sorry, I am poor and dull at English units, fractions and irregular proportions are crazy stuff and quite labor intensive to precious neurons.

The fog is lifting now from last night's binge.

That was a silly simple question.

Note to self:

Never mix corn whiskey with engineering forum.

Are these tubes vented or are both ends sealed?

Can these rubber tubes be used for distillation?

*Hic*

Down of the farm, if we put 50# of force into any two objects and got twice as much energy out of one of them than we did the other, just because it was longer, we would have called it a miracle.

Maybe I should never have left there for the big city.

Your reasoning is flawed. Force is not energy.

You need 50 lbs of force to lift a 50 lb rock to a height of ten feet. You also need 50 lbs of force to lift another 50 lb rock to a height of 20 feet. Which one has the most stored potential energy? When they are dropped, essentially all of the potential energy becomes kinetic energy.

Convervation of energy:

(mgh + 0.5mv^2) at the top equals (mgh + 0.5 mv^2) at the bottom. (i.e., total energy at the top equals total energy at the bottom)

The same principle applies with rubber bands, except potential energy is force x distance. Force is variable with a rubber band. Gravitational force is essentially constant.

This may call for a practical experiment.

Nope. One cannot raise a 50 lb rock to any height with only 50 lbs of force. You need more.

Now, we're confusing work with force.

This may force me to buy some rubber tubing and work this out for myself.

OK, 50lb. (wt) rock, 50lb 1oz. forceto lift. Everyone is basically hitting right near the nail.

W=∫F·ds (where d represents in the infinitesimal and s represents displacement.

We have case 1 where (short one, assume l₀=10 "), is drawn with an approximate force equation of F=ks where by definition, s= l-l₀, and s(max)= 0.5(10")=5", so that W₁=∫₀⁵ksds =1/2k(5)². We have that k=50/5=10#/". Thus W₁= 125 in.-lb.

In case 2, we have a similar integral to evaluate for work input: (in this case I₀=20"), so that s(max)=10", and with k=50/10, or 5#/". W₂ = 0.5(5)(100)=250 in.-lb.

100% conversion of Potential to Kinetic energy will be 0.5mv². and

v₁ =(125/0.5m)^1/2, while v₂ =(250/0.5m)^1/2

(the same mass projectile is used in each case), and the velocity ratio becomes

v₂/v₁ = (2)^1/2 by cancellation of factors. The long tube wins hands down.

GA.

The stupid system here tells me that I have voted this option too often, even though I have done it only once today so far.

The maximum velocity of the projectile will be achieved when it is released and loses

contact with the cup.

The cup will be slowed by wind resistance,and thus will not follow or propel the

projectile for the entire length.

Any remaining travel of the cup has no effect on the

projectile,it is merely follow through,and since the contraction force is becoming

smaller as the tube becomes shorter this will also tend to make the cup lag behind

the projectile.

So it appears to be that all of the instantaneous energy is delivered to the projectile

in a very short portion of the stroke.The rest of the energy is wasted.

A super high speed video of both scenarios would be nice, but I do not have access to one.

HTRN,

You are too old for homework, so what gives?

Curiosity.

I realize a bow and arrow are different, the arrow stays in contact for the full length

of the arrow,or a rifle where the expanding gas pushes the bullet for the entire length

of the barrel.

Position the release from the same point,so as not to give the shorter tube an

advantage by being closer to the target.

If you have a tube 1 meter long, in the relaxed state,stretched to 1.5 meters,with an

applied pull force of 50 pounds,versus a 2 meter tube,in the relaxed state, stretched

to 3 meters,with the same applied force, it will take twice as long for the long tube to

fully contract,(1 meter,vs 1/2 meter)thus requiring twice the time to deliver the full

energy that was put into it.

That means the projectile will accelerate at a slower rate than the shorter

tube,although they may reach the same velocity when leaving the pouch, but the

projectile in the longer tube will leave the pouch later.

IMHO,but of course,as always, I could be wrong.

No. You are wrong. Newton's second law of motion very simply states F=ma. By your two given conditions, both tubes are pulled back to an identical force and the objects to be accelerated have the same mass the initial acceleration must be the same. The shorter slingshot will accelerate for less time making for a slower a release velocity.

This is getting embarrassing.

Force x Distance = Work

If the force is applied against gravity or a spring or a rubber band, the work is stored up as potential energy. (Work done against frictional force is not stored as potential energy. It's converted to heat energy and lost forever. But I digress...)

For a rubber band,

Force x Distance = Potential Energy

If you double the distance with the same force... you double the potential energy.

It's that simple.

Potential energy is converted to kinetic energy when the object is dropped or the tension on the spring or rubber band is released.

If you don't believe me, dust off your Physics 101 book and read the chapter on Conservation of Energy.

1. I assume you are talking to me.

2. I never had a Physics 101 book.

3. I remember something about kinetic energy, but I thought they were talking about the state of Connecticut not the state of the energy.

Total work is greater with the long tube,I will not dispute that fact,but the total work

is not delivered to the projectile,and the applied force is not constant throughout the

whole stroke,it is decreasing as the tube becomes shorter,and the pouch is losing

acceleration.The projectile speed is also decreasing,but at a slower rate.

It is no longer in contact with the pouch.

The maximum acceleration is delivered at the instant of release,and the pouch lags

behind the projectile after this point,unable to impart any more energy into the

projectile.This energy is wasted.

Just as it requires more and more energy input into the band to stretch it

further, conversely,less and less energy is output as it contracts.

The shorter tube contracts to a relaxed state quicker,having only half the distance to

travel,thus imparting a greater "A" to the projectile at the instant of release.

A=the rate of change of velocity per unit of time.

If you were stretching the same relaxed length of tube twice as far,it would be a

different situation.

F = ma... The projectile is being accelerated the entire time it is in contact with the pouch because the pouch is exerting a positive force on the projectile the entire time. The force decreases as the stretch decreases, but it's still pushing the projectile forward. It's still accelerating the projectile forward.

The maximun acceleration is NOT delivered at the instant of release. The maximum acceleration is delivered when the force is at its maximum. The velocity is at its maximum the instant of release, because all of the acvailable potential energy has been converted to kinetic energy at this point. Once contact is lost acceleration ends. The projectile becomes a free-falling object with a positive velocity in whatever direction the slingshot pushed it.

More energy input = more energy output. Essentially all of the potential energy imparted to the projectile during the stretch is converted to kinetic energy dcuring the contraction.

This makes sense to me, at least.

Where is Del ? This scenario is akin to a bow string pull, but the archer is absent...

Ok.As you stated,the velocity is at it's maximum at the instant(or shortly thereafter),of release.

Now my question is:How does the pouch maintain contact with the projectile when the velocity of the pouch is decreasing at a faster rate than the projectile.

Wind resistance on the pouch surely is more than wind resistance on the projectile,

add to this the decreasing force (and decreasing velocity) from the contraction of the bands,and pouch lags far behind the projectile.

It cannot impart any more energy to the projectile.

IF the projectile has enough mass,(Say a bowling ball) it could absorb all of the energy from the full contraction of the bands.

In that case,I would agree with you answer.

Technically,I should have stated the mass of the projectile.

MY bad.

Inertia. You really need a refresher in Newtonian kinematics.

Inertia is a measure of mass.So where is you argument?

No, inertia is not mass. Mass is a scalar attribute. Inertia is a vector attribute. Inertia and energy (aka mass) must both be conserved. You really need some kinematics training.

Inertia is a resistance to change in acceleration.

Mass affects inertia,and effectively,are inseparable.

Semantics for the sake of argument have no value.

https://www.youtube.com/watch?v=fFFgWl6t-6c

Now you're saying that the draw back length is the same with different tubes?

Fine, whatever.

Please get your scenario straight that you want answered.

I may or may not be back to this chaotic thread.

No,that is not what I am saying.

I said in the original post,that each of the tubes were

stretched 50% of their relaxed length to get the same force at the end of the pull

stroke.

The pull force is not constant, but increases with length when pulled,and decreases

with length when released .

This means that the maximum force is applied to the projectile at the instant of

release,and decreases thereafter as the tube contracts.

Since the force is decreasing,and the projectile has already received the maximum

velocity at the instant of release,the projectile is free to leave the pouch and

continue on it's way,while the pouch lags behind as it decelerates due to friction in

the tubes and wind resistance,etc.

This energy is wasted and not delivered to the projectile.

Of course,the heavier the projectile the longer it takes to accelerate to max

speed,and the more energy it can absorb.

A shorter tube will return to the relaxed state quicker,because it has only half the

distance to travel from full stretch length.

Therefor,with the projectiles of the same weight,the shorter tube will deliver a faster

acceleration to the projectile( a greater A).

The longer tube may deliver the same final speed,but if released at the same instant,

the shorter tube will arrive at the target sooner for two reasons:

1. Faster acceleration,and

2. It is closer to the target if the fixed point of the tubes are the same.

Even if the release points are the same, the shorter tube will launch the projectile

first.

It only has to travel 1/2 the distance to deliver the full amount of power required to

launch the projectile.

You are mixing kinematic terms and concepts.

Since the force is decreasing,and the projectile has already received the maximumvelocity at the instant of release, NO! The acceleration is at a maximum at the moment of release. The velocity is at a minimum at the moment of release. The velocity is at a maximum when it leaves the slingshot. The acceleration is at a minimum immediately after leaving the slingshot. The velocity of a bullet is highest when it leaves the muzzle.

There are secondary concerns like; spring mass, spring efficiency, tensile strength of the water balloon, coefficient of friction through air, alignment of force and center of mass that should be considered once one grasps the fundamentals but you are confusing velocity and acceleration. These are the most fundamental terms of kinematics.

Be careful with that Georgia Moon. It appears to have a lingering effect.

Ok.I should have said " the instant (undefined time interval) after release when the projectile leaves the pouch." instead of "at the instant of release.

I just presumed that my intent was clear,but sometimes I am not too good at

explaining what I am seeing in my mind.

I am seeing the projectile leaving the pouch after the maximum acceleration an instant(undefined time interval)after release.

If you break down the time instant small enough, you will see that for a VERY short

instant, the projectile does not move at all after release.

The pouch deforms slightly as it accelerates due to the inertia of the projectile.

As the energy is transferred to the projectile from contact with the pouch,it begins to

accelerate until it reaches maximum velocity,and the velocity from this instant is

decreasing,but the pouch is falling behind due to frictional and other losses,which are

greater than the losses of the projectile.

The projectile cannot accelerate after it leaves the pouch,obviously.

At this point in time, it has not passed the anchor point of the tube ,however.

The point being that the shorter tube will launch the projectile sooner than the longer

tube,due to a quicker contraction of the short tube than the longer tube.(hic!)

Now you're just messing with me. You bring up the crux of the matter and then misapply it. The projectile in the short tube releases the projectile in less time than the long tube. The long tube projectile is accelerated for a longer period of time than the short tube. Longer acceleration time results in more velocity. Δv=a*Δt

Peak velocity may be greater with the longer tube, but by the time the projectile leaves

the pouch of the longer tube, the shorter one is already closer to the target while the

longer tube projectile is still in the pouch.

The longer tube has to contract twice as far to deliver full power,and if the tubes have

the same contraction velocity,it will take twice as long,and the projectile will feel less

initial "G" force after release,it will have a longer time to accelerate to the same speed.

Remember,they are starting with the same amount of pull force,and the contraction

velocity is the same for both tubes.

The longer tube does not have twice the pull force or twice the speed of the shorter

tube.

Once the projectile loses contact with the pouch, the acceleration is over,and I don't

think it is in contact for the full stroke because the force is decreasing(slowing down)

as it contracts.

It would be interesting to see a high speed video of both cases.(hi hi hic!)

On my phone.

You've made a lot of assumptions.

But, the short tube projectile will begin deceleration sooner the long tube projectile.

I am sure you have seen a 1/4 mile race,where the faster car lost to the slower one because of the "hole shot",or faster reaction time from the start.

Higher final speed,but longer Elapsed Time.

I think this case may be similar.

Take that analogy a step further.

I get a 6 inch hole shot on you off the line, but my motor blows up 6 inches sooner than yours.

Now, what we haven't established............................. is the distance to the finish line.

Nor the actual advantage you have with a longer engine life.

Care to nail down all the variables, before we go on?

Remember, on the strip, we start at the same place.

What about the time dilation from the change in G due to the desires of the Cthulu riding a tachyon burst into a molting blackhole? Can this be compensated by a thiotimoline trigger mechanism on the turboencabulator?

Waaaa!!! What can I say about this fabulous elixir!

Yes, but not yet.

First, I'll have to get:

1/4-Inch Surgical Tubing, Amber, 3-Feet

There's an outdoor outfitters store down the street who sells this.

Much work to be done, such as determination of pull back force required to achieve 50% elongation.

Some calibration of equipment, (fish scale) and the decision to use two tubes vs one for ease of launch. Mechanism for release of projectile, etc.

I don't expect results until after the first of the year.

I'm cleaning out my lab and making room for this experiment after completing work on the last cold fusion project. The results of that trial were not conclusive, but there's all this slime on the floor that I'll have to clean up.

I was using Thera-Band therapy tubing for this discussion,and regular surgical tubing may not give the same results.

Finish line is 6 inches from the start.!!

If you both have the same top speed potential you can never catch the hole-shot guy.

Seriously,the farther it is to the finish line, the more the separation distance.

A 1 second advantage at 60 mph is only 88 feet,but at 120 mph, it is twice that.

Let us assume a 200 fps velocity,which is common for a bean shooter,measured at the exit side of fork of the bean shooter.

In 1/400th of a second,the shorter tube projectile is 6 inches from the point of release, and has reached maximum speed of 200 fps.

In the same time interval, the longer tube has only reached 100 fps,assuming a linear power delivery.

In the next 400th of a second, the shorter projectile travels another 6 inches and is now 1 foot from release point.

In the same interval,the longer projectile has reached it's maximum speed of 200

fps,but it is 1 foot behind the shorter one.

Add to this the extra six inches that it is starting from behind the shorter tube.

Since they both have the same maximum speed,the second one cannot catch the

first one,because the same forces will be acting on it as on the shorter one,

decelerating at the same rate,subject to the same friction,assuming the friction is a

result of the distance traveled through the air.The wind resistance to the pouch is

also greater, on the longer tube,having traveled twice as far.

And remember,the long tube is also starting 6 inches behind the short tube,(we are not on a drag strip).

and the internal friction of the tube contracting twice as far has to be considered.

Time starts when tube is released.

Time interval to target is quicker with the short tube.(hic)

Please correct me if I have my math or logic wrong.

Correct you?

You've hole shot me and overwhelmed me with fast talk.

I'll think more about this after I've had a nap.

If you can't dazzle 'em with brilliance......................................................

#48<reply>

Another variable???????????????????????????????????????????????????????????

Can you PM a sample to me?

Do not be baffled by my bull sh-t.

It is the Georgia Moon talkin'.

(Just got another refill from Uncle Bill's still on the hill!)

I noticed a couple of buzzards dropping out of the sky above the still,too drunk to fly.

If I am talking too fast,read slower.

"buzzards dropping out of the sky"

And, that's just from the smell.

Remember, I grew up in Arkansas. My cousin hauled shine in his '55 chevy.

This is not entirely elementary: Impulse is what actually leads to final velocity,

J=∫F dt, and since F=ma, we have also F=dp/dt, thus J = ∫dp = mΔv, and since v(initial) = 0 (at rest, maximum draw), we have J = mv. However, the integral is over time. If the time for case 1 (half-length tube 50% elongation) is half the time for case 2 (full length tube 50% elongation), then very simply mv1<mv2. I am having brain lock all of a sudden. I must defer to those familiar with actual data on rubber tube draw - such as, if F=kd/d₀ (modified Hooke's law) is obeyed, then it will be possible to solve rather easily, but if F= f(d), then not so easy?

Assumption: F=kd, thus a =kd/d₀m, thusly we know that at draw back we have equilibrium of kd/d₀ = 50# (weight, not mass). 50# weight is a force of 9.80*50/2.2 in Kg-m/sec². 222.7 to be precise. and we know that at drawback in both instances, d/d₀ = 1.5. Upon release from hold, the forces are no longer in equilibrium, and d/d₀ →1.0 as time increases, and F drops off also. we need the time to travel back to d/d₀=1.0, but somehow I believe this to be the same. Is this not why bows are specified by pull weight more than the length of the draw?

You can do it that way if you prefer:

mΔV = FΔt

ΔV = Δt(F/m)

The masses (m) of the projectiles are equal.

The force (F) needed to stretch the long rubber band to 150% of its original length is equal to the force needed to stretch the short rubber band to 150% of its original length.

The time required for the long rubber band to contract (Δt) is greater than the time required for the short rubber band to contract.

Therefore, the change in velocity (ΔV) is greater for the long rubber band than the short one.

I think it's easier to use the Conservation of Energy method:

The Potential Energy stored in the projectile by the stretched rubber band = Fd

The Kinetic Energy of the projectile after the release = (1/2)mV²

Potential Energy is converted to Kinetic Energy:

The longer rubber band provides greater Fd, which results in greater (1/2)mV²

IF the projectile was attached to the pouch,it would deliver all of the energy to the

projectile,but since it is free to leave the pouch,it will leave when the pouch begins to

slow down.

If the maximum velocity of the pouch is reached very soon after the release,then it is

slowing down from that time on,and has lost contact with the projectile.

Therefor a lot of the input energy is wasted propelling an empty pouch.

Since this is a hypothetical problem, can't we assume that the pouch has no weight and no cross section?

And, if the projectile was attached to the pouch, it would put your eye out when it came back at your head.

OK, this is my final comment here. Experimental data trumps all theory. Why not set up an experiment with two sling shots with different tubing lengths and test.

I agree with Redfred. This isn't rocket science. It's classic Physics 101. If you want to verify or disprove the theory, make two or three slingshots of varying length and do an experiment. To really illustrate the concept, build one ridiculously long and see what happens.

So long as any tension at all remains in the tube, the pouch will continue to accelerate.

But at some point, the projectile will leave the pouch even though the pouch is still traveling forward.

It has already received sufficient kinetic energy to "set it free" from the pouch.

Wrong; not until the pouch stops accelerating, i.e., no remaining tension.

Ok,so you are saying that the longer tubes can contract faster than the short tubes?

From 50% stretch on each one, with 50 pounds of pull on each one?

Equal pull,twice the length will travel faster than 1/2 the length with same pull?

In effect,you are saying that the longer tube can contract a 1 foot distance faster

than the shorter tube can contract 6 inches?

The heart of the matter is where does the pouch reach maximum velocity?

I am not speaking of stretching the same tube twice the length;that would have an

obvious answer.

This situation is not as obvious at it appears at first glance.

If you stretch a dead horse to 150% of its original length, how many pieces will you have?

We don't stretch dead horses around here,I don't know about Arkansas.

But I have to give an Arkansas inventor credit for inventing the tooth brush.

Anywhere it it would be called a teeth brush.

And,I just recently read that Arkansas just passed a new law:If you divorce your

wife,she can still be your sister.

I will take all of the "Tar Heel" or Red Neck jokes you've got.

In this thread, the Gran Patron and Georgia Moon aficionados seem actually to be drinking Pure Old Panther Piss, which is/was sold in Mason jars in some areas of the countryside.

I'm going to drink mine with some surgical tubing, after stretching it to twice its length.

Slam dunk! The best kind.

In conclusion,

• Natural rubber tubes can be used for flippers.
• Moonshine causes hallucinations.
• Gas operated autoloaders have less muzzle velocity.
• Don't leave rubber tubes laying around for junkies to use.
• The tube ends can be open or closed without effect.
• You can't kill a dead horse with a Wham-o to the touchhole.
• 

I did the calculations. My assumptions are listed below the table.

Case 1 - 12" rubber band stretched to 18" (0.5 feet stretch)

Case 2 - 24" rubber band stretched to 36" (1 ft stretch)

Case 3 - 4 ft rubber band stretched to 6 ft (2 ft stretch)

Assumptions:

mass of 1/4" steel ball bearing = 1.0524 grams

maximum force = 50 lbs

average force = 25 lbs

All of the potential energy (Fd) is converted to kinetic energy (½mv²)

This only works if the ball bearing remains in contact with the pouch and accelerates

for the full length of contraction.

Once the ball loses contact,energy input into the ball bearing ceases.

Any further movement of the pouch is wasted.

Also,what you are saying is that the longer the tube, the faster it can contract to the

relaxed state,even though it has twice the distance to go?

It does not begin at 50 pounds and maintain it for the full length,it decreases as it

contracts,from 50 to zero.

"This only works if the ball bearing remains in contact with the pouch and accelerates for the full length of contraction. Once the ball loses contact,energy input into the ball bearing ceases."

Correct. In case 1, the ball bearing is in contact with the pouch for 6 inches. It is only being accelerated during this 6 inch span. The force exerted on the ball at the beginning is 50 lbs. The force exerted at the end of the 6 inches is 0 lbs. That's where my assumption of 25 lbs average force came from. Likewise, in case 2 the ball bearing is accelerated forward only during the 12" contraction span. The average force during this 12" span is 25 lbs.

"Also,what you are saying is that the longer the tube, the faster it can contract to the relaxed state, even though it has twice the distance to go?"

No. The average force is the same, the distance is longer, the time to contract is longer. The time and distance the ball is being accelerated forward is longer.

"It does not begin at 50 pounds and maintain it for the full length, it decreases as it contracts, from 50 to zero."

Correct. That's where my average force of 25 lbs came from. I assumed the force per unit of stretch distance is linear. Probably not a bad assumption as long as you remain in the elastic range for the rubber band. Even if it's a bad assumption, the error is about the same for all 3 cases.

I agree with Lyn (for once or 200 times), that now is the time to put this to the mythbusters, lay it out on the line, put your money down, and apply the projectiles to the ballistic pendulum. I presume that HTR can hit a ballistic pendulum in the base with a butt fiddle, or was that butt with a base fiddle?

It looks like a challenge. If tribefan is right, those little 1 gram BB's are gonna smart at over 1000 fps! Practically, I do not believe natural rubber will contract that fast at all, ever. I think it gets in its own way.

Yes, but before we order the chronometer and the high speed recorder, we'll need to establish the specific type of rubber tubing under discussion.

Surgical, rubber, silicone, latex, what will we agree to be the standard.

HTR did specify natural rubber in the OP. or whatever is available at the local Ace Hardware. Hevea Brasiliensis it is. As to the design of the punkin' chunker, I don't care, and I really don't care if you have to "shoot" it from the hip to keep from putting your eye out. It's Christmas, of course you want a BB gun.

HTR needs to rejoin the conversation, and specify once again, the precise conditions of the test:

(1) actual lengths of tubing to be used

(2) mass of the projectiles (and shape if not spheres), density (matters)

(3) single band length with straight (more or less) draw, or double band as in slingshot?

(4)distance to target, target composition (prefer ballistic gel), and mass of target, along with suspension length for ballistic pendulum tests. (or high speed bullet chronometer specification).

(5)the number for emergency services for dummies who just shot their eye out right before Christmas

In the original post,I posed two questions:

One: Velocity of the projectile

Two:Time to target.

The material used is Thera Band-Gold Tubing

The specifications are listed on this link.

http://www.thera-band.com/faqs_consumer.php#faq_1

1.The test involves 2 tubes , like a slingshot

The different lengths of tube were arbitrary,used to simplify the math for the

question, the specifications on the link show that regardless of length, the same

percentage of stretch will give the same increase in pull force,and the pull strength

was rounded off for same reasons stated above.The actual lengths of tube do not

matters as long as the same ratios can be maintained The projectile is a 5/16 inch

steel ball, wt: 2.052GThe distance to targets is 10 feet from the front side,the anchor

point,of the sling shot .

Targets are two,each made of the same solid material, 3/4 inch

plywood is suggested, free to swing,and break a photocell at the slightest movement,

positioned below each target,and a timer configured to each photocell to determine

which beam broke first,and thus which target was struck first.The projectile launch

points will have a trigger mechanism mechanically attached to each one,configured

so that each projectile is released at the same time and will also include photocells to

detect when the trigger was pulled,and when the projectile has begun to move.Tests

should be conducted in a closed structure to eliminate any wind fluctuations.This

should provide all the information required to determine velocity and time to

target.

Any other information should be able to be calculated from this information,unless of

course,I have forgotten something.(likely).

And lastly,I do not put slingshot near eyes when shooting,but if you use that

method,goggles or safety glasses are recommended,of course.

Good luck!

I thought 1000 ft/sec seemed a little high, too. But the assumption was rubber tubing that would have 50 lbs of force after being stretched to 150% of its relaxed state. That seems pretty high, too. If anyone wants to measure the actual force and stretch distance for some tubing, give me the data and I will revise my table. Also, the 1000 ft/sec case was 4 feet long relaxed; 6 feet stretched. Pretty big slingshot. If the tubing gets in the way of the little ball during launch, all bets are off. We gotta design it somehow so the tubing doesn't get in the way.

This thread has gotten me interested in slingshots again. I had one many years ago and was a pretty good shot. Not as good as Redneck, but not bad. I'm going to build one this weekend! Ha! Ha!

I think I have stretched this conversation out to the point where the pressure has

increased exponentially,so while you guys hash it out,I will,like the signifying

monkey,who started trouble between the lion and elephant,sit back and watch from a

distance.

Oh great launcher of metal spheres into tiny bird box holes, pray tell, why not participate in the launchery? Or better still - further specify the design of the launcher. I have a few suggestions for the ballistic pendulum.

I applied a 50% elongation as stated in the initial post.

Great chart!

The only problem I see is that you are considering the 50 pounds of force to be

constant from end to end of the tubing,.

It is really decreasing at a linear rate.

At 50% elongation, there will be 50 pounds of force at time of release,thereafter,it

will be less until the relaxed state.

If you look at the force per inch of tubing,the shorter tubing has the same force

concentrated into a shorter length,therefor the "energy density" so to speak, of the

tubing is twice that of the longer tubing.

It should contract quicker,delivering the same amount of thrust at a faster rate,thus

faster Acceleration.

Add to this the 1 foot further from the target In case 2) of the longer tube

(remember,the timing starts when the projectile clears the anchor point of the

tube,which is the same distance from the target for all tubes. )

About a 5% longer distance to target, in case 1, 10% in case 2, and about 33% in

case 3.

The longer tube cannot contract as quickly as the shorter tube,but it applies a gentler

force over a longer distance, end result is same total energy input into the projectile.

Time-to-target should be quicker with the shorter tube.

Where am I going wrong with this train of thought?

Since the % elongation is the same for the long one and the short one, the force is the same. The long one is exerting that force over a longer distance, so the energy is greater. The acceleration is the same in both cases since the force is the same. The longer one is accelerating the ball over a longer distance (and for a longer time), so the final velocity is greater for the longer one.

W=1/2k s² s≠ length, it is displacement elongation. I also used the 14.2 # at 100% rating for the tubing you specified, and I also used 50% elongation is mentioned in the OP, at least I think I did. By now my head is spinning on its axis, and my brain is getting inertial overload.