# 27 Joule slingshot question.



## boyntonstu (Jul 16, 2010)

What force is the draw requirement to achieve 27 Joules?


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## Darb (Sep 14, 2010)

boyntonstu said:


> What force is the draw requirement to achieve 27 Joules?


Mandatory info: Using the simplest math, muzzle energy is a function of mass and velocity, so at a minimum you'd need to provide a figure for the mass of the bullet ... otherwise the velocity cannot be computed.

Optional info: Also, assuming the joules figure is kinetic energy, you'd also need to supply the specs on the type of sling used, in order to be able to reverse guestimate the potential energy needed to achieve the desired kinetic energy (i.e., the sling's efficiency). The mass, and kinetic energy, could then be used to approximate speed and potential energy (drawing force) needed.

We can guestimate the latter, but you need to supply mass for the former.


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## smitty (Dec 17, 2009)

This is Torsten and Joerg territory, but if I were going to give it a try, I would start with the long "albatross" draw length and Theraband gold rubber. Then I would experiment with lead ammo size (weight) and rubber length, width and taper to get as close as I could get. Another question is: At what distance is the force to be delivered ?


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## Darb (Sep 14, 2010)

smitty said:


> This is Torsten and Joerg territory, but if I were going to give it a try, I would start with the long "albatross" draw length and Theraband gold rubber. Then I would experiment with lead ammo size (weight) and rubber length, width and taper to get as close as I could get. Another question is: At what distance is the force to be delivered ?


Muzzle energy, by definition, is the inflection point of the acceleration/deceleration curve ... i.e., the point it stops acceleration (when it leaves the pouch) and begins decelerating (due to air drag and/or impact). It'd simplify things to assume point blank range.

Meanwhile, we cannot proceed without a mass number.


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## boyntonstu (Jul 16, 2010)

smitty said:


> This is Torsten and Joerg territory, but if I were going to give it a try, I would start with the long "albatross" draw length and Theraband gold rubber. Then I would experiment with lead ammo size (weight) and rubber length, width and taper to get as close as I could get. Another question is: At what distance is the force to be delivered ?


I see an add above the forum for a 27 Joule slingshot.

Advertised as the most powerful commercial slingshot.

I thought that you might understand what I was asking about.

Fish's slingshot.


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## smitty (Dec 17, 2009)

Ha Ha Ha! OK, so, the question was asked for what reason, on a new topic ? I don't get it .


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## Darb (Sep 14, 2010)

Example: if we assume a 1/2" steel ball bearing (mass = 8.37 grams), in order to have a kinetic energy of 37 joules, it'd need to be travelling at 308 fps.

The drawforce needed for an 8.37 gram ball to reach 308 fps is a bit more complicated ... it'd be a function of draw length and the efficiency of whatever type of bands are involved. That's definitely in JoergS's area, not mine.

EDIT: crossed posts ... I will revise this post shortly.


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## Darb (Sep 14, 2010)

> The hunter catapult is at this time the most powerful commercially available catapult on the market. It can shoot a steel ball at 210 FEET PER SECOND, that's an amazing 27 JOULES of energy equating to over 18ftlbs!


This is just simple math, not an opinion or comment about the product in question (about which I know nothing):

Plugging 210 fps and 27 joules into a standard kinetic energy calculator, we can derive that the projectile must have a mass of roughly 13.18 grams. That's a pretty hefty sling bullet ... probably just over 1/2" if it were lead, and around 5/8" if it were steel.

Getting a bullet that size going 210 fps would take some pretty hefty bands, and would probably require butterfly style length. I don't have the tables needed to interpolate draw force (potential energy) for a given type and length of band and the desired kinetic energy.


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## boyntonstu (Jul 16, 2010)

smitty said:


> Ha Ha Ha! OK, so, the question was asked for what reason, on a new topic ? I don't get it .


An advertiser on this Forum makes a claim, I am asking what it takes to achieve it.


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## smitty (Dec 17, 2009)

OH ! OK. I was just trying to help out. I didn't realize the nature of the question. I question ALL advertising ALL the time, so you aren't alone.


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## Darb (Sep 14, 2010)

If you scroll up, I partially answered it (i.e., I derived bullet mass, but not drawing force).


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## Darb (Sep 14, 2010)

BTW, just as an aside ... it's a bit strange to see all these slingshots comparing their power to one another. Kinetic energy imparted is largely a function of the type and length of bands used, and the projectile they fire. It's a bit like stock car drivers arguing about who's got the fastest chassis.


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## boyntonstu (Jul 16, 2010)

Darb said:


> If you scroll up, I partially answered it (i.e., I derived bullet mass, but not drawing force).


Thanks.

I want to know the draw (and the draw length BTW).


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## boyntonstu (Jul 16, 2010)

boyntonstu said:


> If you scroll up, I partially answered it (i.e., I derived bullet mass, but not drawing force).


Thanks.

I want to know the draw (and the draw length BTW).

Wouldn't it be nice the know the draw force and length as part of a slingshot specification in addition to the speed and the energy?
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## torsten (Feb 9, 2010)

Using 12mm steelballs I need 5 kg draw. With heavier ammo I got more joules with the same band.






Regards


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## boyntonstu (Jul 16, 2010)

torsten said:


> Using 12mm steelballs I need 5 kg draw. With heavier ammo I got more joules with the same band.
> 
> 
> 
> ...


Your video sets the standard. It is great!

My question, in part, is based on your data.

Is the commercial 27 Joule slingshot overdrawn for a single shot?

That is one reason for my question.

You pull 160 cm. How many of your TB Black band lengths go into 160?

I can build a very nice 125 cm slingshot on a broom stick or a 2.54 cm wide slice of plywood.

In your opinion, is 125 cm long enough to get into the efficiency ball park of your 160 cm butterfly?


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## torsten (Feb 9, 2010)

125 cm is not as efficient as 160 cm. But better than 80, 90... cm. The longer the draw and the band, the more speed or joule you can get with the same force.
When I remember right, I was using in this video 30 cm long tbb, 3 layers per side 20mm x 10mm.


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## JoergS (Dec 17, 2009)

Stu, the 27 Joules for the Fish Hunter have been measured by me, in this video:






The relevant part starts at about 3:25.

I got 67 m/s with the 15 mm steel ball, weighing 12 gramms.

The draw was about 90 cm, and 12 kg.

There are other ways to get to 27 Joules, a longer draw (butterfly) and heavier ammo for example.

Jörg

PS: Watching these old vids is kind of odd. I lost about 25 pounds since then.


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## boyntonstu (Jul 16, 2010)

JoergS said:


> Stu, the 27 Joules for the Fish Hunter have been measured by me, in this video:
> 
> 
> 
> ...


Thanks for going back to your video.

How did you lose the weight???

"The hunter catapult is at this time the most powerful commercially available catapult on the market. It can shoot a steel ball at 210 FEET PER SECOND, that's an amazing 27 JOULES of energy equating to over 18ftlbs!"

I would like the draw to be under 9 kg. (20 pounds)

15 pounds would be even better.

The challenge is to get 27 joules in 125 cm with 9kg.


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## boyntonstu (Jul 16, 2010)

torsten said:


> 125 cm is not as efficient as 160 cm. But better than 80, 90... cm. The longer the draw and the band, the more speed or joule you can get with the same force.
> When I remember right, I was using in this video 30 cm long tbb, 3 layers per side 20mm x 10mm.


I calculate 5.33 times the original 30cm length. 160/30 = 5.33

Is that a 5.33 x or a 4.33 x stretch?

Do the tbb bands last at that stretch?


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## Darb (Sep 14, 2010)

Perhaps give the heated bands trick a try ?


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