# String Jig Tensioners



## T_well (Jan 10, 2010)

Thanks guys. these pics are great.


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## T_well (Jan 10, 2010)

2nd post was fecetious, Buck. I don't have any.


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## nuts&bolts (Mar 25, 2005)

T_well said:


> 2nd post was fecetious, Buck. I don't have any.


Yellowstone Micro Stretcher


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## nuts&bolts (Mar 25, 2005)

If you want to use a die spring (very precise compression spring),
insert the spring between the angle and the knob.

400 lb per inch spring,
will be 200 lbs of compression at 1/2-inch shorter
than the starting spring length
and
will be 300 lbs of compression at 3/4-inch shorter
than the starting spring length.


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## jlnel (Dec 22, 2009)

they look nice, how much were they


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## nuts&bolts (Mar 25, 2005)

jlnel said:


> they look nice, how much were they


http://www.lancasterarchery.com/product_info.php?manufacturers_id=161&products_id=2197

$49.99


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## DannyRO (Apr 17, 2009)

nuts&bolts said:


> If you want to use a die spring (very precise compression spring),
> insert the spring between the angle and the knob.
> 
> 400 lb per inch spring,
> ...


Are you sure of the linear progress of a compression spring tension?


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## nuts&bolts (Mar 25, 2005)

DannyRO said:


> Are you sure of the linear progress of a compression spring tension?


Write up about precision compression and die springs
at McMaster Carr.

yes.

http://www.mcmaster.com/#about-die-springs/=64p1x0


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## kc hay seed (Jul 12, 2007)

any electrical supply house will have L-brackets for mounting unistrut that look just like this jig. the J- bolts you can get at the hardware store. this is how i made mine.


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## Ray.Klefstad (Oct 7, 2006)

You can make your own from Home Depot (or Lowes) parts for Unistrut. I found pictures here on AT from user Automan. He is also very helpful with info if you ask.

The hooks are 3/8" clothesline hooks which were just under $2 each and they come with two nuts each. I also bought wing nuts to make it easier to adjust by hand.

The 3/16" thick L-Brackets were a few dollars each.

A pack of 5 1/2" unistrut nuts were a few dollars.

And another few dollars for some 1/2" bolts to hold the L-brackets to the Unistrut.

The 10' long piece of Unistrut is under $20.

I also bought two 6" long 1/2" bolts to use for my string jig posts.

I got some used Porsche valve springs (free) to use for tensioning. You can measure the movement with a known weight to get the spring constant. And, yes, springs move linearly with the increase in weight as long as you stay within their elastic limit - i.e., don't over stretch or over compress them.

Ray


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## T_well (Jan 10, 2010)

Thanks guys! These ideas are about the same as what I was trying to design. My only problem has been finding a spring with enough resistance without being 10 in. long. Here's another question; if i find springs that aren't strong enough can they be added together? For instance if I had 3 100# springs could they be put in a row lengthwise to make 300#, or could you take 6 50# springs and align them side by side like in a revolver pistol? 

Nuts & Bolts, I really want to hear what you and/or Unk Bond have to say on this as well!


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## TMax27 (Nov 7, 2006)

http://www.mcmaster.com/#compression-and-die-springs/=64rgr6

Here's one. 400lb/inch 2 1/2" Long


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## T_well (Jan 10, 2010)

Another question along the same lines. If you take a spring and set it up to put pressure on it using a spring scale and pull 300# on the scale, is there 300# of pressure on the system or because there are 2 springs counting the scale is it doubled or cut in half possibly?


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## nuts&bolts (Mar 25, 2005)

T_well said:


> Another question along the same lines. If you take a spring and set it up to put pressure on it using a spring scale and pull 300# on the scale, is there 300# of pressure on the system or because there are 2 springs counting the scale is it doubled or cut in half possibly?


Springs are additive.

Let's say I have spring #1,
rated at 100 lbs of compression per inch.

Let's say spring #1 starts at 2-inches uncompressed.


So,
we have spring #2,
also rated at 100 lbs of compression per inch.

Let's say spring #2 also starts at 2-inches uncompressed.

So,
you use both spring #1 and #2,
so the total uncompressed length = 4-inches.

So,
you squeeze the springs to a compressed length
of 2-inches total. You have shortened the combination
2-inches LESS than the starting length.

You have 100 lbs of compression per inch,
and you have compressed 2-inches,
so you have 200 lbs of pressure.


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## T_well (Jan 10, 2010)

nuts&bolts said:


> Springs are additive.
> 
> Let's say I have spring #1,
> rated at 100 lbs of compression per inch.
> ...



SO...let's leave compression out of it because I can explain this better with a pull type spring. Say I take a trampoline spring and hook it to an immovable object and hook a spring scale to the other end and pull until the scale reads 50#; does that mean I actually exerted 100# of force on the working end given that the trampoline spring and the scale spring have the same #/in. ratio?


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## nuts&bolts (Mar 25, 2005)

T_well said:


> SO...let's leave compression out of it because I can explain this better with a pull type spring. Say I take a trampoline spring and hook it to an immovable object and hook a spring scale to the other end and pull until the scale reads 50#; does that mean I actually exerted 100# of force on the working end given that the trampoline spring and the scale spring have the same #/in. ratio?



Ok.

A concrete anchor that weighs 100 tons,
or 200,000 lbs.

That's basically an immovable object.

Now,
I hook up a braided wire cable
that is 10 inches thick.

Basically zero stretch.


Now,
hook up the spring scale.

The 100 lb mark is 1-inch away from the zero line.

The 50 lb mark is 1/2-inch away from the zero line.

This means that the spring rate = 100 lbs per inch
(that's why the 100 # mark is one inch away or 1 inch of stretch)
(that's why the 200# mark is two inches away or showing 2 inches of stretch)


So,
you pull until the scale reads 100 lbs,
when the spring scale is attached to the 10-inch diameter braided steel rope.

You stretch a 100 lb per inch spring,
a total of ONE INCH,
then you have applied 100 pounds of pulling force.



Now,
hook up a 100 lb per inch spring
to the 200,000 pound anchor.

This is spring #1.


Now,
attach your spring scale 
to Spring #1.


Now,
Spring #1 is an exact match
to the internal guts of your spring scale.


Remember,
your spring scale has the 100# mark at 1-inch away from the zero line
and
your spring scale has the 200# mark at 2-inches away from the zero line
and
your spring scale has the 300# mark at 3-inches away from the zero line.


Sooooo,
you apply 100 lbs of pulling force...

you setup a frame and a pulley
and attach 100 lbs of dumb bell weights on the end of the pulley rope.


So,
you look at the scale.

Wait just a second.


THE SCALE only reads 50 pounds?

Why?

Cuz,
we have two springs that are mounted in line with each other.


Spring #1 has no shell
and is a raw spring,
and spring #1 has STRETCHED 1/2-inch.


Spring #2 has a shell,
and it is a metal box with lots of lines,
that show 100 lbs, 200 lbs, 300 lbs, etc.

So,
when you apply 100 lbs of pulling force,
and
you have TWO springs installed in a line...

then,
BOTH springs share the work.


100 lbs of pulling force

Spring #1 does half the work.
Spring #2 does half the work.


Therefore,
Spring #1 stretches from 50 lbs of force
and
Spring #2 stretches from 50 lbs of force,
and that's why Spring #2, only shows 50 lbs of force.


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## DannyRO (Apr 17, 2009)

nuts&bolts said:


> Write up about precision compression and die springs
> at McMaster Carr.
> 
> yes.
> ...



no offence, but no compression spring have a linear rate of tension: http://www.engineersedge.com/spring_comp_calc.htm


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## T_well (Jan 10, 2010)

Alright N&B I knew you'd come through! So I put a spring in the above stretcher between the nut and the base/holder; in order to find where to mark the bolt for 300# using a spring scale I actually need the scale to read more depending on the rate of the springs. So that rules out one of my "calibration" techniques. Like the Guiness commercial--BRILLIANT :darkbeer:


As with most mechanical tools you rally only want to be using the "sweet spot" anyway. the middle 60% in most cases. Like a torque wrench; if you need 50ft/lb of torque, it's best to use a 100ft/lb wrench instead of the 50.


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## nuts&bolts (Mar 25, 2005)

DannyRO said:


> no offence, but no compression spring have a linear rate of tension: http://www.engineersedge.com/spring_comp_calc.htm


"...spring rate tends to be constant over the central 60% deflection range.."


For the purposes of building a bowstring tension device,
where folks are trying to stretch a bowstring enough,
so that the bowstring bundle does not rotate,
while applying a serving (end serving, center serving)...

die compression springs will suffice.


So,
if we are looking for a range of 100 to 300 lbs of tension...

100 lbs of tension to measure bowstring length, per AMO spec

and

300 lbs of tension to prevent bowstring twist, while applying a tight serving with a serving tool....

and we want to stay within the 60% central deflection range...

which means we want to avoid the 0-20% deflection range
and
which means we want to avoid the 80-100% deflection range (spring bind)...

then,
the idea of using multiple springs is an excellent idea.


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## nuts&bolts (Mar 25, 2005)

T_well said:


> Alright N&B I knew you'd come through! So I put a spring in the above stretcher between the nut and the base/holder; in order to find where to mark the bolt for 300# using a spring scale I actually need the scale to read more depending on the rate of the springs. So that rules out one of my "calibration" techniques. Like the Guiness commercial--BRILLIANT :darkbeer:
> 
> 
> As with most mechanical tools you rally only want to be using the "sweet spot" anyway. the middle 60% in most cases. Like a torque wrench; if you need 50ft/lb of torque, it's best to use a 100ft/lb wrench instead of the 50.


Well said.


When I use my Viking 300 lb scale,
the thing is sooooo long,
it takes up a lot of room on my piece of Unistrut.


So,
use a 300 lb scale to calibrate your spring compression amount,
to figure out how much compression is needed to get 100 lbs of force
and
to figure out how much compression is needed to get 300 lbs of force.


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## Dthbyhoyt (Dec 4, 2004)

Here is one I made for my old Apple string jig , it slides over the square tube , then uses a compression spring rated at 300#..

One end hooks to the loop of the string and the other to the eye bolt .


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## petrey10 (Oct 24, 2008)

that link to the spring didn't really help me cause I don't have the specific spring I just get to search through the site for the spring.. I have no clue what to look for


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## nuts&bolts (Mar 25, 2005)

petrey10 said:


> that link to the spring didn't really help me cause I don't have the specific spring I just get to search through the site for the spring.. I have no clue what to look for


http://www.mcmaster.com/#catalog/116/1210/=65976a

400 lbs per inch.

At 20% of deflection,
you should get 160 lbs of force.

Spring is 2-inches overall length....(starting uncompressed length).

This fits over a 1/2-inch diameter rod.

So,
this spring will fit over any threaded rod smaller than 1/2-inch.



This one is rated at exactly 300 lbs of force,
at 15% of compression.

http://www.mcmaster.com/#about-die-springs/=6596a7

This spring fits a rod diameter of 3/4-inch.

Overall all length is also 2-inches.

This one is extremely stiff,
and probably not one you want.


Whatever spring you get,
use a 300 lb spring scale
to calibrate how much you need to squeeze the spring,
to get to 300 lbs of force.


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## nuts&bolts (Mar 25, 2005)

The compression spring arrangement
would look like this one.











This is a string jig stretcher built by Bowtech182.

See post #7.

http://www.archerytalk.com/vb/showthread.php?t=464240&referrerid=22477


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## T_well (Jan 10, 2010)

*Got it finally!*

I went to an engine shop. The guy had a machine that measured spring resistance. We found a valve spring that would hold up to 600#. Best part is we could pull 100# and it would give us the length, 200# and the length, and the same for 300#. Now I can simply build my stretcher and calibrate it with a micrometer for the set spring length. This way it doesn't matter if I'm in the sweet spot or not because I know the length at the given #'s. Life is good again.


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## Purka (Sep 8, 2004)

This is what I use...no springs, a set of scales and a boat winch.. These days I hardly use the scales, I just go by the sound or feel of the string under tension.


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## nuts&bolts (Mar 25, 2005)

T_well said:


> I went to an engine shop. The guy had a machine that measured spring resistance. We found a valve spring that would hold up to 600#. Best part is we could pull 100# and it would give us the length, 200# and the length, and the same for 300#. Now I can simply build my stretcher and calibrate it with a micrometer for the set spring length. This way it doesn't matter if I'm in the sweet spot or not because I know the length at the given #'s. Life is good again.


Perfect.


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