# how much tension a compond bow puts on its cable???



## bernardinifan (Dec 12, 2010)

Does anyone of you know how much tension is in your cable for a 60-70 ponds bow?


Thanks for your answer


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## Vortex69 (Jul 8, 2007)

At brace or full draw?

With the cables so close together it would be hard to make a cable that would allow most scales, at least the ones that I own, to fit on the bow for a test like that. There was a very knowledgeable fella round here a while back (think his name was copterdoc) who was loosly floating the unchallenged number of 300#, but like I said, the number was unchallenged.


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## flag (Oct 4, 2009)

i dont know about when they are on the bow but i have put 450lbs on mine in the stretcher


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## Deer Eliminator (Jan 21, 2010)

I have put them to 500lbs no problems. But I read on here somewhere they did 800lbs and well they didn't survive.


Hutch


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## Vortex69 (Jul 8, 2007)

I may be mistaken but I believe the OP is asking for the amount of tension the BOW is putting on the cables..not the streacher.


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## bernardinifan (Dec 12, 2010)

Vortex69 said:


> I may be mistaken but I believe the OP is asking for the amount of tension the BOW is putting on the cables..not the streacher.


this i what i meant


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## nuts&bolts (Mar 25, 2005)

bernardinifan said:


> Does anyone of you know how much tension is in your cable for a 60-70 ponds bow?
> 
> 
> Thanks for your answer


Roughly,
300% of the peak draw weight.

So,
when you are at full draw,
about 210 lbs of tension in the cables,
if you have the bow set at 70 lbs.


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## Vortex69 (Jul 8, 2007)

nuts&bolts said:


> ...about 210 lbs of tension in the cables...


Is that approx 105 per cable at full draw or ... 210 total for the cable group at full draw?

Gettin kind of curious myself.


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## Andy. (Jan 12, 2011)

Vortex69 said:


> Is that approx 105 per cable at full draw or ... 210 total for the cable group at full draw?
> 
> Gettin kind of curious myself.


Cable is toting almost all the load at full draw. The string just have holding weight poundage. 


Andy


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## Vortex69 (Jul 8, 2007)

Vortex69 said:


> Is that approx 105 per cable at full draw or ... 210 total for the cable group at full draw?
> 
> Gettin kind of curious myself.


That was dum....

What I ment was is that 210 per cable or 210 across both cables?


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## nuts&bolts (Mar 25, 2005)

Vortex69 said:


> Is that approx 105 per cable at full draw or ... 210 total for the cable group at full draw?
> 
> Gettin kind of curious myself.


I drew a force diagram,
and made some assumptions about
the horizontal travel for the axles,
from brace to full draw,
as well as limb pocket angle.

I come up with about 300% of draw weight,
as the total cable load,
when at full draw.

So,
yes,
70 lbs of draw weight,
results in a cable system load of roughly 210 lbs,
so
with two cables,
each cable would be at 100-ish pounds of load.


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## Vortex69 (Jul 8, 2007)

Pretty sure that will take care of the OP's question....

I know it works for me.

Thanks.


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## bernardinifan (Dec 12, 2010)

thanks for the answer


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## copterdoc (Oct 9, 2005)

I have measured it, using a draw board and cable tensiometer, but it's easier to calculate it, by figuring out your cam's total mechanical advantage.

The way you calculate the TOTAL mechanical advantage of the cam, is by measuring the ratio of string/fed-to-cable/reeled.

All cams transmit load through the axles.

The load is always applied on the plane of the cable/string. So, in order to find the starting and stopping points, you need to "draw a line" from the string, to the axle at 90 degrees to the plane of the string.

Mark the location at that point on the string, and draw the bow back with a draw board. Repeat the same measurement at full draw and measure the distance between the two marks. The difference in ATA from brace to full draw, tells you how much cable was reeled in. That will give you the total ratio of mechanical advantage.

That ratio, will be the inverse of the peak draw weight-to-peak limb load. For most modern bows, the cams feed out 3-4 inches of string, for every inch of cable reeled in. That means a 70# bow, has about 245# of total load from the limbs at full draw.

Now, that seems like a lot, but it doesn't even factor in let-off!

We all want let-off right? Well, everything you take from the string, goes right to the cables. If you have 80% let-off, that means that while the string is under 15# of load, the cables are under 230.


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## copterdoc (Oct 9, 2005)

BTW, true dual cam systems, do not mechanically maintain equal cable loading. You can actually place ALL of the cable load on one cable, and let the other go slack.

Binary cam systems, do not allow that to happen. Both cables in a binary system, are ALWAYS under equal load.

Single and hybrid cam systems, only have one cable.


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## Vortex69 (Jul 8, 2007)

So would a 60lb bow be (60/70)230=197lb?, and how much load is in the cable(s) at brace?

By the way, I was only partially correct when I dropped the 300# number. Just found it (post 53):

http://www.archerytalk.com/vb/showthread.php?t=1364232&page=2


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## copterdoc (Oct 9, 2005)

Vortex69 said:


> So would a 60lb bow be (60/70)230=197lb?


No. But you would be close enough.
You would multiply the peak draw weight by the mechanical advantage of the cam. If it was 3.5" to 1", like the 70# example, it would be 210#. Then you would subtract the holding weight of 12#, for a total cable load of 198#.


Vortex69 said:


> ....how much load is in the cable(s) at brace?


It depends on cam geometry and timing. The closer the cable tracks to the axle, the less leverage it has. The farther the string tracks from the axle, the more leverage it has.

As the bow is drawn back, the cable tracks closer to the axle, and the string tracks farther from the axle. You need to measure the entire draw cycle, to calculate the cam's mechanical advantage.


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## dwagoner (Sep 27, 2007)

OK so lets go on the other end, when a bows shot and it comes back to rest how much force is that on the cables??? I think the force on cables after the shot and bow coming back to rest is prolly higher than a bow at full draw.


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## copterdoc (Oct 9, 2005)

dwagoner said:


> .....I think the force on cables after the shot and bow coming back to rest is prolly higher than a bow at full draw.


I don't think so.

All of the inertia and momentum, is "trying" to break the string at that point. The cables are "opposite" the string. That's a good way to think about it.

I *do* believe, that each cable should be at least twice as strong as it "needs to be", particularly with a true dual cam system.


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## monte_arrow (Aug 30, 2006)

bernardinifan said:


> Does anyone of you know how much tension is in your cable for a 60-70 ponds bow?
> 
> 
> Thanks for your answer


Why don't you calculate it by yourself, for your particular bow? It is not that difficult.
First you split the opposite draw weight vector at full draw in 2 vectors, each directed as the upper and lower part of the string.
Than you do some measurement on your cam, of the normal distance between the direction of the cable forces and the cam axle,
since the strings exert the torque moments equal to the string load x this length. 
On the upper cam you will have two equal torques, oposing each other: from the string and from the control cable.
On the lower cam, you will have the 3 torque moments: from the string and the control cable in the drawing direction,
and from the buss cable in the opposite one. First compute the load in the control cable using the upper cam torque balance.
Then compute the load in the buss cable, using the lower cam balance, and this will be the higher one of the two, since
the buss cable takes most of the load at full draw.

PS. this is valid for a cam 1/2 hybrid, but for any cam system it's the same principle.


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## Vortex69 (Jul 8, 2007)

monte_arrow said:


> ....It is not that difficult.....


Well then, vector this:

If it takes a chicken & a half a day & a half to lay an egg & a half...

How long does it take for a brass monkey with a wooden leg to kick the seeds out of a dill pickle?


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## copterdoc (Oct 9, 2005)

What Monte said, isn't "wrong". It's just a wee bit too convoluted, and the measurements are more difficult to accurately obtain.

Also, the string below the loop, going to the bottom cam, does the "lion's share" of operating the cam system. There is not equal tension on the string above the loop, going to the top cam.


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## Trailerdog (Jul 24, 2008)

Vortex69 said:


> Well then, vector this:
> 
> If it takes a chicken & a half a day & a half to lay an egg & a half...
> 
> How long does it take for a brass monkey with a wooden leg to kick the seeds out of a dill pickle?



Now that's funny!


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## kicker338 (Nov 30, 2008)

Vortex69 said:


> Well then, vector this:
> 
> If it takes a chicken & a half a day & a half to lay an egg & a half...
> 
> How long does it take for a brass monkey with a wooden leg to kick the seeds out of a dill pickle?


About as long as it takes for the dog to take a dump in the yard and for you to step in it haha


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## Vortex69 (Jul 8, 2007)

kicker338 said:


> About as long as it takes for the dog to take a dump in the yard and for you to step in it haha


With a name like "kicker" I'm sure any vector reference to said object would lead me to your neighbors yard.


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## kicker338 (Nov 30, 2008)

Vortex seriously thinking about giving my neighbors yard a try. The old coot goes south every winter and while he's gone i'm tempted to fertlize his pretty lawn. Got 2 pugs and they'r manufacturing machines LOL


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## whitebuck (Oct 17, 2003)

This answer only works using statics .... dynamics in the shot cycle is not linear, the loading is a series of peaks and troughs

I think this pic shows dynamics are involved, look at difference between the top and bottom limbs


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## monte_arrow (Aug 30, 2006)

Vortex69 said:


> Well then, vector this:
> 
> If it takes a chicken & a half a day & a half to lay an egg & a half...
> 
> How long does it take for a brass monkey with a wooden leg to kick the seeds out of a dill pickle?



Well, vortex, it's not anybodys fault but your's if you missed your elementary school counting chicken egs, so you find it too
complicated to multiply F x L, and so on. If you don't know &$&& about something - you should not mix into it, you will be
smarther like that.


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## Vortex69 (Jul 8, 2007)

Thanks for the advice!!......I'm feeling smarter already.


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