Author Topic: Trapping?  (Read 2624 times)

Offline Longcruise

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Trapping?
« on: May 01, 2022, 07:19:47 PM »
I didn't want to sidetrack the bow swap topic so I'm starting this one.

I've only trapped board bows so I don't know if it's much different when getting into glass.  On board bows i trapped before final tillering at the belly.  What I'm wondering about is the ratio of the width of the trap.

Since reducing the sides of the limb has a linear effect on draw weight and reducing thickness is to a factor of eight,  how should the trap  e adjusted in ratio Since both width and thickness is being reduced.

Clear as mud, eh?

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Offline Crooked Stic

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Re: Trapping?
« Reply #1 on: May 01, 2022, 09:42:14 PM »
Do you trap to the belly or the front and witch is best. What degree do you use ----
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Offline Longcruise

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Re: Trapping?
« Reply #2 on: May 01, 2022, 11:12:18 PM »
I trapped the back on the board bows.
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Online EvilDogBeast

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Re: Trapping?
« Reply #3 on: May 02, 2022, 06:07:46 PM »
Would definitely be interested in a trap-along because I have never done it.

Online Pat B

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Re: Trapping?
« Reply #4 on: May 02, 2022, 07:37:21 PM »
Trapping a wood bow is used to equalize the tension and compression strain between the back and belly. For tension strong woods the trapped bow has a narrower back and wider belly. Generally most woods are stronger in tension than compression so the belly is wider than the back.
 With a glass bow I doubt there is much if any performance value to trapping. Probably mostly cosmetic.
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Offline Longcruise

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Re: Trapping?
« Reply #5 on: May 02, 2022, 09:08:17 PM »
What I see in ASLs with their thick cores is a tendency to follow the string (just as a board bow or bow from a stave).  I've   not seen this in recurves and RD longbows with their greater glass to core ratio.

So most, especially died in the wool Hill aficionados,  will dismiss that string follow or,  just as often, extoll it's virtues.  I don't fault that but as the bowyer I see it as something to be remedied. 

I've tried using thicker glass on the belly and that helps,  but given the thick cores of the ASL,  reaching an acceptable glass/core ratio while keeping belly glass thicker becomes difficult on bows over about 35# draw.

I did several with .030 core tuff under the belly glass and that's helpful but it seemed a little doggy 🐕

So now I'm looking to trap the back.   That brings us up to the concept of the ratio or amount of trap that it will take.
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Offline Mad Max

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Re: Trapping?
« Reply #6 on: May 03, 2022, 09:02:34 AM »
Trapping a wood bow is used to equalize the tension and compression strain between the back and belly. For tension strong woods the trapped bow has a narrower back and wider belly. Generally most woods are stronger in tension than compression so the belly is wider than the back.
 With a glass bow I doubt there is much if any performance value to trapping. Probably mostly cosmetic.

Trapping will remove poundage from the bow :thumbsup:
Any glass removed from the limbs end to end will remove weight.
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Online mmattockx

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Re: Trapping?
« Reply #7 on: May 03, 2022, 10:25:27 AM »
What I'm wondering about is the ratio of the width of the trap.

It can be from soup to nuts, there is no single number or approach you can take. I can calculate how much weight you will take off and how much stress/strain you will add to the back and remove from the belly for any given trapping shape, but there isn't one number that is always the magic solution.


Mark

Offline Longcruise

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Re: Trapping?
« Reply #8 on: May 03, 2022, 11:19:44 AM »
What I'm wondering about is the ratio of the width of the trap.

It can be from soup to nuts, there is no single number or approach you can take. I can calculate how much weight you will take off and how much stress/strain you will add to the back and remove from the belly for any given trapping shape, but there isn't one number that is always the magic solution.


Mark

So, are you suggesting that the only answer is empirical data addressing a huge range of materials and applications?
"Every man is the creature of the age in which he lives;  very few are able to raise themselves above the ideas of the time"     Voltaire

Online mmattockx

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Re: Trapping?
« Reply #9 on: May 03, 2022, 12:19:17 PM »
So, are you suggesting that the only answer is empirical data addressing a huge range of materials and applications?

No, just that if you want a specific result you need to design for the specific details of that bow. I can predict how much any particular trapping shape will drop the weight and change stresses but only for that one trapping shape. I am tied up at the moment, but can work up a couple sample examples later today and post them.


Mark

Online Kirkll

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Re: Trapping?
« Reply #10 on: May 03, 2022, 12:57:56 PM »
On glass long bows i always trap the belly side in a spiral shape. At the tips i get more aggressive with the tip section from the tip notch down about 6" at 20-25 degrees, then i gradually change the trap angle to 5-10 degrees back to the fades. This is mostly for lowering tip mass weight purpose, and the lower portion is weight reduction.... Often times I only trap the tip section. I just eyeball the whole thing, and they are all a bit different.... So i have no secret formula or magic numbers.

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Re: Trapping?
« Reply #11 on: May 03, 2022, 08:17:49 PM »
  Kinda do the same thing as Kirk... If the tips are a litle on the stiff side I trap about a third of the limb and maybe go half way...  Hardly ever more than that...

Online mmattockx

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Re: Trapping?
« Reply #12 on: May 07, 2022, 02:43:22 PM »
I promised a trapping calculation example and here it is:

First off, trapping is easier to calculate for a selfbow because the material is homogeneous and the stiffness the same throughout the limb thickness. I can calculate one cross section change and it will hold for the full length of the limb. FG limbs do not do this because the fibreglass is ~2.5-3 times as stiff as the core wood and the ratio of FG to wood changes down the length of a thickness tapered limb. If I really want a precise answer for a FG bow I need to calculate the section at several spots along the limb length to get an idea of how the results change along the limb.

This is the cross section I used for the example (belly side is the wider side):




The trap is given in terms of percentages of the width and thickness so that it applies along the limb as the width and thickness change. If you simply trap a 1/4" (or whatever) off the back width that width change is a different percentage of total width in every spot on the limb and the stiffness and stress changes vary much more than with a constant cross section based on percentages.

Because Longcruise asked about ASL and HH bows I chose a width of 1.5" and a stack thickness of 0.450" as the basic limb section dimensions. The results are specific to that combo and will not apply to other width and thickness combinations. I assumed 0.040" FG lams and hard maple as the core wood because I have a decent handle on how stiff it is (the modulus of elasticity, that is). Even better is to measure the MOE of a sample taken from the same board you made the lams from.

I like to leave a small flat on the sides because it helps prevent handling damage that can occur if you taper to a sharp point on the belly side. Rounding the belly corner is another way to keep that edge less delicate.

For the given cross section the limb will be 12.1% weaker than the full rectangular section. This means if you followed this section full length then your bow weight would drop by 12.1%. Stresses on the back will increase by 6.6% and be reduced by 6.6% on the belly. This may not sound like much, but often this sort of shift is enough to significantly reduce set. The more you reduce the back width, the greater the reduction in weight and shifting of stresses.

If anyone wants I can show the math that I used to get the results, but I figured very few are that interested in the fine details.


Mark

Offline Longcruise

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Re: Trapping?
« Reply #13 on: May 07, 2022, 05:22:19 PM »
I promised a trapping calculation example and here it is:

First off, trapping is easier to calculate for a selfbow because the material is homogeneous and the stiffness the same throughout the limb thickness. I can calculate one cross section change and it will hold for the full length of the limb. FG limbs do not do this because the fibreglass is ~2.5-3 times as stiff as the core wood and the ratio of FG to wood changes down the length of a thickness tapered limb. If I really want a precise answer for a FG bow I need to calculate the section at several spots along the limb length to get an idea of how the results change along the limb.

This is the cross section I used for the example (belly side is the wider side):




The trap is given in terms of percentages of the width and thickness so that it applies along the limb as the width and thickness change. If you simply trap a 1/4" (or whatever) off the back width that width change is a different percentage of total width in every spot on the limb and the stiffness and stress changes vary much more than with a constant cross section based on percentages.

Because Longcruise asked about ASL and HH bows I chose a width of 1.5" and a stack thickness of 0.450" as the basic limb section dimensions. The results are specific to that combo and will not apply to other width and thickness combinations. I assumed 0.040" FG lams and hard maple as the core wood because I have a decent handle on how stiff it is (the modulus of elasticity, that is). Even better is to measure the MOE of a sample taken from the same board you made the lams from.

I like to leave a small flat on the sides because it helps prevent handling damage that can occur if you taper to a sharp point on the belly side. Rounding the belly corner is another way to keep that edge less delicate.

For the given cross section the limb will be 12.1% weaker than the full rectangular section. This means if you followed this section full length then your bow weight would drop by 12.1%. Stresses on the back will increase by 6.6% and be reduced by 6.6% on the belly. This may not sound like much, but often this sort of shift is enough to significantly reduce set. The more you reduce the back width, the greater the reduction in weight and shifting of stresses.

If anyone wants I can show the math that I used to get the results, but I figured very few are that interested in the fine details.


Mark

Thanks for your effort and I am interested in your math.
"Every man is the creature of the age in which he lives;  very few are able to raise themselves above the ideas of the time"     Voltaire

Online mmattockx

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Re: Trapping?
« Reply #14 on: May 08, 2022, 11:48:45 AM »
Thanks for your effort and I am interested in your math.

I am trying to find an easy way to explain it and I can't. You really need an engineering background to understand where stuff comes from and how to calculate the cross section properties, then the changes are easy to calculate. What you are comparing is the moment of inertia (also called the second moment of area) of the original rectangular cross section and the MOI of the trapped cross section.

I use a section properties calculator built into Autocad to calculate the MOI of each section. It will calculate the section properties of any closed shape, so it is very versatile and fast to use. You can calculate the MOI of complicated shapes by hand but it is very tedious and slow. On top of that, the FG is much stiffer than the core wood so you also need to account for that in the section calculations. I tried to find a free online section properties calculator that will do the same thing but had no luck. There are lots of calculators online, but they only do simple shapes like rectangles, circles, triangles, etc.

Below is how things are calculated once you have the section properties for the untrapped and trapped limb sections.

For the above limb dimensions:

Ioriginal = 0.01898in4

Itrapped = 0.01668in4

For the original rectangular cross section the neutral axis (the axis which the section bends about) is mid thickness because the section is symmetrical.

For the trapped section the neutral axis shifts towards the belly. My calculator gives me the location of the neutral axis. In this case it is 0.01493" off of the mid thickness point. The percentage shift in the neutral axis gives the percentage change in the belly and back stresses. Because the NA shifts towards the belly in this case the belly stresses go down and the back stresses go up.

% change in stiffness = % change in MOI = [0.01668/0.01898] x 100 = 87.9% of the original stiffness

% shift in NA = [0.01493/0.225] x 100 = 6.64%  (the 0.225" is half the limb thickness, where the original NA is located)


I'm sure that is clear as mud without the calculations for the section properties, but those are not something I am going to show here. The concepts required to understand how to calculate sections and how to solve for composite beams (those with multiple different materials in them) really do require a number of engineering classes to have a good grasp of.


Mark

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Re: Trapping?
« Reply #15 on: May 08, 2022, 12:33:45 PM »
Exactly as I thought... :laughing:
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Online Kirkll

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Re: Trapping?
« Reply #16 on: May 08, 2022, 01:17:12 PM »
Trapping a wood bow is used to equalize the tension and compression strain between the back and belly. For tension strong woods the trapped bow has a narrower back and wider belly. Generally most woods are stronger in tension than compression so the belly is wider than the back.
 With a glass bow I doubt there is much if any performance value to trapping. Probably mostly cosmetic.

You are right on the money Pat. There isn't really any significant, or noticeable performance differences with the exception of limb mass reduction at the tips. But even then, it's just a few FPS, and possibly some vibration reduction on some limb designs.
It's mostly used for draw weight reduction on glass bows, and they are typically are trapped to the belly side on long bow and hybrid long bow limbs. With RC bows i only trap the tip section to follow the tear drop shape of the tip notches and reduce a bit of mass weight.

For the engineers in the crowd here that love the math, i'm glad to see you having fun with it. But i seriously doubt that many bowyers are going to get into such a precise analogy of something that has such a minimal effect on the bow. Your time would be better spent calculating energy storage, and how to successfully transfer said energy to the shaft..... Food for thought...     Kirk
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Offline Longcruise

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Re: Trapping?
« Reply #17 on: May 08, 2022, 09:05:44 PM »
Thanks for your effort and I am interested in your math.

I am trying to find an easy way to explain it and I can't. You really need an engineering background to understand where stuff comes from and how to calculate the cross section properties, then the changes are easy to calculate. What you are comparing is the moment of inertia (also called the second moment of area) of the original rectangular cross section and the MOI of the trapped cross section.

I use a section properties calculator built into Autocad to calculate the MOI of each section. It will calculate the section properties of any closed shape, so it is very versatile and fast to use. You can calculate the MOI of complicated shapes by hand but it is very tedious and slow. On top of that, the FG is much stiffer than the core wood so you also need to account for that in the section calculations. I tried to find a free online section properties calculator that will do the same thing but had no luck. There are lots of calculators online, but they only do simple shapes like rectangles, circles, triangles, etc.

Below is how things are calculated once you have the section properties for the untrapped and trapped limb sections.

For the above limb dimensions:

Ioriginal = 0.01898in4

Itrapped = 0.01668in4

For the original rectangular cross section the neutral axis (the axis which the section bends about) is mid thickness because the section is symmetrical.

For the trapped section the neutral axis shifts towards the belly. My calculator gives me the location of the neutral axis. In this case it is 0.01493" off of the mid thickness point. The percentage shift in the neutral axis gives the percentage change in the belly and back stresses. Because the NA shifts towards the belly in this case the belly stresses go down and the back stresses go up.

% change in stiffness = % change in MOI = [0.01668/0.01898] x 100 = 87.9% of the original stiffness

% shift in NA = [0.01493/0.225] x 100 = 6.64%  (the 0.225" is half the limb thickness, where the original NA is located)


I'm sure that is clear as mud without the calculations for the section properties, but those are not something I am going to show here. The concepts required to understand how to calculate sections and how to solve for composite beams (those with multiple different materials in them) really do require a number of engineering classes to have a good grasp of.


Mark

Well,  I thank you very much for doing all that.  With that,  I admit to having no ability to comprehend it.   And, that of course,  is on me,  not you. 

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Online mmattockx

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Re: Trapping?
« Reply #18 on: May 09, 2022, 01:33:32 PM »
Well,  I thank you very much for doing all that.  With that,  I admit to having no ability to comprehend it.   And, that of course,  is on me,  not you.

No worries. I was expecting to be able to find an online section calculator that worked like the one I have in Autocad. If I had found one then I could teach you how to do it without too many problems. It is the background knowledge to understand the section property calculations that is the big stumbling block. A calculator would make all of that unnecessary and let you just draw the shape and then plug the property numbers in to see how the new section compares to the old one.

If I ever come across such a calculator I will revisit this topic as I think it would be helpful for those tech inclined bowyers to be able to predict the changes. It would let you make changes with a decent idea of the outcome instead of just guessing and hoping.


Mark

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Re: Trapping?
« Reply #19 on: May 10, 2022, 12:56:49 AM »
  Questions, mm...

   Not to be a stickler but trying to comprehend as to what is exactly going on here... Since the limb width did not change I would assume the stresses on each side of the limb did not change... If this is so would it not be more correct in saying the back is weaker by 6.6 percent compared to it's limb thickness and of the previous rectangular section?? Or percentage wise the back has to handle the same amount of stress but at a decreased strength of 6.6 percent??

  Another question... Since the back is more stronger in tension vs the belly in compression wouldn't the numbers be different?? Wouldn't there be some kind of curve??  Lets say the back is 6% weaker and the belly is 2% stronger compared to thickness of limb.. Or could be the back is 2% weaker and the belly 6% stronger... I am not sure about numbers but you get the idea... At least, I hope...

  Another question...  Unidirectional glass, since tension trumps compression in this case wouldn't the neutral zone be located slightly towards the back in a rectangular section and moved more towards the center when the limb is trapped and material removed from the back??

 I hope I worded this correctly... I am not the best in writing down what is in my head...

  I also hope I read your posts correctly...

   

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