Primitive Archer
Main Discussion Area => Bows => Topic started by: DC on May 16, 2017, 01:21:13 pm
-
If I have a 40# bow, would each limb measure 20# separately? So for testing components like joints and siyahs I should stress them to half the draw weight? Plus a large safety margin of course.
-
yes
-
No, I'd test 'em to 40#
If you pull to 40# there is 40# of tension in the string and it's pulling on both limbs...
I'm not saying I'm right and I'll be only too happy to be shown to be wrong.
But there was a similar sort of question on another forum about if pulling 40# is there 40 on your fingers and 40# on your bow hand or 20# on each.... which is the wrong way to look at it.
I think you have to consider the entire system and not break it into parts.
Del
-
Del, it's mechanics, or physics. Two 20# forces added together equal 40#.
-
Yes and no.lol I guess ask a stupid question get a stupid answer. Just kidding DC. I would guess awfully close but that's just a guess. It seems like if it didn't it would rock in your hand but like I said I'm just guessing
Bjrogg
-
Del is right. If you want to test a handle joint on a 40# bow then that joint is holding 40#, not 20#.
-
OP did not say HANDLE joints. He said joints and siyahs.
Just to provide more reason to scratch one's head, if you string a bow with its string behind the string of a strung bow (><), each bow being 60#s pull, then pull them away from each other to full draw, it will only take 60# pull to do it. >:D
-
Touche', Mr. Davis
-
It's not as simple or obvious as it seems and it depends on how you do your testing.
If we consider a symmetrical bow pulling 40# the tension on the bow string either side of the hand is not 40# nor is it 20# because of the angles.
Thus you can't just hang a weight of 40# or 20# on the joint.
I've had a quick look at the physics, but can't be bothered to work it through.
I still don't claim to know the detail, but I do feel it would be prudent to test to full draw weight.
it's a bit like the long string vs braced bow thing.
I expect the full analysis of the mechanics taking into account the leverage on the siyahs and angle of string and limb etc would be quite tortuous.
For anyone who doubts that it is complex. Consider the case at brace... there is zero draw weight, yet the joints will be under considerable stress... so it's not simply down to draw weight.
To summarise:- I don't know, but I'm smart enough to realize that I don't know ;) ;D
Del
-
i wouldnt hurt to test at 40,,, I used to test the tips till they would break so I knew the limit,, :OK
-
Actually I was going to test to 60#. Three times load like strings. I just wanted to get a minimum. This is like the loop on a Flemish twist string. It looks like it would be half the strength but it isn't.
It also is like stacking scales. I put my small gram scale on my bigger gram scale and zeroed them both out. I put a 100 gr weight on the small one. I really didn't know what to expect. They both read 100 gr.
-
not stupid at all DC. I am surprised at some of the answers posted above. In theory, tension changes with string angle when the nocks are a fixed distance apart. Since a bow bends and the NTN distance changes, so perhaps the easiest way to find out wothout too much math, would be to put a scale in bow string and make a test.
http://www.williamhackett.co.uk/phpmedia/graphics/79acc09728a9669320a3de67d0db3d88.jpg
-
I've just some rough calculation assuming the string forms a 90 degree angle on the fingers and pulls on the tips at 90 degrees.
That would give the string tension and thus force on each tip of about 0.7 times the draw weight.
So for a 40# bow there would be 28# on each tip.
Now this doesn't even start to look at the leverage onto the siyah....
Del
(I still reserve the right to be wrong ;D )
-
Actually I was going to test to 60#. Three times load like strings. I just wanted to get a minimum. This is like the loop on a Flemish twist string. It looks like it would be half the strength but it isn't.
It also is like stacking scales. I put my small gram scale on my bigger gram scale and zeroed them both out. I put a 100 gr weight on the small one. I really didn't know what to expect. They both read 100 gr.
Don't you mean an endless loop?
-
Oh yeah :-[ :-[ :-[
-
Del, you are certainly right about the angles affecting the forces. And, like you, when it gets real complicated, I won't bother to try to work the math. I just go try something to see if it works.
-
You've never asked a stupid question DC. You ask stuff that I wish I could ask, but can't find the right words to put it. :)
-
It's tough to find the right words. My Mom was and old school English teacher. Maybe she did drill something into me. She would be pleased to hear that :D :D
-
I have clamped 1 limb (with siyahs) on the tiller tree and use 2 ropes and pulleys to keep the center of the string going down the center of the tiller tree.
For checking siyah angles, and glue joints.
That would be 20#
-
The one limb may be exerting 20# of force on the string but so is the other limb, I say go with 40
-
Stupid question two. What is siyahs? I would think that if you hang a bow string on a scale and pulled the bow to 40# . The scale would be holding forty pound and you would be pulling the same forty pounds. Is that correct? Arvin
-
Oh ! Siyah is one of those jacked up Japanese bows. Arvin
-
I would think that if you hang a bow string on a scale and pulled the bow to 40# . The scale would be holding forty pound and you would be pulling the same forty pounds. Is that correct? Arvin
Yes
-
A siyah is kind of a glued on recurve. It's mostly used in composite(horn) bows I believe. I just stole the idea to make a bow longer and recurved. It's been done before.
-
I have no clue how you will do the testing. Clamp the bow in a vice and attach weight vertically on limb or tip?
This is not how the real forces are working on a bow. In fact you get a changing momentum in the handle when a bow is drawn.
When you attach a scale between the two halves of a string, you will notice that string tension is highest at brace height (or near brace height). It is about 2,5 as high as draw weight and will decrease when the bow is drawn!
I would suggest: make all your parts as exact as possible and use good well mixed epoxi and your fine!
-
I was thinking I would make a mock-up of a siyah and it's "V" joint and do just as you suggested. Clamp it in a vice and hang weight on it. OS is incredibly strong and I thought that by doing this I could get a very rough idea how small I could go with these parts. At the moment I have no idea how small they could be. They may be 3 or 4 times too big and they may be entirely the wrong shape. I would rather break a couple of mock-ups than a bow :D
-
To do your math modeling, sketch a bow of the type you want to build at full draw. Fill in the major forces. The sum of all forces should equal zero unless the bow is being thrown. The force always acts on the bow in a direction. There are basically two forces on the bow, the bow hand force pushing the bow away from the archer at 40lbs and the string hand force pushing the string towards the archer. Now draw a free body diagram of the point where the string is being drawn. You have a 40 lb force pulling at zero degrees. You know that strings can only pull in the direction of the string so you have two other forces at other angles. For the sake of illustration let's say the angle is 60 degrees from horizontal top and bottom. So, the horizontal component of these forces has to be equal to the 40 lbs going the opposite way. The string leaves the draw point in two directions symmetrically which makes it mathematically easy. 2*20-40=0. However that is not the total force on the string. That is the force on the string in the horizontal direction. To find the force on the string draw a triangle with the horizontal leg being 20, the vertical leg being unknown. The hypotenuse is unknown in length as well however, the angles are 30,60,90 with the 20 lb leg going from the 60 to 90. From this we can use trig functions on the calculator to find out that the magnitude of the tension in the string is 40 lbs. now draw a new free body diagram at the tip of the bow to calculate the forces on the tip. Once you have these you can draw a free body diagram of the segment and work your way around the bow. The technique for doing this is a full quarter in engineering school. If you want to do this kind of research pick up an old book on statics at a university library and dig in. You'll want to have at least trig under your belt to get started. Also, the numbers I used came out even for illustration. A real bow would be far more difficult. I have the math to do this. I but never do. Archery is fun not school.
-
wow ???
-
wow +1
I understand about that kind of math but not how to do it. Would love to see you write an article for Primitive Archer working over the forces at work in the launching of an arrow from a primitive bow (no shelf please and not cut near center.) Article should be as plain spoken as the math allows and replete with illustrations. Or maybe write a follow up to Klopsteg's "Archery, the Technical Side."
-
I have the math to do this. I but never do. Archery is fun not school.
the math is interesting, insomuch as bows are a fascinating study in statics and strength of materiels. but as said above, not too much fun to do
now if Jim or anyone else, were to come up with an really interesting challenge for an article........any other suggestions out there?
Article should be as plain spoken as the math allows and replete with illustrations.
that's actually the hardest part.
-
We need an Isaac Asimov. There's a test for you all ;D ;D
-
I've just some rough calculation assuming the string forms a 90 degree angle on the fingers and pulls on the tips at 90 degrees.
That would give the string tension and thus force on each tip of about 0.7 times the draw weight.
So for a 40# bow there would be 28# on each tip.
Now this doesn't even start to look at the leverage onto the siyah....
Del
(I still reserve the right to be wrong ;D )
Del has this correct by my reckoning. The 40# is the horizontal leg of the triangle and the string is the hypotenus [spl?] So, you divide 40# by .717 to get ~57# and half of this comes from each string going away from the arrow so you get ~28.5# at the tip (This is also the Shearing force at the nock). If the string is at a 90deg angle with the tip, then the stress going down the siyah (which is effectively straight) is based on the moment within the siyah which increases the further down the limb you go so M = F x L. Then, the bending stress (V) is based on the moment and the section modulus ( S ) and V = M / S or combined V = F x L / S. Now, you also have the shearing stress (U) from the shearing force U = Fs /A. Finally the maximum stress is the combination of the bending stress and the shearing stress ( based on Mohrs Circle) stress max = (V^2 + U^2)^ .5 ( this is read "the square root of the sum of the squares").
For glued on siyahs you would want your glues bond strength at least 4 times as strong as the max stress to account for the shock load aspect.
Side note : section modulus is very specific to shape. For rectangular sections, S = width x height^2 / 6
Ken
-
wow +1
I understand about that kind of math but not how to do it. Would love to see you write an article for Primitive Archer working over the forces at work in the launching of an arrow from a primitive bow (no shelf please and not cut near center.) Article should be as plain spoken as the math allows and replete with illustrations. Or maybe write a follow up to Klopsteg's "Archery, the Technical Side."
If you mean after the string is released And the arrow begins to move and accelerates around the handle, that is beyond me. Engineering dynamics involves vector calculus. I can say that the energy is not released along a straight line and there is a lot of movement in directions other than what does a hunter or warrior any good.
-
WOW +2. (--) I use to be pretty good at math but gave it up when I started building bows. ;) :) :)
Pappy
-
Wow +3. The buck stops for me somewhere around 11th grade. LOL. Lots of props to you math guys though. ;)
-
What happened to "primitive"?? If I get my draw weight/length, and the bow doesn't break, and the arrow flies true, the world is right! And perhaps, there is meat tonight (SH) :NN :-D! Hmm" Apple smoked venison! I think someone is overthinking the process here, on the other hand, I am mathematically challenged and don't have a clue what has been said in that field. Good theory, though.
Hawkdancer
-
hang 2 weights from a string, you'll be pulling their weights added together. so the siyah is feeling the equivalent of it having a loop put on it and being pulled 20#. or I think it does... since they're at an angle it may be different? just need 2 scales to test it. tie a string to a scale that's tied to something, the other end somewhere else solid. pull on it on the center as if it was a bow with another scale. that or make a bowstring with a scale in between it (ie tied to a loop going on the bow's notch, and the hook tied to the rest of the string), and pull it with another scale.
but maybe it'll be more than that when the bow is released and then reaches brace
-
So... Just for fun, I grabbed a pencil and paper... This is what I came up with:)
I believe that when testing each limb separately, the only way to correctly simulate the internal forces acting on the wood fibers in that limb is to pull the tip in the exact same direction (at the same angle to the bow) as it would be pulled with when normally drawn with a regular length bow-string, and this is nearly impossible to keep track of by hand. On the first page there is a quick break-down of the applicable forces, and on the second an apparatus that I am of the opinion would be required to correctly test individual limbs by subjecting them to the same force they would undergo if the bow was normally drawn. Note that this device will only work on bows with non-bending handles.
Oh, and of course: I am not by any means an expert, there may be serious errors, I just took twenty minutes and put down some scribbles...
Enjoy!
-
The testing tool suggested is definitely on the right track. I would suggest being careful how the clamp is used because the wood could bruise or the jig could flex and give a false reading. It seems the key on the clamp would be to keep forces per unit area (stress) on the test piece similar to that experienced under duress from the bow hand. Perhaps a pair of rubber coated pegs an inch in diameter set up as a fulcrum and balance pin would make a good clamp.