Bowyers Question of Physics/Geometry

I think you're going bonkers mate.....

So, with the short string fitted the curve of the bow would be as in 'B' wouldn't it?

Extending the string would bend the bow further than point 'B' and possibly into the bow's non-elastic range????

Other than that I'm discombobulated..... But just call me John.

You are missunderstanding the Q.
Let me re phrase it.
Take two cases where the bow tips are bent to exactly the same dflection.
First with a v long string and spring balance.
Second with a short string (almost straight across the tips) and the spring balance.
Will the spring balance read the same in each case, my belief is it won't, but I cant to the analysis.

For reference B is the bracing height (e.g the normal strung state of the bow before the arrow is put on the string and the string drawn back.)
Del

It's the same phenomenon as if you had a taut steel wire 3' long clamped at each end between two solid supports, you would still be able to deflect it by a small amount. Whereas if the wire was 9' long clamped between the same two supports, once you took up the slack it would feel rock solid.
It's definitely to do with the angles...I'm sure pythagoras could expain it.
For small angle the cosine is approx 1 so the deflection doesn't move the bow tips as much as the same deflection would at the greater angle caused by the longer string but how does that show up on the spring balance...Ah ha, the balance is actually measuring force for a given deflection...oh my furry brain hurts.
I can see it intuitively but can't quite put my finger on it (story of my life)

Blimey I'm answering my own thread, should I give myself a GA?
Del

Oooohhhhhh yes I see what you mean now.....

To move the bow a fraction past its 'B' point with the SS only just moving the balance will only read a tiny amount due to the angle of the dangle....

Yes that's it.... you get a GA Del.....

PS I thought Bowyers was a maker of pork pies???

Arrrgh, it be Vectors, lad.

Vectors.

Nice graph, tad small but nice...
Ta
Del _{(note to self, buy bigger monitor and new reading glasses)}

Use the zoom tool in internet exploder... Del.... LOL

The same effect can be found in regulations related to lifting chains, the most efficient pull will always be at 90 degrees to the thing you are pulling

"Marking of accessories (reg.7(c))

67 The SWL of a multi-leg chain sling varies with the angle between the legs. This is one of the characteristics affecting safe use. Multi-leg slings are currently tagged to show the 3 SWLs corresponding to an angle of 45°, 90° and 120° between the legs, although developing European standards are likely to refer to the angle between a leg and the vertical."

http://www.hse.gov.uk/lau/lacs/90-4.htm

Anything beyond 90 degrees included angle results in a compressive force between the lifting points on the load, which increases the force on the chains. Very large angles can see the chains having to cope with more than the weight of the load.

As has been said, it's described with vectors, resultant forces etc.

Other applications are with cantilever brakes on bicycles, - and brake compensator arms on trucks I remember.

Consider the lines A and B to be in the X axis and the line of force to be in the Y axis. The force that bends the bow is in the X axis but the pull is in the Y axis.

Consider a taut string, as you pull it a small distance, the triangle formed by the string and the axes is a force diagram, the Y direction component is small compared with the X component, the resultant is the force in the string.

The more you pull, the greater the Y component for the same X component, until you reach the same angle and get the same 60# pull as the original.

Ta, I get it, but I s'pose what I really want some sort of rule of thumb (a petite archery joke there) which would let me predict draw weight from the longer string measurements.
Mind, bow length would come into it and umpteen other factors.
I s'pose it's what's called experience, and why making bows is easy in theory but tricky in practice.

I've just finished a silly bow which is what raised the question, the darned thing is a very very long, longbow 79".
It was almost impossible to get the darned string on it, but once it was on the draw weight was dissapointingly low.
Mind you still wouldn't want to try to stop an arrow from it! I shall post a little blog on it soon.
Del

This looks like an all British thread, am a Londoner with dual citizenship. I once read a lament that the old Longbows needed a pull of 150lbs, I think those Cheshire Archers at Agincourt must have been way strong.

The spring rate of the bow will determine the X component in the string, the angle of the string will determine the Y component, and that Y component is the initial force on the arrow. The length of the bow will determine the distance through which the force is applied to the arrow for a given angle. All this remembering that there are two sides to the bow.

Hope that makes sense.

Do you really want such a programme to compute the performance as a bow from the "long string" bending?

It can be done on EXCELL but it depends very much of the geometry of the wooden part. Its not very complicated but quite complex.

Let me know.

Thanks for the offer,
Now I'm convinced it's a real effect and I'm not going bonkers I can minimise it by keeping the 'long string' as short as possible in the early stages. (I can shorten it progressively by threaing it though a wooden toggle (like a wooden 8).
Also each bow is very different so the practical approach is prob' better.
Cheers
Del

I do not know whom you meant but I try to give an explanation as short as possible and since a picture is worth a 1000 words here is one on which I shall base the explanation:

Have first a look at the left part. At the start you deal with a STRAIGHT future bow so tha if you want to bend it you are obliged to have forces acting arallel to the symetry axis. Since you pull with one force you use two strings making an angle "α". The bow is loaded with 2 forces (components or projections of the string force) one act vertical and has a lever =h and the other acts horizontal and has a lever "v". At bthe start you MUST have a big "h" since "v"=0 thus the angle is big. and the forces are important.

Now situation No 2 you raech the good bend and put the final string it is the second situation on the left.

You put the string and now we go to the right part of sketch. On the string you will apply a force in the middle to further load the bow. At the very beginning the string and the force are perpendicular so that the angle between string and applied force is almost 90° so that the force induced in the string is very high as the sketch shows it.

If at the beginning (phase 1 left) you generated the bending with axial forces mainly now you obtain the moment mainly with the help of horizontal forces (components) multiplied by the "v" lever. Due to the angle even at small axial forces the component is VERY high and generate same moment with a definitely smaller force. The vector graph shows it.

I hope it is clear, if not I can add equations but I intended to make it simple at least at first.

Ta, good helpful diagrams.
Del

Thank you for the appreciation but I have the feeling that it did not satisfy your expectations. Why? What do you expected as information if not what I send?

Your explanation is fine and has helped convince me that I'm not imagining it.

I was sort of hoping for a condensed answer like 'the force shown on the spring balance will be directly proportional to the cosine of the angle of the string from the horizontal'.
Maybe that is the answer I dunno, I'm also sort of after an explanation which would be palatable to a layman at an archery club...they are not all from an engineering background (One nice old lady who keeps bees is having a go at bowmaking).

It's fundamentally my whole problem as an engineer, my instinct, visualisation and relatively methodical way of experimentation is good but my maths is poor.
Still not bad for a cat
Del

That is exactly the reason I did not use any equation of force projections or force combination, this is the reason I made a sketch because as is said a picture is more than 1000 words worth. I wanted to avoid any mathematical presentation since I remember that you already mentioned not liking those sophisticated presentations very much.

I thought that you were not satisfied since you did not consider it to be either a good answer or not coming from whom you expected to get the answer.

Now if you prefer bla-bla I can put it this form:

Due to the fact that at start you bend a straight piece of wood it can be done ONLY if the force is giving an important projection normal to the future bow axis. This is the reason you are obliged to work with strings making a small angle with the line perpendicular to the rod in its middle.

After the string is put in place, the bow being already bend, any force applied on the before mentioned spring will generate due to the small angle when the spring is deformed an important force in the string and further bend the rod via forces which are parallel to the chord and use as lever arm the distance between the fastening point of the spring and the fulcrum of the arc. In fact the bending force orientation and relative magnitudes are rotated with pi/4 (90°).

Is that what you expected? Is a picture not better?

You have helped to further my understanding which is all I ask _{but you sound cross with me...I shall go and hide in my secret cat nest (it's behind the curtain)}
Del

Del, for the layman, a picture is worth a thousand words, but a model is worth a thousand pictures.

Hang a weight (W) from a hook. Run another string horizontal from the weight to the spring balance. Pull the spring balance side ways, keeping that string horizontal. The force shown on the spring balance is the weight multiplied by the tangent of the angle made by the first string with the vertical. Tan 0 = 0, Tan 45 = 1.0. At 90^o the force is infinite.

This is a simplification of the real thing, because of the curvature of the bow, but it should demonstrate the principle.

Drawing a "free body diagram" is another way to understand it. Wikipedia has a short article on free body diagrams. They allow us to see all the forces on an item resolved into their x and y co-ordinates. You would draw one for each string length and possibly for pulled and non pulled positions. Then you could compare how the x and y components of the forces on each point changed under different circumstances.

I would like to address the issue of the draw force being low. You have an exceptionally long bow. If you have followed "normal proportions" (like tapering thinner away from the handle part) then you have reduced the stiffness at the ends too much and the portions of the limb closer to the handle are not storing energy.

If you want a good bow with a higher arrow velocity then you need shorter limbs wwith low limb weight. In short a laminated limb that combines materials that give you high energy storage with lightness.

Hello, hello, what's all this then?

I've never worked with longbows, but a long time ago with re-curves. We were shown a way of stringing , where you lock one limb at your ankle, so you can flex the opposite tip. Is it different on a longbow or is this an initial stringing that's part of the construction process, that doesn't need to be done every time?

Packdrill

It's the string angle. The extreme angle of the long string causes the bow to "stack" while the short string gives a mechanical advantage in early draw as each inch of draw only advances the tips a fraction of an inch. Same effect of a long draw on a short bow versus a long bow. When at full draw, the increased angle of the string, each inch of draw advances the tips an inch.

Reference "The Traditional Bowyer's Bible, Volume I" page 53.

"The Traditional Bowyer's Bible, Volume I" page 53.
I must read my copy more carefully.
Del

Just think in terms of mechanical advantage.

How far do you need to move the centre of the long string in order to get a 1" change in B? How far do you need to move the centre of the short string in order to get a 1" change in B?

Excellent, simple concise, easilly digestible, gives me a glossy coat and a nice cool nose
GA
Del

You may have a cool nose and a glossy coat, but I don't understand the explanation given...not that it is important that I do. Meow!

You were the 'you know who I mean' But the boat has sailed !
I triedto PM you but couldn't remember you user name cos of that daft slash in the middle
That's the odd thing about explanation isn't it? Sometimes you can spend ages trying to explain something then a smack round the head some odd comment or a different tack will make it all clear
Cheers
Del

In the diagram above, the only support is a fixed support at point C. A load P is applied to the long string as shown. In its final position, the string makes an angle α with the horiizontal.

The reaction at B is:

V = P/2 and H = P/2tanα

Moment at C = PL/4 + Ph/2tanα (where L is the length of the bow end to end)

This shows that the angle alpha has a huge impact on the moment at C (and the moment at any point between B and C).

If you bend the bow with a long string, then fit a short string between opposite ends, then release the long string, the short string will be pre-tensioned causing strain energy to be stored in the bow. The end of the bow will move upward and outward from point B as the short string stretches elastically. But the tension in the string will cause strain energy to be stored in the bow and the force P required to bring the end of the bow back to B will be a mere fraction of the original force used to bend it.

In fact, if you now apply a jack at the end of the string and increase its tension, the end of the bow will go back to B with no force at all applied to the string.

Now, isn't that PRRRR-ty?

Ta,
(but I think you've put that big red arrow the wrong way on the string _(joke)the pointy end goes away from the archer )
Del

Force [lbs] Distances[in] as origin the same as you did thus curve goes not through zero cross of both axis.

Dear Dell,

With the different informations I got from your descriptions I build up a programme to compute the force versus displacement as you presented in your graphs.

This is the result but not yet as good as I would like since I do not have the full geometry of the bows.

I would highly appreciate if you could give the dimensions so that I can check if my programme works within a correct limit of uncertainty/error. As you see the curve is not far from the one you gave but I would like to make it even better. When it will be ready I shall compute the possible shooting distances if you could give me indications about the arrow as mass and form.

The different presentations you made gave me the feeling that it could be interesting to make a bow myself and I would like to pre-design it based on your experiences.

I hope you do not mind to share your experience and expect the data:

- total length as straight

- length where it is thinner

- width at root and at point where the string is fastened

- thickness at root and at end

Of course it is not possible to compute all but I think that I could come near enough to the experimental result and avoid a too cumbersome adjustment or a failure.

Thank you for the kindness to send me the data,

Nick Name

Ok, I'll get some figures for my longbow.

This whole thing intrigues me, as you'd expect that as the draw increases, the geometry should increase the poundage and the limbs should increase it too, so you'd expect the weight to go up in some sort of exponential or square law, but it actually is surprisingly linear.

Here are the figures for the longbow, the first figure is at 8" draw, that's 8" from bow back to string, the bracing height is 7" (centre of bow to string)
Straight length70" , length of string 67.75" (difficult to give accurate figures as the ends are rounded)

I shall change to metric for bow cross section for convenience, it's difficult to measure accurately as it's an organic shape.
At limb root 35mm wide 30mm thick.
At the tip 18mm wide 16mm thick.
Arrow weight ~ 1.25ounces length 29" dia 11/32" range 220yards

Draw vs poundage figures

Del

FYI, here's a plot of your data:

I've added the point (7,0); data points have ±1 error bars. Trendline in red is linear, black one is 2nd order poly.

Ta, I have done it myself, but not the curve fitting, the prob with taking accurate measurements is two fold.
1. The parallax in the rig trying to read the scale, the spring ballance and hold 70 pounds!

2. If you hold a longbow at full draw it will lease a few pounds in a minute as the wood starts to do stuff (stretch, compress, relax...I'm not sure that anyone knows exactly what happens with wood.
Just for the sake of completeness, here's the longbow on a simple stick tiller.

The pic nicely shows the natural composite, sapwood heartwood.
BTW the central 4" or so is much thicker as there is a huge knot there, otherwise the bow would be built to flex along the entire length.
Del

Del, is that yew?

It most certainly is.
One of natures finest materials, alongside Bamboo, silk, linen, flint.
Del

You forgot alcohol Del......LOL

We have some form of yew here in Florida, also referred to as podocarpus?, but it's used as a shrubby ornamental or hedge. I notice that out west in the Ozark area, they have a tree, osage orange, aka bois de' arc that was used for bow making.

you mentioned the natural laminations of the sapwood and heartwood in the longbow. I don't know much about the long bow, can it be strung either side io counteract the wood taking a set or do the layers have to be oriented a certain way in relation to the pull?

How does the long bow compare to a recurve bow, in range , accuracy and rate of fire( flight? ,flinging?) ?

The white sap wood is on the back (side away from the bowyer).
Osage is recognised as a fine bow wood, but doesn't seem to be grow over here, well it's not a native. Almost any wood will make a bow of sorts, the longbow design is good for warfare when made of Yew, the native Americans used many different woods ,usually of a shorter flatter design (which is more efficient) often backed with sinew.
If you want to see some fine bowmaking with natuaral materials, sign up to primitive archer website.
Del

Thanks, will do.