Bow Index

[inksoup]

it is closely related to MOE and MOR values of woods.

MOE is modulus of elasticity: this number should be as low as possible. low value means wood can be easily bended.

MOR is modulus of rupture: this value should be as high a s possible. high value means wood can NOT be broken easily.

so the formula:

Bow Index = (MOR / MOE)*1000

(multiply with 1000 since the division number will not be easy to read    )

Example:

osage orange:

MOR: 18,650
MOE: 1,689,000

(18650 / 1689000 )*1000 = 11,042036708 ~ 11.05

any value between 9 - 13 is ok for the bow.

so...

for the bow wood choosing this information can give you an idea.

up to here i summarized from  http://www.wood-database.com/wood-articles/bow-woods/


now My Theory for the laminated wood bows...

each of the lamination level should be calculated as in the formula and take the percentage of the wood on the bow and sum them up.

let see:

so as an example i will use this  http://tradgang.com/noncgi/ultimatebb.php?ubb=get_topic;f=125;t=013128

it is ipe and hickory. i do not know which ipe and which hickory so i just used more common ones.

so ipe bow index with the formula is 8.01
hickory bow index with the formula is 9.35

assume that 75% ipe, 25% hickory used
8.01*0.75 + 9.35*0.25 = 9.01

so this combination is just ok for the bow making.

so friends... go and grab your calculator and start some math...  :D  because both math and bow making are fun!


have fun...

[Elison]

Very nice!

thanks!

[Crittergetter]

Dang it Ink, now I have a headache!   :confused:

[inksoup]

hahahaha crittergetter... get a beer yourself, then you will start feeling better  :D


there some more variables in laminated bows that i was planning to put in the formula but i guess that will do the job for a while....

[Buemaker]

Thanks but bowmaking is more fun than math.    

[LittleBen]

OMG I can't believe I just deleted all that! I'll try again in pieces.

Those who don't like math should close their eyes.

OK, first of all, I'm going to discuss calculating energy storage per unit mass in woods. The premise being that the best bow woods store the most energy with the least weight. This may not be the only concern, but this is in my opinion the most critical factor.

First lets see what variables we have to use. From wood database we can obtain:

MOR - Modulus of Rupture; a measure of strain at failure. units of force per unit cross sectional area.

MOE - Modulus of Elasticity; a measure of the stiffness on a material in the region of elastic deformation

Density - measure of mass per unit volume, SG is normalized to the density of water.

OK so before we go farther, let me mention that wood is not a simple engineering material for the following reasons at least:
1)non-isotropic (mechanical properties depend on the orientation of the material)
2)wood exhibits significant creep and stress relaxation
3)wood deforms plastically before failing, therefore MOR is always measured after the wood has taken maximum set!

We will use the following assumptions:
1)all data from wood database is taken in the longitudinal axis(along the grain)
2)quasi-static; creep and stress relaxation are ignored
3)MOR is proportional to stress at the elastic limit (shown as proportionality limit in the stress strain curve below). In other words, higher MOR always means higher stress at the elastic limit. Elastic limit and proportionality limit are not identical but close enough for this discussion.

With these assumptions we can make a ******qualitative estimate of energy storage *******

So what has Ink calculated with MOR/MOE? well given the assumptions I have made, Ink has basically estimated (only qualitatively, not quantitatively) the strain at the elastic limit.

So what do we need to calculate or rather estimate? We need to estimate the energy stored by the wood during elastic deformation. The energy is stored is roughly 1/2 (stress at the elastic limit) * (Strain at the elastic limit).

Since we already said that MOR/MOE is roughly the strain at the elastic limit, and that MOR is an estimator of stress at the elastic limit, the equation for estimated energy storage is :
Energy = 1/2 (MOR/MOE)*(MOR)

since our estimate is only qualitative (only for comparison between woods) we can ignore the 1/2.
Therefore:
Energy ~ (MOR/MOE)*MOR

to get Energy/mass ~(MOR/MOE)*(MOR)*(1/density)

so there it is ...

Where does this fall short? Well MOR may not be a proportional estimator of the stress at the elastic limit. It's probably lose enough that it will give you some info, but may not be accurate enough to determine the better of two woods which rate very closely.

Stress strain curve to be added momentarily.

[LittleBen]

[Buemaker]

I got it, or-- Think I'll have a double Mojito.  

[LittleBen]

[Roy from Pa]

I'm just gonna stick wif Osage:)

[takefive]

Wow, if I was still a drinking man it would take at least 3 vodka gimlets to get through this.  And I still wouldn't understand it.
Guess I'm stuck with the old math:  Osage=Good.  

[takefive]

Ya beat me to it, Roy.

[Mad Max]

I just have one question


It is possible to synthesize excited bromide in an argon matrix ?   :cool:

[Roy from Pa]

I do it all the time, Mark..  

[Mad Max]

Littleben's occupation is a Mad Scientist

   

[mikkekeswick]

The real problem I see is that all wood no matter what species varies so much from piece to piece that trying to nail down and give specific values to MOE, MOR, stress, strain etc isn't really going to give you meaningful numbers at the end. Too many givens!
A simply bend test will tell you all you need to know about any potential bow wood....granted you need to have a piece to test but still  

[Roy from Pa]

X'2 Mike.. Sometimes even the best looking piece of Osage has given me a loud cracking surprise on the tillering tree:)

[canopyboy]

Originally posted by mikkekeswick:
The real problem I see is that all wood no matter what species varies so much from piece to piece that trying to nail down and give specific values to MOE, MOR, stress, strain etc isn't really going to give you meaningful numbers...

Yup.

It's easy for us engineering types to overthink these things. Just read some I my older posts.

[Roy from Pa]

Never met an engineer I trusted...    Had one here a couple years ago to help me with my first glass takedown build. He had it between his ears, and talked a good game plan, but couldn't transfer it to his hands. I ended up trashing that bow.

[LittleBen]

I think Mike and Dave both make good points. And I wouldn't waste my time to determine a chart of "bow index". Also, the simple fact that Steve Gardner's mass formula works suggests that most wood is pretty close in terms of energy storage per mass. So my guess is if we bothered to calculate them all out, they'd all be remarkably close.

I do think, however, that there is meaningful information you can take from the empirical data we can get our hands on.

An example, let's say i built a bamboo backed osage 1.25" wide, and 62" long. The bow turned out well, and I liked it. I have a piece of clean yew and I want to make a bow of basically the same length and design and weight as I did with the osage, but using my nice new piece of pacific yew.

I would log onto TG and say, hey mike, how wide should my awesome yew bow be? Since no one has bend tested my osage or my yew, we're basically just going on people's experience. And probably if 10 people responded I would get 4-5 different answers.
I could alternatively log onto wood database and I could see that osage and yew  end about the same amount before breaking, and osage is about 20% stiffer than yew.
So I could just start building the yew bow 20% wider than my osage bow which would be 1.5" wide.

Is it perfect? No. But there's definately meaningful information to be taken.

Osage and Yew are also well known, so what do I do when I wanna build a bow with a shedua belly? I've got to start somewhere, and there may be no one to ask.

So bottom line, if you can bend test ... Great, but if you've got nothing to spare, better to start from the wood database data than a total guess.