Author Topic: a tillering formula ?  (Read 1695 times)

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Offline dragonman

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a tillering formula ?
« on: July 13, 2009, 03:59:24 pm »
Here is a formula that I use  when making bows ,that I discovered my self!!! It is only a rough guide but I have found it quite usefull. It comes in after the floor tillering stage. If you make a string the same length as your bow (ntn distance) and start bending it on the tillering tree, then the weight needed to pull it far enough to put a real string on (at a 6 inch brace height), will roughly equal the bows weight at full draw with the real string, with a few pounds extra, to give room to adjust the tiller. I find this stage critical.  This helps me to avoid  bending the bow when it is too heavy thus not over straining the wood and to get the bow dimensions as close to final  before bending the bow. I would hate to be responsible for anyone ruining a bow so please dont rely on this (  not that anyone probably would ) But I would be interested to get some feed back if it has any value or not and maybe I could have  added something  usefull to the bow making world
'expansion and compression'.. the secret of life is to balance these two opposing forces.......

Offline Pat B

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Re: a tillering formula ?
« Reply #1 on: July 13, 2009, 04:30:18 pm »
Sounds interesting. I'll check it out on my next project!  Little tricks like this can make it easier for the newbies and us old timers alike.
Make the most of all that comes and the least of all that goes!    Pat Brennan  Brevard, NC

Offline woodenwonder

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Re: a tillering formula ?
« Reply #2 on: July 13, 2009, 05:47:35 pm »
With the string as long as the ntn bow, you could develop a formula for the long string pull Lenth that would equal a 6 inch brace. I'll see if I can lay that out on my CAD program.

Offline knightd

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Re: a tillering formula ?
« Reply #3 on: July 14, 2009, 08:18:54 am »
I find this to be true as well..