Deflex

[Eric Garza]

To answer the earlier question regarding whether a bow with a 20# brace vs. one with a 30# brace would impart more energy to an arrow, here are my thoughts.  Folks are encouraged to pick this apart.

I would guess that both the 20# at brace and the 30# at brace bow would impart the same energy into the arrow.  

If we assume that two bows that pull 60# at full draw store the same amount of potential energy, they have to release the same amount of energy, since energy cannot be created or destroyed (i.e. First Law of Thermodynamics).  There are three places for this potential energy to go:  (1) Limbs, (2) arrow, (3) dissipated as heat due to friction.   When you draw the bow, most of the energy will be imparted into the arrow early in the release (i.e. the limbs move fastest soon after release, and are moving more slowly as they return to brace height).  Since the limbs are moving fastest earlier, they are turning the maximum amount of potential energy into kenetic energy, and this is where the arrow gets most of its momentum.  

Since we're assuming that the limbs in both bows have identical mass and if we assume an identical brace height, the amount of potential energy absorbed by the limbs should be the same.  Even though the limbs of the bow with a 20# brace height will be moving more slowly as the string hits home, the arrow already has most of its resulting momentum from the earlier stage of release, so this slower ending stroke doesn't matter.  Unless there are meaningful differences in friction, we are forced to conclude that the arrow's resulting kinetic energy must also be the same, or so similar that we probably wouldn't be able to see a difference with your typical backyard chronograph.  

Others' thoughts?

-Eric Garza

[Keenan]

 Thats the kind of disecting discusion that I was hoping this would spark.
 Kegan; I hope this is not view as a hijacking of your thread I am hoping this will really pull in some answeres and thoughts and theories.    Keenan

[JackCrafty]

Hmmmm....What, explain what energy is?  

If you're familiar with physics, this stuff is a piece of cake....but if you're using your intuition it's hard to understand.

The easiest way to think about how a bow works is to think about the draw in small increments.  If a bow is 30# (at brace) and you draw it back 1" then the energy is 30 (30x1).  If you then draw it back another inch the energy might be 33, etc.  If you add all the pieces together you'll see that a bow with a higher brace weight stores more energy.

[Keenan]

 Ok Jack so if that were the case, A bow with zero at brace would be (0x1) and (0x2) and so on   Sorry but It's my nature to think to the extream of theories.

 "If you add all the pieces together you'll see that a bow with a higher brace weight stores more energy"  I agree (stored energy) but is it transfered energy?

[JackCrafty]

You're right, a bow with 0# at brace will have near zero energy when drawn 1" (0x1)....but at 2" you would start to feel some resistance, right?  Lets say it's 1# @ 2"....so the energy would be 2 (not zero).  Lets say it's 2# @ 3", so the energy is 6 at that point, etc, etc.

Stored energy becomes transferred energy when you release the string.  Not all of the stored energy goes to the arrow because the energy has to move the bowstring, bow limbs, and even the air around the arrow, string, and bow.  There are lots of things that affect how the energy is transferred to the arrow....but again, the physics is a bit complicated.  Basically, MOST of the stored energy is transferred to the arrow (about 70%).

[JackCrafty]

When you think of energy in "pieces" you will also begin to understand that longer draw lengths also store more energy (and vice versa).   

[Gordon]

Keenan, Patrick is correct. The stored energy is expressed as the sum of each minute movement of the string.

[Keenan]

 Great explination Jack.
Now;  
 "There are lots of things that affect how the energy is transferred to the arrow....but again, the physics is a bit complicated.  Basically, MOST of the stored energy is transferred to the arrow (about 70%)."
 What if any difference is there if I shot a bow with a slack string  (unbraced) (what amount of absorbed energy) vrs. a braced bow at roughly 70% absorbed energy.  

[Keenan]

 Guys I totally agree with the thoughts and theories I'm just challenging our minds to examine things from a radical prospective.
 I wonder if I strung a bow and shot at 30 lbs. (normal brace) then took the same bow and shot it with a slack string (unbraced) but again pulled to 30 lbs. what would the difference in cast, speed and transfered energy?

[JackCrafty]

Hmmmm....a bow with a slack string is not a bow....it's a fishing rod.  That's the difference.

Actually, I'm not sure I understand the question.

I wish I could devote more time to this discussion but I've got to go. I'll be back later tonight.    Keep asking those questions....

[Eric Garza]

Ok, so after reading Jack's response I feel like backing my claim up a little better (i.e. doing a little self-dissection).

I'm not a physicist, but I do study energetics of natural ecosystems and human societies, so I'm familiar enough with energy to explain it a little more thoroughly than I did before.  What we're interested in is potential energy, because the potential energy stored in the limbs of a bow will end up as the kinetic energy embodied in the arrow.  There are several types of potential energy, including gravitational, electrical, chemical, and elastic.  We're obviously interested in elastic potential energy.  Elastic potential energy is calculated:

E = (λ χ²) /2 l

Where:
λ is the elastic material's modulus of elasticity,
χ is the elastic material's extension beyond resting, and
l is the elastic material's natural length or length when at a resting state

The bow's natural length represents the horizontal length of the bow from the point on the string where the arrow is nocked to the front of the bow's shooting shelf.  This is, effectively, the bow's brace height.  The extension beyond resting is the distance from the string at the arrow's nock when the bow is drawn to where that point sits in space when the bow is braced but not drawn.  These points equate to the braced bow's setup envisioned as if it were a spring, with the braced bow being the resting state.

Note here that the only variables that influence the potential energy at full draw are these two distances.  The wood's modulus of elasticity is constant (or at lease we're assuming it to be).  So this means that the potential energy generated by drawing the bow will be the same in both bows, and again since most of the arrow's energy will be given to it very soon after release the fact that the bow's tension at brace is different doesn't matter.  It doesn't factor into the equation.

Alternatively, one might argue that the bow's resting state is actually as an unbraced bow.  In this case there would be a difference in potential energy stored if the deflexed bow's tips were not reflexed enough to being them to the same plane as a bow without deflex.  I'm not convinced that this is the right way to look at the problem, though, as the websites I looked at to put this together all showed a braced bow when describing its resting state.  Just for argument's sake...

Overall, though, I'll stick to my guns.  There's no reason why a deflexed bow with lower tension when braced won't transfer just as much kinetic energy to an arrow as a bow that isn't deflexed, given the same limb, string and arrow mass, brace heights and draw lengths.  

All the best,

-Eric Garza

[Keenan]

 Very good food for thought  Eric. hopefully others will engage

[JackCrafty]

Can't resist....There is one main problem with Eric's theory:

Bows with different limb cross sections produce different amounts of energy (when shot) even if things like limb mass, draw length, final draw weight, etc. are identical between the bows.  This is the basis for the science of bow design.

[tom sawyer]

Eric, the flaw in your math is based on the fact that the equation doesn't account for the bow being braced.  A deflexed bow will generate less poundage at 1" of draw, than a reflexed bow.   The arrow gets pushed the whole time its on the string, not just at the instant you let go.  Thats why Jack is right about adding the area under a F/D curve to calculate stored energy.  It all adds up to the arrow speed at the moment it comes off the string.

Now a deflexed bow might have a similar efficiency to a reflexed bow, if its limbs are about the same geometry and mass.  But it won't store as much energy, so the arrow won't travel as fast as the one coming off the reflexed bow.

[Badger]

      The way I see it is that the limb cross section is only a design feature dictated by the design of th ebow. A bow with a lot of bending limb can use a lot deeper limb cross section than a bow with a short bending area. Bows should always be built at a high level of strain without going past their elastic limits. Wide flat bendy handle bow of long lengths woud not make much sense unless they were made from very light woods. Just as narrow thick cross sections would not make good sense used in a stiff handled, stiff tipped, r/d design. The limb cross section should always be dictated by the design and tiller of the bow to give the working part of the limb at least an 80% of maximum strain value. Steve