Stresses on a bow, reflex and deflex, wood capabilities

[beast]

I have been thinking a lot recently and want to share my thoughts, and what you know about the following topic.
Right so this is hard to phrase but il try.

lets for all of these examples assume the bows won't stack and all have pretty linear force draw graphs. also lets say they draw 50 pounds because that's a nice round number.

1. Say we have a normal d style bow, no reflex, no deflex, and no recurve. and we compare this to a reflexed bow, say reflexed 10 inches. you might say that is a lot of reflex but my question is why.

The reflexed bow will be doing a hella lot more work, as we know because of early draw weight. but because its drawing 50 pounds like the d style bow, the limbs will have to be thinner, again because its reflexed.

Now, we say the reflexed bow is under more stress, but arent both bows drawn out under the same 50 pounds? Doesn't that mean they are both under the same amount of tension forces and the same amount of compression forces at the belly?

SO if I'm correct, then both bows are now stressed the same. the only difference is that the reflexed bow is under more stress at brace. but i don't think that changes the performance much, does it? besides maybe making the bow take some extra set?

Let me know what you think.
oh, and I'm new to this forum so..

[RyanY]

I think it’s incorrect to say that the bow is stressed more as I understand stress as being how close it is to failing. If both designed for their respective designs and characteristics both bows could be stressed equally. The reflexed bow will be storing more energy through its draw as at each increment of draw length it will be a higher draw weight so even if the FD curves are both linear, the reflexed bow will have a shallower slope compared to the straight bow. If designed well the reflexed bow may not take more set. The trade off to get higher energy storage is needing more mass or a wider limb to achieve this. If my understanding is correct, there may be other factors of increased vibration or hysteresis as well. This is why massively reflexed bows don’t always outperform more mildly reflexed ones. At least that’s my understanding.

[beast]

but that is what i am saying ryan. why should we need wider limbs on reflexed bows if their draw weight is the same, because that would mean the tension forces on the bow are the same. comparing a d bow of 50 pounds draw weight and 50 pounds reflexed bow, the reflexed will be thinner limbs, and thinner limbs will be able to bend further and stack less, thus being able to be reflexed without problems. so my point is reflexing a limb doesn't mean we will make the bow stack, or at least if my logic is correct.

[RyanY]

Think of it as comparing amount of total tip movement between the two bows. So you your d bow and another bow that, instead of being reflexed, just pulls another 5” of draw. The longer draw bow will need to be thinner to accommodate the tighter bend and wider to have enough wood to store the same amount of energy. The longer draw bow will be heavier overall because it has to be disproportionately wider than thinner for enough wood to do the same amount of work. Reflexed limbs experience this in the same way except that extra movement is experienced when the bow is braced as opposed to at the end of the draw.

[diliviu]

Stress depends also (and more) on the degree of bending, regardless the poundage. I mean both bending degree and poundage influence stress individually, but bending degree more than poundage (like a 10% increase in bending degree increases stress of wood much more than a 10% increase in poundage). For the same 50 lb, at a higher bending degree the actual tension and compression forces on back and belly will be higher (even if poundage is the same).

[willie]

Sometimes technical discussions get difficult if specific terms are used that don't mean the same to  all.

Stress is understood in scientific and engineering as pressure. PSI, pounds per square inch in the US, is a common unit.

The reflexed bow will be doing a hella lot more work, as we know because of early draw weight. but because its drawing 50 pounds like the d style bow, the limbs will have to be thinner, again because its reflexed.

Now, we say the reflexed bow is under more stress, but arent both bows drawn out under the same 50 pounds? Doesn't that mean they are both under the same amount of tension forces and the same amount of compression forces at the belly?

For the same bend:   the tension and compression stress is reduced when the limb is made thinner. 

If you had chosen to make the recurve limb narrower to keep it at 50#, the stress would be higher because the original thickness will have had to bend further.

[bradsmith2010]

the reflexed bow is more likely to blow up,,

[beast]

thanks bradsmith.
dilivou, u said the stressses on a material are more affected by the degree of bend rather than the poundage. but like, how would that make sense. If I'm pulling any bow back 30 inches , 50 pounds of force, the bow is also dealing with 50 pounds of force.

[beast]

ALso, im not increasing the poundage here. like ryan said,  well think about reflexes like we would a normal bow with extra draw length.
Right then:
o and Ryan, you keep talking about the width of the bows. "and wider to have enough wood to store the same amount of energy."
brother we already have enough wood to store the energy, that's why where thinning it out so it would reduce the draw weight back down to 50.
a normal 50 lb bow at 30 inches, drawn to 35 inches, would reach 60 or 70 lbs or something like that in poundage.
so we would have to thin it back out.
where does the width come into play here

[beast]

Sometimes technical discussions get difficult if specific terms are used that don't mean the same to  all.

Stress is understood in scientific and engineering as pressure. PSI, pounds per square inch in the US, is a common unit.

The reflexed bow will be doing a hella lot more work, as we know because of early draw weight. but because its drawing 50 pounds like the d style bow, the limbs will have to be thinner, again because its reflexed.

Now, we say the reflexed bow is under more stress, but arent both bows drawn out under the same 50 pounds? Doesn't that mean they are both under the same amount of tension forces and the same amount of compression forces at the belly?

For the same bend:   the tension and compression stress is reduced when the limb is made thinner. 

If you had chosen to make the recurve limb narrower to keep it at 50#, the stress would be higher because the original thickness will have had to bend further.

willie mate, im not reducing the width of the bow, just the thickness. so then, the stress will be the same, because now the reduced thickness will have to bend the same distance??

[beast]

i dunno what just happened up there, I just tried to quote willie and ended up quoting myself.

[RyanY]

That description of stress sounds confusing to me. Really we have to think of wood as stretching on the back and compressing on the belly. Wood can only stretch to a certain degree before breaking. If the wood is thicker the back and belly will be forced to stretch/compress further than if thin. Optimally built bows are as thick as possible without taking much set as that cross section is less massive than a bow that is made wider and thinner.

Beast, it sounds like your question can be distilled down to, if we are pulling bows to the same draw weight, why do they have to have different masses? The reason is energy storage. 50# draw weight between two different bows can have totally different amounts of energy storage. Simply speaking we need more wood as a place to store that energy. This is probably the end of my understanding before doing more reading on these concepts. 

[beast]

thanks for replying ryan.
but lol
no that isn't the question I'm trying to get at.
look.
say we have a short bow and a longbow.
the shortbow will be thinner to reach the same draw weight because of how short it is, but wider so we can have in increase in draw weight without making the wood undergo more compression and mention forces. but in this situation im not trying to mess with width or length, because changing to many factors gets confusing.
now as I understand it all woods have modulus of rupture wich is how much tension a back can take before, rupturing.
now, we have a D bow that is 65 inches. it pulls 50 pounds. at 30 inches.
Now, without changing the width, we make the wood thinner. the draw weight at 30 inches decreases. right? this decreases the stress on the bow.
Ok well now I want a 50 pound bow, not some thin stick, so I would steam in some reflex, such that the draw weight bumps back up to 50 pounds at 30 inches. NOw, is this final bow going through more or less or the same amount of stress as the bow we started with

[willie]

Now, without changing the width, we make the wood thinner. the draw weight at 30 inches decreases. right? this decreases the stress on the bow.  correct, because you decreased thickness
Ok well now I want a 50 pound bow, not some thin stick, so I would steam in some reflex, such that the draw weight bumps back up to 50 pounds at 30 inches. NOw, is this final bow going through more or less or the same amount of stress as the bow we started with

it could be more and it could be less. it could also be the same if your thinning was done in a proper amount related to how much additional reflex you added

[Digital Caveman]

Here's how my intro to engineering class explained these terms: Stress = force per area = pressure, as mentioned above.  Strain refers to the change in length of a material as it is loaded.  Stiffness refers to how much force causes a unit change in length (the resistance to bending)  Strength = stress at breaking point.

I hope this helps.