We list our own tests and you say "Read Ashby's reports." You sound like a broken record.

ORIGINAL: Straightarrow

ORIGINAL: Straightarrow

You call what you did a test? That is simply a joke, pure and simple. Ashby is the only one who has done relevant tests with penetration on animals and scientifically recorded the results over a long period of time with hundreds of shots. Yet you want us to believe your attempt at shooting foam is a significant test that overrides his - well it ain't going to happen.

Well, I've beaten this horse enough and it's past dead and decomposed. I'll bow out of this and leave you thinking your fast arrows are just as effective. You can continue to think Ashby is a joke who has discovered nothing with his many tests - with no evidence to the contrary.

ORIGINAL: davepjr71

You crack me up.

You have no proof that the heavier arrow is going to do more damage but from one guy's experiment and swallow it whole like it's from God himself. Dead is dead, period. bottom line. End of story.

ORIGINAL: Straightarrow

READ ASHBY'S REPORTS!!!!
READ ASHBY'S REPORTS!!!

Why don't you get out and do some testing yourself instead of sitting in the lab relying strictly on someone else's experiments. We list our own tests and you say "Read Ashby's reports." You sound like a broken record.

ORIGINAL: quiksilver

So, I guess when you consider that, if you're shooting two otherwise identical arrows (same head, same speed, same weight, same friction coefficient on the shaft, same fletchings) - and arrow A is a lincoln log (diameter .5"), while Arrow B is pencil thin (diameter .3") - you're probably going to get better penetration with Arrow B, simply b/c the force is transferred over a smaller surface area.

Yes.

Same goes with broadheads. If you have two identical arrows (same speed, same weight, same fletchings, same diameter, same friction coefficient on the shaft) - and you put a 125 grain muzzy on shaft A,while on shaft B, you put a 125 grain mechanical head with a cutting diameter of 2.75" - you're bound to get much better penetration from arrow A (provided that the blades are the same thickness/sharpness).

Yes.

So, while a KE calculation is nice and handy, there are hiddenpenetration killers lurking (fat shafts, shafts that "drag" more than others once moving through the target, broadhead selection, etc..) Unfortunately, these are things that can't be easily quantified by a quick KE or Momentum calculation. Instead, we resort to field testing.

Yes.

Good post and good discussionthough.

ORIGINAL: gators37

There is another item that I am interested in (which was briefly mentioned in some of the previous post)andwill be specific to a particular bow and that is if the bow "transfer to arrow" efficiency is indeed a curve (to a point where the slope changes which should indicate max. efficiency) or does the energy efficiency in the transfer just keep going up as you make the arrow heavier and heavier.

ORIGINAL: doublelunginem

Bowhunter44, Thanks for the terminology...it's all coming back to me...lol
So the KE would have to be a parabola since it is the integral of the draw force curve or the area under the curve.

Actually the graph of KE is parabolic because it is a quadratic function (the velocity squared). The integral of the draw force curve (yes the area under the curve) represents the total potential energy stored in the bow at full draw.

Just trying to refresh here, so how would the momentum graph reach a vertex (which we both agree that it must) since it is a linear function. I guess my thoughts are leading me to the fact that in this situation of dealing with bow performance, momentum would have to be graphed in relation to KE. As I stated before, when we get to the vertex of the KE graph, we are dealing with the maximum efficiency of the bow in the test.

Where the maximum effeciency of the bow is realtive to the vertex of the parabola of the KE graph, I honestly don't know. I don't have an intuative feel foreither. Great questions! Arthur, can you help out?

Once this is fixed, velocity decreases exponentially as weight goes up. So would this mean that the graph of the momentum be a " / " with half of an upside down parabola following it and connecting at the highest point?

I doubt it - unfortuantely. I suspect if that were to happen it would be serendipitous and not out of necessity.

The 2nd half would resemble y= 1/x ? Or would it be y = -x? Dude, you ask great questions!!

What I am attepting to prove with out the math proplems or actual test results is that momentum and KE are so closely related that they would maximize at the same point. If infact, velocity is decreasing "exponetially" once the bow has maximized efficiency and KE is maximized, then momentum would have to decrease exponentially at that point as well.

Again, I doubt that is necessairly true. Instead of the decrease of momentum being a graph of P = 1/mv it could be P = -mv, simply a negative slope of the linear momentum graph (P = mv). But again, you ask good (tough to answer) questions.

P.S...I had a physics and chemistry teacher simular to you in HS....he's the one that got me so interested in this stuff. We had a lab every week and it was all real life scenerios. We would spend Monday learning the physics and the rest of the week performing the physics that we learned...Lab report was due by end of class on Friday for a grade. He never taught from a book and never helped with the Labs...you had to listen on Monday very closely and figure it out on your own. He was the BEST DAMN TEACHER I ever had!!!!

Well, I doubt I'm the best damn teacher around. But it sure is nice to hear that some of us do make a difference. Obviously your physics teacher did (make a difference) as your recall of physics is great and you analytical skills are great as well!

ORIGINAL: bow_hunter44

Actually, the source of the information is irrelevant. Or in other words, even if the source of the information is god, the fact of the matter remains that the laws of physics are the laws of physics. End of story.