Wow, we're on this subject again. Must be the off season[&:]. I really don't understand how anyone can say that heavier arrows will always increase momentum and KE. When one keeps increasing the arrow weight, the closer that velocity approaches 0. There has to be a breaking point to where arrow weight will start to harm penetration. Momentum and KE just can't increase into infinity.

I am just thinking that because it is highly unlikely for the KE/momentum to peak, maybe efficiency will. I certainly expect this to be the case. If you look at the HYPOTHETICAL data on my previous post, you would see that at 435 grains, the gain in KE Peaks at 5# KE. That is how I would define efficiency. I hope this helps.

I think that they can. In fact, I am sure of it. But, you make an excellent point about penetration, so I have slightly changed the question.

WHAT IS THE MOST EFFICIENT ARROW WEIGHT FOR MY SET UP?

How can momentum increase infinitely? Momentum simply means "mass in motion". If the velocity is 0, the object's not moving-that object has 0 momentum. Your revised question, however, is a great one.

A few observations.

Kinetic energy and arrow mass. It has been stated that KE ALWAYS increases with arrow mass (weight). That statement didn’t make sense to me, so I did some math on the idea. A relatively straightforward calculation was to write an inequality that states the KE of a 400-grain arrow is greater than the KE of a 500-grain arrow (KE400 > KE500). Substituting the value of ½ mv2 in for KE one gets ½ mv2400 > ½ mv2500. Then solving the inequality for the velocity of the 400-grain arrow, one obtains v400 >root 5/2v500. Or in other words, if the velocity of the 400-grain arrow isroot 5/2times greater than the velocity of the 500-grain arrow, the 400-grain arrow will have a greater KE than the 500-grain arrow. In more practical terms – if the velocity of a 500-grain arrow were 280 fps and the velocity of a 500-grain arrow were 313.05 fps (280 times root 5/2) the 400-grain arrow would have more KE than the 500-grain arrow. End of story.

A similar argument can be made for momentum. However, the results are considerably different. For a 400-grain arrow to have more momentum than a 500-grain arrow, the 400-grain arrow would have to have a velocity 5/4 times more than that of the 500-grain arrow to have a greater momentum. In other words, the same 500-grain /400-grain arrow comparison – the 500-grain arrow with a velocity of 280 fps would have more momentum than the 400-grain arrow until the velocity of the 400-grain arrow is 350 fps!

Momentum and penetration. The importance of momentum relative to penetration cannot be overstated. From a pure physics standpoint, momentum is penetration. Momentum is, by definition, the resistance to the change on the motion of an object (actually that definition is for inertia – but momentum is inertia in motion). The more momentum an object has, the more difficult it will be to stop. The more difficult the object is to stop, the greater will be its penetration. Again, end of story, at least from a pure physics perspective. To say the momentum is the only argument to consider when developing a hunting rig would be silly. Parameters such as trajectory, broadhead, matching the arrow to the bow, FOC, bow tuning, the list goes on and on (this multitude of variables is one of the things that makes archery so much fun) are all significant considerations.

So, to answer the original question of what arrow will produce the most momentum is, to say the leaset,difficult to answer. Even answering the modified question of which arrow will be the most efficient is going to be difficult to answer, but perhaps easier than the momentum question. It seems the finding that ‘sweet spot’ for the bow, in terms of KE, is as reasonable of an approach as any I have heard.

Excellent post. About the most balanced perspective on the subject I've seen thus far.

KE goes up with the square of velocity. The 20# arrow, the velocity would go down compared to the 2# arrow. Velocity is the significant factor in KE, where momentum, it does not go up by the square of velocity, and mass is the significant figure. This is statics and dynamics 101 sylvan. Fairly simple stuff. Surprised your having such a hard time getting a handle on it.

O.K., here's an excerpt that explains why ke goes up with arrow mass. It's not my opinion, its archery mechanics. Look it up for yourselves. Bigcountry, you might want to pay close attention, there will be a quiz at the end of the class to see if you have grasped the construct.

Note the expression m / (m + v) used in the determination of bow efficiency. Note that mass and virtual mass are summed in the denominator thus forcing efficiency and of course ke to increase as mass increases.

You candisagree all you want but don't argue with me. I'mmerely giving you the established physics regarding this. Argue with the scientists and engineers that havedeveloped the knowledge over the last 400 years or so.

O.K. here goes....


BOW EFFICIENCY AND THE CONCEPT OF VIRTUAL MASS
Defining Virtual Mass[align=left]Starting with a stationary bow at the brace height position, as you draw it back you store energy in the bow. The area under the force draw curve equals the total energy stored (Et). When the arrow is released the bow ends up stationary again at the bracing height. The total energy stored has gone somewhere. Most of this energy ends up where you want it as arrow linear kinetic energy (Ea) but the rest (Ew)is wasted.[/align]The energy equation for the bow is thus:
Et = Ea + Ew
The energy efficiency of the bow (F) is defined as the ratio of the arrow energy to the total stored energy. i.e.
F = Ea/Et
Supposing the arrow leaves the bow with a particular speed (S) then you can write the total stored energy in the bow as being equal to some imaginary mass (M) travelling at the same speed as the arrow. i.e.
Et = MS2/2
The value "M" includes the mass of the arrow (m) with the remaining mass (v) called the Virtual Mass. i.e.
M = m + v
This gives you that Et = MS2/2 = (m + v)S2/2 = mS2/2 + vS2/2
mS2/2 is just the kinetic energy of the arrow and vS2/2 is the wasted energy Ew in tems of the kinetic energy of an imaginary Virtual Mass.

[/align]The bow energy efficiency F = Ea/Et thus becomes:
F = (mS2/2)/((m + v)S2/2) = m/(m + v)
The bow energy efficiency can be defined in terms of the arrow mass and the value of the virtual mass. What makes this useful is that it is found by experiment that for a given bow the value of the virtual mass is a constant over a sensible range of arrow mass.

Note that this expression for bow energy efficiency and the value of v being constant indicates that the heavier the arrow then the more energy efficient the bow becomes. We'll look at the practical effects of this later on.

================================================== ===================

This is a pretty abridged version. If anyone has a question or needs further explanation I'd be glad to help if I can but I'm not going to argue about this. The physics of archery has been well understood for the last 80 years or so, with vortex shedding being one of the last areas to explore. It can be discovered and understood by most everybody but debating fundamental concepts that have been well understood and established for many years is a waste of time.

Have a ball!

Oh yea, I should have added that the above is ALWAYS true. If the bow is capable of launching the spectrum of arrow masses you want to test, the efficiency will ALWAYS increase as the mass increases. The equation F = (mS2/2)/((m + v)S2/2) = m/(m + v) will always apply.

Again, I am really surprised your having so much trouble with physics and dynamics.

Why not just post the link to the work that someone else has done instead of stealing it as your own?

http://www.archery-engineering.co.za/conversions/boweffencyandvertialmass.htm

Steal it? I said it was an excerpt.

Here's the link I took it from






http://www.tipsforarchery.com/long20bow/BOW%20EFFICIENCY.php

here are some more






http://www.koniaris.com/archery/weight/






http://www.hesston.edu/ACADEMIC/FACULTY/NELSONK/PhysicsResearch/Archery/paper.htm

and one of my favorites...






http://www.goarchers.org.uk/mechanics/

They all explain basicall the same thing I've been trying to explain here. I'll give up on you though bc, I don't think you have the kind of attitude that will allow you to learn anything.Take some time and learn and spend less time telling others they don't understand. I'll bet thepublishers ofthe information I've presented here would love to be enlightened by you. I believe one of the sights has email so give it a shot and let us all know how you made out. LOL, they deserve a good laugh!


bc, please explain the effect of m on Fin the following equation.

The bow energy efficiency F = Ea/Et thus becomes:
F = (mS2/2)/((m + v)S2/2) = m/(m + v)

If you understand the math you will recognize easily that as m increases so does F. If you disagree with that you don't understand the math, if you disagree with the equation itself, you're not arguing with me your arguing with the established physics you claim to understand.