Sport Archery » Archery Arrows » Trade off between mass and velocity

Trade off between mass and velocity

Question:

Maybe I’m an empiricist at heart, but I would have thought that the *only* accuarate way to settle the issue would be to put a statistically significant number of shafts into a suitable target medium… I can’t think of any practical way to do this without laying out serious money, though. A web survey, perhaps? Have hunters details the equipment used for pass-through and non-pass-through shots? But you have the trouble of the demographic (you’re pre-selecting for web-anabled hunters who give a sh*t :->)… My comment about resistive forces was more in the way of viewing the target as medium a shaft passes through, rather than a surface it hits… The point (no pun intended) being that if the shaft passes through, it doesn’t give much of an impulse to the target… so the energy loss can be viewed as frictional (or drag through the medium?). So the projectile passes from one medium (air) into another (the target)… this starts to sound a little like optics… refracting arrows anyone…? – Hide quoted text — Show quoted text ->I think it’s a matter of elasticity and the impulse needed to break >something. > Indeed. The characteristics of the target are very important. >If (in the example) you don’t overcome the breakage energy >of the glass, it doesn’t break (ever thrown a stone at a window when >it didn’t break? It just bounces off…). > That’s right.  But when you give the stone the proper combination of > mass and velocity … bye-bye window pane. > Notice the requirements:  mass and velocity.  Same requirements for > both kinetic energy and for momentum. >On the other hand: grains of >sand don’t need to break for you to penetrate a layer, > Hmmm.  What’s a "layer" of sand grains? > I could also say that you don’t need as much energy/momentum to chip a > bit of glass from the window pane as you would to shatter the thing. > Would that be an equivalent "layer"? >you just need to push them out of the way… > Oh, yeah.  As you would the shards of shattered glass … after you’ve > applied enough energy to the pane to pulverise it. >So a projectile could have enough >energy to push through a layer of particulate material, where it would >bounce off a solid layer. > The particles in a sand pile aren’t so tightly bound to each other.  A > bit of weak electrostatic attraction, perhaps, or the pressure of the > sand above. >Just to complicate matters, fluid layers can behave differently >depending on circumstances (bullets bouncing off the surface of >water… that sort of thing). > Indeed.  This is where the quality of "vector" comes in, but we’ve > pretty much ignored that in the discussion and assumed a head-on > collision between arrow and target. >I guess a practical example of the effect of the mechanical properties >of the target material is backstop netting. Hang it taught and arrows >pass through. Hang it loose and you don’t get pass-throughs. Um, >energy re-distribution over a larger area? > Yep.  Impulse.  Target moves a bit and extends the impulse time. >One question I haven’t seen addressed is the resistive nature of the >target medium. The forces produced will (I would assume) be varying >with the velocity, or maybe even the square of the velocity. So a >slower moving shaft, will experience less resistive force, and also >has a greater inertia to overcome in bringing it to a stop…. > The inertia of the target is the same regardless of the speed or mass > of the arrow.  The inertia of the arrow doesn’t depend on the target. >Didn’t someone (Andrew Middleton?) do something with shooting into >sides of beef as an experiment…? > I can’t imagine that it would be an accurate experiment, since a side > of beef has bones and meat in varying proportions in varying places. > How could he guarantee that each arrow he placed met with exactly the > same type and density of material as the previous one?  Might be a > great practical experiment too see if a particular bow/arrow setup has > "the right stuff todo the job", but it wouldn’t be a reliable means of > measuring something like the penetration of light arrows vs. heavier > arrows or accurately calculating penetration based on different KE and > momentum figures. > Cheers, >> I LOVE it when someone takes up discussing with Joe in a logical and intelligent way – we get to read a VERY good exposition and see thought >> processes in action from both sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>. >> Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart enough to make a comfortable decision on some of >> this stuff so it’s fun to sit in the peanut gallery and try to follow along. >> Joe and Wayl, thanks!  🙂 >> hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his point, I also tend to the Wayl arguement about the >> displacement issue being identical for both materials.   Joe, what am I missing? >> thanks… >> > >> If a light arrow and a heavy arrow are launched from a bow with >> > >> identical energy-transfer characteristics for both, the lighter arrow >> > >> will travel faster.   Of course it will decelerate faster when it >> > >> hits. Going from 200 fps to zero is "decelerating faster" than going >> > >> from 100 fps to zero in the same time interval. >> > >> But if both arrows have the same kinetic energy and cutting properties >> > >> (blade sharpness, surface area, ruggedness and shape of the broadhead >> > >> and shaft), why would a lighter arrow penetrate less than a heavier >> > >> arrow? >> > >Wayl >> > >What’s needed here is to understand the difference between momentum and >> > >kinetic energy re arrow penetration. Although they are related they are >> > >not the same thing anymore than say volume and mass are not they same >> > >thing (though related). >> > Not a good analogy.   Volume exists regardless of mass.  Volume is a >> > measure of space, alone.  Any amount of mass may occupy that volume — >> > or not (there can be a complete vacuum within a volume).  I suspect >> > that you are thinking about one of the physical properties of a gas. >> > Regardless of the mass of the gas, it will "occupy" the available >> > volume of a container. >> > Kinetic energy and momentum are both measures of mass and velocity, >> > nothing more.  They are just calculated differently. >> > >Suppose you have a target made up of a whole series of glass sheets with >> > >air gaps between them and you shoot an arrow at it. >> > Which pretty much describes the pile of sand you mention later:  glass >> > sheets with air gaps in between.  The primary difference is the size >> > and shape of the grains (sheets) and the volume of the air gaps.  Even >> > the material (silica) is the same for both! >> > >The arrow hits the >> > >first sheet – what determines whether the arrow breaks through this >> > >sheet is the arrow momentum, not it’s kinetic energy. >> > Wrong. If the arrow has enough kinetic energy to continue moving >> > forward, then it also has enough momentum, or vice versa.  Drop the >> > KE, and you’ll also drop the momentum. >> > Here’s the verbal definition of momentum:  "the property of a moving >> > body that determines the length of time required to bring it to rest >> > when under the action of a constant force or moment". >> > A mass of two kilograms moving along at five meters per second will >> > have a momentum of ten kilogram-meters per second. >> > This means only that in order to bring the object to a full stop, an >> > opposing force of ten kilogram-meters must be applied for one second. >> > That force may be the instantaneous or cumulative effect of friction. >> > Instantaneous — arrow hits a dense, fixed object.  Cumulative — >> > arrow penetrates into or through a sand pile or a torso. >> > >If the arrow has  enough momentum it goes through if not it doesn’t. >> > Same with the sand.  If the arrow has enough KE (or momentum) to move >> > the first grain of sand, it gets to try it again on the next grain, >> > and the next, and so on until the kinetic energy is dissipated >> > (transferred to the glass or the grains of sand). >> > >Let’s say it does. In breaking through the sheet the arrow loses energy. >> > Same with the sand pile.  The arrow loses some energy when it hits the >> > first grain of sand (or hair, skin, muscle, bone and other tissue >> > cells, to bring the analogy closer to the NG theme).  It continues >> > transferring energy to each subsequent particle (cell or grain) it >> > encounters until it comes to rest. >> > Perhaps you’re thinking that the arrow’s act of "breaking" through the >> > glass sheets is fundamentally different from "penetrating" the sand >> > pile (or mass of tissue).  Not really.   Some of the kinetic energy of >> > the arrow is transferred through friction into breaking the molecular >> > bonds of the glass sheet.  In the case of the sand pile, molecular >> > bonds may not be broken, but the arrow’s kinetic energy, through >> > friction, becomes the kinetic energy of the sand particles (they are >> > moved aside). >> > >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops. >> > Exactly. >> > >The arrow hits the  second sheet –

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Response:

>uhhh, what do fig cookie manufacturering equipment repairers have to do with it?

Well, the maintenance staff of this purveyor of fine confections earn wages or pay taxes that fund the purchase of authorised and qualified textbooks on Newtonian mechanics, which their children may then ignore in favour of this week’s Top Ten Computer Games. – Hide quoted text — Show quoted text ->("And I’d better shut up, anyway, before this turns in to an >unauthorised, somewhat unqualified and certainly unsolicited textbook >on Newtonian mechanics.")

Response:

– Hide quoted text — Show quoted text ->> If a light arrow and a heavy arrow are launched from a bow with >> identical energy-transfer characteristics for both, the lighter arrow >> will travel faster.   Of course it will decelerate faster when it >> hits. Going from 200 fps to zero is "decelerating faster" than going >> from 100 fps to zero in the same time interval. >> But if both arrows have the same kinetic energy and cutting properties >> (blade sharpness, surface area, ruggedness and shape of the broadhead >> and shaft), why would a lighter arrow penetrate less than a heavier >> arrow?

Maybe it helps to consider what momentum & kinetic energy are "for". Conservation of momentum: http://theory.uwinnipeg.ca/physics/mom/node3.html As far as collisions go, an arrow striking a target is a pretty inelastic collision (bouncers not included). http://theory.uwinnipeg.ca/physics/mom/node5.html So, the momentum of the combined arrow+target after the collision will be conserved (until it hits the problem of moving the world to which the target stand has been pegged, which had been equally affected by the launch of the arrow in the first place). So, the energy delivered to the target will be greatest when the KE is highest. a good e.g. is at: http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Energy/KENOT… As Wayl points out, essentially the only force that we need to deal with penetration is friction. Since this is a force F=m.a=kg*m/(s*s) What we are interested in is the displacement of the arrow after the force begins to work on it… Definitions: KE=0.5*m*v*v = 0.5*kg*m*m/(s*s) Work=Force*displacement*cos(theta)=kg*m/(s*s)*displacement*cos(theta) Since the arrow generates frictional force at 180 degrees to it’s direction of travel we can ignore the cos(theta) bit. Comparing units you can see that the units for KE are the same as the units for work. KE=F*m Since we are going to arrive at rest KE will equal work done. In arbitrary numbers: For a KE of 1, with a Frictional force of 1 1=1*displacement i.e. penetration is 1 if you double the velocity KE=4 4=1*displacement i.e. penetration distance is 4 There are much more technical desciptions of the physics at: http://www.physicsclassroom.com/Class/energy/u5l1a.html Does that help?? Andy — # Microbiology: staph only. Your numbers are: 1 1 2 2 10 75  (+-/*) Target:603

Response:

uhhh, what do fig cookie manufacturering equipment repairers have to do with it? (" And I’d better shut up, anyway, before this turns in to an unauthorised, somewhat unqualified and certainly unsolicited textbook on Newtonian mechanics.") – Hide quoted text — Show quoted text -> I LOVE it when someone takes up discussing with Joe in a logical and intelligent way > – we get to read a VERY good exposition and see thought processes in action from both > sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>. > Thanks for the compliment.  It’s fun to argue rational points in a > rational manner.  If I’m proven wrong about certain points at issue > (which is quite possible, anytime), then I get to learn something new. > Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart > enough to make a comfortable decision on some of this stuff so it’s fun to sit in the > peanut gallery and try to follow along. > Unnecessary modesty.  If you’re having fun "watching from the peanut > gallery", then you most certainly are smart enough to make a decision, > even if it’s the wrong one! :] >Joe and Wayl, thanks!  🙂 > My pleasure. >hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his >point, I also tend to the Wayl arguement about the displacement issue being identical for >both materials.   Joe, what am I missing? > Joe will surely give a better explanation for his choice of > hypothetical targets, but I’d like to take a guess, and also take the > chance to elaborate on my point that kinetic energy and momentum are > equally "important for penetration". > Being a rigid material, the glass panes will have a measureable, > discrete figure for the impact energy required to make it shatter. > So, for a certain sheet of glass, we can say that the energy from an > arrow that has a momentum of 1000 gram-centimeters per second will > shatter the glass.   To get this figure, we can choose a hypothetical > arrow of 100 grams travelling at 10 centimeters per second.  100 grams > (mass)  x  10 centimeters per second (velocity) = 1000 > gram-centimeters per second (momentum). > Double the velocity of the arrow to 20 centimeters per second, and you > get twice the momentum, 2000 gram-centimeters per second.   If an > arrow travelling at 2000 g-cm/sec clears the first sheet of glass and > isn’t disturbed by flying shards or any other obstruction, then it > will have enough momentum to break the second sheet of similar glass. > The calculation follows a nice, linear path.  Twice as much velocity > equals twice as much momentum equals sufficient force to break TWO > sheets of glass.   But wait!  What happens when you use the basic > kinetic energy equation?  (Just for simplicity, I’ll drop the fraction > "1/2" in the equation KE = 1/2 m x v^2 — it doesn’t really matter for > this discussion.) > Double the velocity and you get a quadrupled figure for kinetic > energy. > 100 grams x 10 cm/sec x 10 cm/sec = 10,000 gm-cm^2/sec^2. > 100 grams x 20 cm/sec x 20 cm/sec = 40,000 gm-cm^2/sec^2. > This doesn’t mean that kinetic energy is either less or more important > than momentum, it just shows that the figures used to represent one > label follow a different rule than does the other. > The only thing we’ve done is double the velocity, for BOTH equations. > We’ve done nothing other than double the velocity of the arrow, and we > get two drastically different numbers for momentum and kinetic energy. >  But it’s the single change in velocity that produced those two wildly > different numbers, not the other way around. > People are accustomed to thinking arithmetically, so the geometric > (logarithmic) progression of numbers looks bizarre.   If people were > trained to think only geometrically, the linear (arithmetic) > progression of numbers would look bizarre. > But the arrow behaves the same, regardless of whether you use KE > figures or momentum figures to describes its actual performance. > Why the difference between KE and momentum?   Momentum is a measure of > mechanical force (any action that tends to maintain or alter the > position of a body or to distort it).   The momentum equation answsers > the question "How much force (work) must I apply to this thing to make > it stop moving?" > Kinetic energy is a measure of the  energy (the *capacity* for doing > work) of motion.  Work is done when energy is transferred from one > body to another, moving it or deforming it.  As I’ve said already, > BOTH momentum and KE are functions of mass and velocity, nothing more. >  For any given arrow mass, the potentional energy of the bow gets > transferred into the kinetic energy of the arrow, which is then > tranferred into the motion or deformation of the target.  The kinetic > energy equation answers the question "How much capacity for work do I > have in this moving object?" > The questions look quite different, but in a discussion of arrow > penetration,  the questions and answers are essentially the same; > they’re just expressed in a different dialect. > Now, about the sand — can you imagine how terribly messy it would be > to calculate the energy needed to move each grain of sand out of the > way?  If I were a physics schooteacher, I’d give such a problem to an > unruly student for *punishment*. > I suspect that’s why Joe chose the glass sheet example. > And I’d better shut up, anyway, before this turns in to an > unauthorised, somewhat unqualified and certainly unsolicited textbook > on Newtonian mechanics. > Cheers, >thanks… >> >> If a light arrow and a heavy arrow are launched from a bow with >> >> identical energy-transfer characteristics for both, the lighter arrow >> >> will travel faster.   Of course it will decelerate faster when it >> >> hits. Going from 200 fps to zero is "decelerating faster" than going >> >> from 100 fps to zero in the same time interval. >> >> But if both arrows have the same kinetic energy and cutting properties >> >> (blade sharpness, surface area, ruggedness and shape of the broadhead >> >> and shaft), why would a lighter arrow penetrate less than a heavier >> >> arrow? >> >Wayl >> >What’s needed here is to understand the difference between momentum and >> >kinetic energy re arrow penetration. Although they are related they are >> >not the same thing anymore than say volume and mass are not they same >> >thing (though related). >> Not a good analogy.   Volume exists regardless of mass.  Volume is a >> measure of space, alone.  Any amount of mass may occupy that volume — >> or not (there can be a complete vacuum within a volume).  I suspect >> that you are thinking about one of the physical properties of a gas. >> Regardless of the mass of the gas, it will "occupy" the available >> volume of a container. >> Kinetic energy and momentum are both measures of mass and velocity, >> nothing more.  They are just calculated differently. >> >Suppose you have a target made up of a whole series of glass sheets with >> >air gaps between them and you shoot an arrow at it. >> Which pretty much describes the pile of sand you mention later:  glass >> sheets with air gaps in between.  The primary difference is the size >> and shape of the grains (sheets) and the volume of the air gaps.  Even >> the material (silica) is the same for both! >> >The arrow hits the >> >first sheet – what determines whether the arrow breaks through this >> >sheet is the arrow momentum, not it’s kinetic energy. >> Wrong. If the arrow has enough kinetic energy to continue moving >> forward, then it also has enough momentum, or vice versa.  Drop the >> KE, and you’ll also drop the momentum. >> Here’s the verbal definition of momentum:  "the property of a moving >> body that determines the length of time required to bring it to rest >> when under the action of a constant force or moment". >> A mass of two kilograms moving along at five meters per second will >> have a momentum of ten kilogram-meters per second. >> This means only that in order to bring the object to a full stop, an >> opposing force of ten kilogram-meters must be applied for one second. >> That force may be the instantaneous or cumulative effect of friction. >> Instantaneous — arrow hits a dense, fixed object.  Cumulative — >> arrow penetrates into or through a sand pile or a torso. >> >If the arrow has  enough momentum it goes through if not it doesn’t. >> Same with the sand.  If the arrow has enough KE (or momentum) to move >> the first grain of sand, it gets to try it again on the next grain, >> and the next, and so on until the kinetic energy is dissipated >> (transferred to the glass or the grains of sand). >> >Let’s say it does. In breaking through the sheet the arrow loses energy. >> Same with the sand pile.  The arrow loses some energy when it hits the >> first grain of sand (or hair, skin, muscle, bone and other tissue >> cells, to bring the analogy closer to the NG theme).  It continues >> transferring energy to each subsequent particle (cell or grain) it >> encounters until it comes to rest. >> Perhaps you’re thinking that the arrow’s act of "breaking" through the >> glass sheets is fundamentally different from "penetrating" the sand >> pile (or mass of tissue).  Not really.   Some of the kinetic energy of >> the arrow is transferred through friction into breaking the molecular >> bonds of the glass sheet.  In the case of the sand pile, molecular >> bonds may not be broken, but the arrow’s kinetic energy, through >> friction, becomes the kinetic energy of the sand particles (they are >> moved aside).

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Response:

>I think it’s a matter of elasticity and the impulse needed to break >something.

Indeed. The characteristics of the target are very important. >If (in the example) you don’t overcome the breakage energy >of the glass, it doesn’t break (ever thrown a stone at a window when >it didn’t break? It just bounces off…).

That’s right.  But when you give the stone the proper combination of mass and velocity … bye-bye window pane. Notice the requirements:  mass and velocity.  Same requirements for both kinetic energy and for momentum. >On the other hand: grains of >sand don’t need to break for you to penetrate a layer,

Hmmm.  What’s a "layer" of sand grains? I could also say that you don’t need as much energy/momentum to chip a bit of glass from the window pane as you would to shatter the thing. Would that be an equivalent "layer"? >you just need to push them out of the way…

Oh, yeah.  As you would the shards of shattered glass … after you’ve applied enough energy to the pane to pulverise it. >So a projectile could have enough >energy to push through a layer of particulate material, where it would >bounce off a solid layer.

The particles in a sand pile aren’t so tightly bound to each other.  A bit of weak electrostatic attraction, perhaps, or the pressure of the sand above. >Just to complicate matters, fluid layers can behave differently >depending on circumstances (bullets bouncing off the surface of >water… that sort of thing).

Indeed.  This is where the quality of "vector" comes in, but we’ve pretty much ignored that in the discussion and assumed a head-on collision between arrow and target. >I guess a practical example of the effect of the mechanical properties >of the target material is backstop netting. Hang it taught and arrows >pass through. Hang it loose and you don’t get pass-throughs. Um, >energy re-distribution over a larger area?

Yep.  Impulse.  Target moves a bit and extends the impulse time. >One question I haven’t seen addressed is the resistive nature of the >target medium. The forces produced will (I would assume) be varying >with the velocity, or maybe even the square of the velocity. So a >slower moving shaft, will experience less resistive force, and also >has a greater inertia to overcome in bringing it to a stop….

The inertia of the target is the same regardless of the speed or mass of the arrow.  The inertia of the arrow doesn’t depend on the target. >Didn’t someone (Andrew Middleton?) do something with shooting into >sides of beef as an experiment…?

I can’t imagine that it would be an accurate experiment, since a side of beef has bones and meat in varying proportions in varying places. How could he guarantee that each arrow he placed met with exactly the same type and density of material as the previous one?  Might be a great practical experiment too see if a particular bow/arrow setup has "the right stuff todo the job", but it wouldn’t be a reliable means of measuring something like the penetration of light arrows vs. heavier arrows or accurately calculating penetration based on different KE and momentum figures. Cheers, – Hide quoted text — Show quoted text -> I LOVE it when someone takes up discussing with Joe in a logical and intelligent way – we get to read a VERY good exposition and see thought > processes in action from both sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>. > Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart enough to make a comfortable decision on some of > this stuff so it’s fun to sit in the peanut gallery and try to follow along. > Joe and Wayl, thanks!  🙂 > hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his point, I also tend to the Wayl arguement about the > displacement issue being identical for both materials.   Joe, what am I missing? > thanks… > > >> If a light arrow and a heavy arrow are launched from a bow with > > >> identical energy-transfer characteristics for both, the lighter arrow > > >> will travel faster.   Of course it will decelerate faster when it > > >> hits. Going from 200 fps to zero is "decelerating faster" than going > > >> from 100 fps to zero in the same time interval. > > >> But if both arrows have the same kinetic energy and cutting properties > > >> (blade sharpness, surface area, ruggedness and shape of the broadhead > > >> and shaft), why would a lighter arrow penetrate less than a heavier > > >> arrow? > > >Wayl > > >What’s needed here is to understand the difference between momentum and > > >kinetic energy re arrow penetration. Although they are related they are > > >not the same thing anymore than say volume and mass are not they same > > >thing (though related). > > Not a good analogy.   Volume exists regardless of mass.  Volume is a > > measure of space, alone.  Any amount of mass may occupy that volume — > > or not (there can be a complete vacuum within a volume).  I suspect > > that you are thinking about one of the physical properties of a gas. > > Regardless of the mass of the gas, it will "occupy" the available > > volume of a container. > > Kinetic energy and momentum are both measures of mass and velocity, > > nothing more.  They are just calculated differently. > > >Suppose you have a target made up of a whole series of glass sheets with > > >air gaps between them and you shoot an arrow at it. > > Which pretty much describes the pile of sand you mention later:  glass > > sheets with air gaps in between.  The primary difference is the size > > and shape of the grains (sheets) and the volume of the air gaps.  Even > > the material (silica) is the same for both! > > >The arrow hits the > > >first sheet – what determines whether the arrow breaks through this > > >sheet is the arrow momentum, not it’s kinetic energy. > > Wrong. If the arrow has enough kinetic energy to continue moving > > forward, then it also has enough momentum, or vice versa.  Drop the > > KE, and you’ll also drop the momentum. > > Here’s the verbal definition of momentum:  "the property of a moving > > body that determines the length of time required to bring it to rest > > when under the action of a constant force or moment". > > A mass of two kilograms moving along at five meters per second will > > have a momentum of ten kilogram-meters per second. > > This means only that in order to bring the object to a full stop, an > > opposing force of ten kilogram-meters must be applied for one second. > > That force may be the instantaneous or cumulative effect of friction. > > Instantaneous — arrow hits a dense, fixed object.  Cumulative — > > arrow penetrates into or through a sand pile or a torso. > > >If the arrow has  enough momentum it goes through if not it doesn’t. > > Same with the sand.  If the arrow has enough KE (or momentum) to move > > the first grain of sand, it gets to try it again on the next grain, > > and the next, and so on until the kinetic energy is dissipated > > (transferred to the glass or the grains of sand). > > >Let’s say it does. In breaking through the sheet the arrow loses energy. > > Same with the sand pile.  The arrow loses some energy when it hits the > > first grain of sand (or hair, skin, muscle, bone and other tissue > > cells, to bring the analogy closer to the NG theme).  It continues > > transferring energy to each subsequent particle (cell or grain) it > > encounters until it comes to rest. > > Perhaps you’re thinking that the arrow’s act of "breaking" through the > > glass sheets is fundamentally different from "penetrating" the sand > > pile (or mass of tissue).  Not really.   Some of the kinetic energy of > > the arrow is transferred through friction into breaking the molecular > > bonds of the glass sheet.  In the case of the sand pile, molecular > > bonds may not be broken, but the arrow’s kinetic energy, through > > friction, becomes the kinetic energy of the sand particles (they are > > moved aside). > > >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops. > > Exactly. > > >The arrow hits the  second sheet – enough momentum it breaks through not > > >enough and it  doesn’t and so on until the arrow doesn’t have enough > > >momentum to break  the sheet and it stops. With this target the penetration > > >is determined solely by the arrow momentum. The arrow KE is irrelevant. > > >If the target instead is a pile of sand then however small the arrow > > >momentum it will keep going, it doesn’t have to ‘cut through’ anything. > > It depends on what you mean by "cut".  If "cut" means breaking > > molecular bonds, then there may be an arguable distinction between > > piercing a solid target and tunneling through a pile of particles.  In > > the former, the energy of the moving arrow is transferred to the > > target and it both breaks bonds and moves particles aside.  In the > > latter, the energy is transferred almost exclusively to moving > > particles aside; few, if any, chemical bonds are broken in the > > process. In either case, the arrow doesn’t care; friction makes it > > give up energy (and lose momentum) to the target, which will use its > > newly found energy in whatever way it sees fit. > > >The arrow just travels along losing energy to friction until the > > >velocity drops to zero. With this target penetration is determined > > >solely by the arrow KE. The arrow momentum is irrelevant. > > Maybe it’ll look like I’m nit-picking words, but I think the > > distinction is important.  Penetration isn’t "determined solely" by

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Response:

- Hide quoted text — Show quoted text -> hurm…While I am certain that Joe chose the glass and sand analogy > purposely to make his point, I also tend to the Wayl arguement > about the displacement issue being identical for both materials. > Joe, what am I missing? >Ron >The ’sheet’ represents a solid and the ’sand’ represents a fluid. Poke a >  pencil into a fluid and it keeps penetrating as long as you keep >pushing. Poke a pencil into a steel plate and it doesn’t penetrate if >you keep pushing till doomsday. In order to get the pencil to penetrate >the plate you need to supply enough force to stress the plate beyond its >strength limits. The force exerted on the plate by an arrow is related >to the arrow momentum not its KE (I’m sure you remember Newton’s second >law of motion from your schooldays).

Well, pardon me, but I’ll just butt right in … again. What you say above — that the force required to deform a steel plate is related to an arrow’s momentum and not it’s kinetic energy — can be explored through the equations for kinetic energy and momentum. First of all, lets agree upon the equations. Kinetic energy (KE):  One-half the arrow’s mass (m) times the square of the arrow’s velocity.  KE = 1/2 m * v^2.  (I’ll drop the "1/2" constant, because it does nothing for this argument except add a few more keystrokes.)  So, KE = m * v^2.     Momentum (p):   Product of mass and velocity.    p = m * v. Let’s pick some figures: not truly accurate figures for arrows, but numbers that are easy to manipulate (and type).    mass = 100 grams. velocity = 10 cm/sec. Arrow of 100 grams moving at 10 cm/sec has a momentum of 1000 gram-centimeters-per-second. This just happens to coincide with the momentum of an arrow that we’ve already proven can deform the steel plate.   We also tried another arrow (same mass) that was clocked at a terminal velocity of 9.99 cm/sec, and it bounced off the plate leaving nary a scratch.  Ruined another good broadhead.  Now we know:  100 grams at 10 cm/sec is the minimum deal. Arrow of 100 grams moving at 10 cm/sec has a kinetic enerty of 10,000 grams-centimeters^2-per-second^2.   Big number.  Weird units.  Really intimidating. OK, so lets get a number that’ll be closer to the one that popped out of the momentum calculator. Arrow of 100 grams moving at a velocity of 3.16227766 cm/sec has a kinetic energy of almost 1000 g-cm^2/sec^2.  The arrow’s velocity has dropped by over two-thirds. We can do that, can’t we?  After all, the kinetic energy isn’t (presumably) relevant to the arrow’s ability to deform the target.  Or to quote your position more accurately, "The force exerted on the plate by an arrow is related to the arrow momentum not its KE." Or can we?   Don’t the laws of physics compel us be fair?   So if we had to drop the arrow’s velocity to make the KE figure look prettier, must we not also drop the velocity of the arrow in the momentum calculation, out of a keen sense of fairness and because we are Newtonian Law-abiding citizens?  If we do that, how does it affect the momentum figure, not to mention the arrow’s ability to deform the steel plate? So, can you tell me exactly what divorces the arrow’s kinetic energy figure from the effect the arrow may have on the steel plate? Cheers,

Response:

I think it’s a matter of elasticity and the impulse needed to break something. If (in the example) you don’t overcome the breakage energy of the glass, it doesn’t break (ever thrown a stone at a window when it didn’t break? It just bounces off…). On the other hand: grains of sand don’t need to break for you to penetrate a layer, you just need to push them out of the way… So a projectile could have enough energy to push through a layer of particulate material, where it would bounce off a solid layer. Just to complicate matters, fluid layers can behave differently depending on circumstances (bullets bouncing off the surface of water… that sort of thing). I guess a practical example of the effect of the mechanical properties of the target material is backstop netting. Hang it taught and arrows pass through. Hang it loose and you don’t get pass-throughs. Um, energy re-distribution over a larger area? One question I haven’t seen addressed is the resistive nature of the target medium. The forces produced will (I would assume) be varying with the velocity, or maybe even the square of the velocity. So a slower moving shaft, will experience less resistive force, and also has a greater inertia to overcome in bringing it to a stop…. Didn’t someone (Andrew Middleton?) do something with shooting into sides of beef as an experiment…? – Hide quoted text — Show quoted text – > I LOVE it when someone takes up discussing with Joe in a logical and intelligent way – we get to read a VERY good exposition and see thought > processes in action from both sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>. > Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart enough to make a comfortable decision on some of > this stuff so it’s fun to sit in the peanut gallery and try to follow along. > Joe and Wayl, thanks!  🙂 > hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his point, I also tend to the Wayl arguement about the > displacement issue being identical for both materials.   Joe, what am I missing? > thanks… > >> If a light arrow and a heavy arrow are launched from a bow with > >> identical energy-transfer characteristics for both, the lighter arrow > >> will travel faster.   Of course it will decelerate faster when it > >> hits. Going from 200 fps to zero is "decelerating faster" than going > >> from 100 fps to zero in the same time interval. > >> But if both arrows have the same kinetic energy and cutting properties > >> (blade sharpness, surface area, ruggedness and shape of the broadhead > >> and shaft), why would a lighter arrow penetrate less than a heavier > >> arrow? > >Wayl > >What’s needed here is to understand the difference between momentum and > >kinetic energy re arrow penetration. Although they are related they are > >not the same thing anymore than say volume and mass are not they same > >thing (though related). > Not a good analogy.   Volume exists regardless of mass.  Volume is a > measure of space, alone.  Any amount of mass may occupy that volume — > or not (there can be a complete vacuum within a volume).  I suspect > that you are thinking about one of the physical properties of a gas. > Regardless of the mass of the gas, it will "occupy" the available > volume of a container. > Kinetic energy and momentum are both measures of mass and velocity, > nothing more.  They are just calculated differently. > >Suppose you have a target made up of a whole series of glass sheets with > >air gaps between them and you shoot an arrow at it. > Which pretty much describes the pile of sand you mention later:  glass > sheets with air gaps in between.  The primary difference is the size > and shape of the grains (sheets) and the volume of the air gaps.  Even > the material (silica) is the same for both! > >The arrow hits the > >first sheet – what determines whether the arrow breaks through this > >sheet is the arrow momentum, not it’s kinetic energy. > Wrong. If the arrow has enough kinetic energy to continue moving > forward, then it also has enough momentum, or vice versa.  Drop the > KE, and you’ll also drop the momentum. > Here’s the verbal definition of momentum:  "the property of a moving > body that determines the length of time required to bring it to rest > when under the action of a constant force or moment". > A mass of two kilograms moving along at five meters per second will > have a momentum of ten kilogram-meters per second. > This means only that in order to bring the object to a full stop, an > opposing force of ten kilogram-meters must be applied for one second. > That force may be the instantaneous or cumulative effect of friction. > Instantaneous — arrow hits a dense, fixed object.  Cumulative — > arrow penetrates into or through a sand pile or a torso. > >If the arrow has  enough momentum it goes through if not it doesn’t. > Same with the sand.  If the arrow has enough KE (or momentum) to move > the first grain of sand, it gets to try it again on the next grain, > and the next, and so on until the kinetic energy is dissipated > (transferred to the glass or the grains of sand). > >Let’s say it does. In breaking through the sheet the arrow loses energy. > Same with the sand pile.  The arrow loses some energy when it hits the > first grain of sand (or hair, skin, muscle, bone and other tissue > cells, to bring the analogy closer to the NG theme).  It continues > transferring energy to each subsequent particle (cell or grain) it > encounters until it comes to rest. > Perhaps you’re thinking that the arrow’s act of "breaking" through the > glass sheets is fundamentally different from "penetrating" the sand > pile (or mass of tissue).  Not really.   Some of the kinetic energy of > the arrow is transferred through friction into breaking the molecular > bonds of the glass sheet.  In the case of the sand pile, molecular > bonds may not be broken, but the arrow’s kinetic energy, through > friction, becomes the kinetic energy of the sand particles (they are > moved aside). > >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops. > Exactly. > >The arrow hits the  second sheet – enough momentum it breaks through not > >enough and it  doesn’t and so on until the arrow doesn’t have enough > >momentum to break  the sheet and it stops. With this target the penetration > >is determined solely by the arrow momentum. The arrow KE is irrelevant. > >If the target instead is a pile of sand then however small the arrow > >momentum it will keep going, it doesn’t have to ‘cut through’ anything. > It depends on what you mean by "cut".  If "cut" means breaking > molecular bonds, then there may be an arguable distinction between > piercing a solid target and tunneling through a pile of particles.  In > the former, the energy of the moving arrow is transferred to the > target and it both breaks bonds and moves particles aside.  In the > latter, the energy is transferred almost exclusively to moving > particles aside; few, if any, chemical bonds are broken in the > process. In either case, the arrow doesn’t care; friction makes it > give up energy (and lose momentum) to the target, which will use its > newly found energy in whatever way it sees fit. > >The arrow just travels along losing energy to friction until the > >velocity drops to zero. With this target penetration is determined > >solely by the arrow KE. The arrow momentum is irrelevant. > Maybe it’ll look like I’m nit-picking words, but I think the > distinction is important.  Penetration isn’t "determined solely" by > either kinetic energy or momentum. > >Most targets are bit of both, ‘cutting through’ and friction. > "Cutting through" IS friction. > >What  decides the balance between momentum and KE is the nature of > >the target. > There is no "balance" between kinetic energy and momentum.  They are > both expressions of the same three fundamental factors:  mass, > distance and time. > Take a look at the equations: > Momentum —  the product of mass and velocity.   momentum = m * v. > Kinetic energy —   one half the product of mass of the particle and > the square of its velocity.   KE = 1/2 (m * v^2) > Notice the *only* variables in both equations:  mass and velocity. > The relationship — the "importance" — of mass and velocity remain > the same in both equations.  The kinetic energy equation adds a > constant (1/2) and an operation (squaring the velocity), but this only > changes the descriptive quality of the number on the left-hand side of > the equation.  In the case of the momentum, the number changes in a > linear fashion.  In the case of kinetic energy, the number changes in > a logarithmic fashion. > Consider the stucture of the target.   It changes the momentum and > kinetic energy of the projectile  For any given combination of mass > and velocity, that projectile will have ONE specific figure for > momentum and ONE specific figure for kinetic energy.  As the > projectile contacts the target, the energy of the former is > transferred to the latter.  In this case, the mass of the projectile > remains practically the same, and the energy transfer has the effect > only of reducing the projectile’s velocity. > If you could measure the change in velocity from the instant before > the arrow contacts the target until the time it comes to a full stop, > you’d find that the figure for momentum changes in a

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Response:

Wayl Your reply indicates your not stupid hence it seems unlikely that you don’t understand the difference between momentum and KE (or area and volume or foot-lbs and density and any other defined quantities that just happen to contain the same ‘units’). I conclude that your posts are just a wind up. Apologies of course if I’m wrong.

Response:

 >  > hurm…While I am certain that Joe chose the glass and sand analogy  > purposely to make his point, I also tend to the Wayl arguement  > about the displacement issue being identical for both materials.  > Joe, what am I missing?  > Ron The ’sheet’ represents a solid and the ’sand’ represents a fluid. Poke a   pencil into a fluid and it keeps penetrating as long as you keep pushing. Poke a pencil into a steel plate and it doesn’t penetrate if you keep pushing till doomsday. In order to get the pencil to penetrate the plate you need to supply enough force to stress the plate beyond its strength limits. The force exerted on the plate by an arrow is related to the arrow momentum not its KE (I’m sure you remember Newton’s second law of motion from your schooldays).

Response:

>Wayl >Your reply indicates your not stupid

Well, that would be mostly a matter of opinion, wouldn’t it?  :] >hence it seems unlikely that you  don’t understand the difference >between momentum and KE (or area and volume or foot-lbs and density >and any other defined quantities that just happen to contain the same >’units’).

Indeed I do understand the difference between various measurements, units and qualities.   I’ve tried to explain my understanding of some of these difference in my earlier posts.  For example, you claimed that volume and mass are somewhat "related".   I argued and explained that that this just ain’t necessarily so.  (Now *density* is related to both volume and mass, but mass and volume are pretty much independent variables, much like mass and velocity in the KE and momentum equations.)   Consider a volume of one cubic meter.  That’s a segment of space which, if it’s shaped like a cube, would measure one meter in each dimension.  Within that cube – within the volume – could be a mass of *nothing* (a perfect vacuum), a single hydrogen atom or one cubic meter’s worth of pure neutronium (can’t be much denser than that without getting into singularities.  Boy, would THAT do something interesting to KE and momentum figures!). OK. So what’s this alleged relationship between mass and volume? At this point, I may conclude one of the following: 1.   You read the first line or two of my posts, got bored and dismissed the rest.   This is understandable! :] 2.  You read the whole post and are now hastily constructing a sound rebuttal to what you feel are my weak points.  This would be the most challenging option, for me.  You may very well prove me wrong on some or all points.  If so, I’ll survive … I guess.   I’ll just pick up my shattered dignity and perhaps return for a future re-match (or another dose of intellectual abuse). 3.  In spite of your opening sentence in this post, you have indeed concluded that I’m an idiot, and that I’m not worthy of discourse with the truly enlightened. Then again, maybe there’s a fourth, fifth and sixth option… >I conclude that your posts are just a wind up. >Apologies of course if I’m wrong.

Wrong about what?   Me not being stupid, or that my posts are just a wind-up? I believe you’d be wrong about the latter.   I’ll leave it up to you to prove the former. Cheers,

Response:

Hey, All I apparently have waaaay too much time on my hands today.  I have been working to find the best set-up for weight/velocity of my Hunting arrows so as to find the  most suitable combination of trajectory and penetration for hunting.  As I worked on my Kinetic Energy tables, some of my results were: I began to wonder; at what point does the mass of the arrow begin to impact the arrows velocity that you get a negative effect on Kinetic Energy?  The only other useful arrow to experiment with was an aluminum-clad fiberglass ft/sec. Yielding only 45.29 ft/lbs. KE. My question is this:  Is there any formula that can be used to determine where that trade-off begins to occur, or can it only be obtained through experimentation?  Just curious, but would appreciate any knowledgeable responses. Chris

Response:

– Hide quoted text — Show quoted text -> Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting > arrows so as to find the  most suitable combination of trajectory and > penetration for hunting.  As I worked on my Kinetic Energy tables, > some of my results were: > I began to wonder; at what point does the mass of the arrow begin to > impact the arrows velocity that you get a negative effect on Kinetic > Energy?  The only other useful arrow to experiment with was an > aluminum-clad fiberglass fishing arrow with a finished weight of 1305 > My question is this:  Is there any formula that can be used to > determine where that trade-off begins to occur, or can it only be > obtained through experimentation?  Just curious, but would appreciate > any knowledgeable responses. > Chris

I suppose you could write the KE as an equation as K (Kinetic Energy) K = 0.5 * mass * velocity ^2 i.e. 0.5(mv^2) I suppose you could take the differential of this either by mass or velocity.  Where the derivative = 0 this gives the maximum KE.   Or something like that…  its been a while since I had to do math like that….  But I think thats the correct approach to take. The difficulty might be the fact that there are two unknowns, mass & velocity. Interesting problem though! regards — Brian

Response:

– Hide quoted text — Show quoted text -> Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting arrows so > as to find the  most suitable combination of trajectory and penetration for > hunting.  As I worked on my Kinetic Energy tables, some of my results were: > I began to wonder; at what point does the mass of the arrow begin to impact > the arrows velocity that you get a negative effect on Kinetic Energy?  The > only other useful arrow to experiment with was an aluminum-clad fiberglass > ft/sec. Yielding only 45.29 ft/lbs. KE. > My question is this:  Is there any formula that can be used to determine > where that trade-off begins to occur, or can it only be obtained through > experimentation?  Just curious, but would appreciate any knowledgeable > responses. > Chris

You don’t have enough numbers. I’ll throw in some additional stuff to complicate it for you. Mass isn’t the same as weight. Mass is a constant measure regardless of gravity. Weight is mass that’s acted on by gravity. It will vary slightly by altitude. For those who don’t know it, the commonly used formula for KE is wt (in grains) times velocity  squared (in feet/sec) divided by 450240. The figure 450240 contains the conversions from mass to weight and it converts grains to lbs to get  foot-pounds. To further complicate it, consider momentum. The formula for that is mass x velocity. Momentum puts more emphasis on the mass rather than on velocity and, in my opinion, is a more accurate indicator of what an arrow can be expected to do on impact. Dick

Response:

You need to factor in terminal velocity too.  What is drop off at 20 yds or so?  Personally I avoid the real light arrows as they are harder on the bow. A few foot pounds won’t make any difference to a well hit deer. Frank

Response:

> You don’t have enough numbers. I’ll throw in some additional stuff to > complicate it for you. > Mass isn’t the same as weight. Mass is a constant measure regardless of > gravity. Weight is mass that’s acted on by gravity. It will vary slightly by > altitude. For those who don’t know it, the commonly used formula for KE is > wt (in grains) times velocity  squared (in feet/sec) divided by 450240. The > figure 450240 contains the conversions from mass to weight and it converts > grains to lbs to get  foot-pounds. > To further complicate it, consider momentum. The formula for that is mass x > velocity. Momentum puts more emphasis on the mass rather than on velocity > and, in my opinion, is a more accurate indicator of what an arrow can be > expected to do on impact. > Dick

Thanks, Dick. I understand how the factors of weight conversion (7000 grains/lbs.), Gravity (32.16) and the 1/2 multiplier figure in to give 450,240.  I hadn’t, however, considered momentum vs. KE.  If I refigure for momentum instead of KE, the 1305 grain arrow maintained much greater momentum at 10 yards than the others, even though it retained less Energy. Once my brain catches up, that will probably make sense to me. :-0  In the meantime, I assume what that tells me is that even though it might not be carrying more *energy* it will still take more force to slow down, thus delivering better penetration?  Thanks again, Chris

Response:

- Hide quoted text — Show quoted text -> Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting arrows >  so > as to find the  most suitable combination of trajectory and penetration >  for > hunting.  As I worked on my Kinetic Energy tables, some of my results >  were: > I began to wonder; at what point does the mass of the arrow begin to >  impact > the arrows velocity that you get a negative effect on Kinetic Energy?  The > only other useful arrow to experiment with was an aluminum-clad fiberglass > ft/sec. Yielding only 45.29 ft/lbs. KE. > My question is this:  Is there any formula that can be used to determine > where that trade-off begins to occur, or can it only be obtained through > experimentation?  Just curious, but would appreciate any knowledgeable > responses. > Chris > You don’t have enough numbers. I’ll throw in some additional stuff to > complicate it for you. > Mass isn’t the same as weight.

Yes it is, much of the time.  Weight is an ambiguous word, with more than one meaning. > Mass is a constant measure regardless of > gravity. Weight is mass that’s acted on by gravity. It will vary slightly by > altitude. For those who don’t know it, the commonly used formula for KE is > wt (in grains) times velocity  squared (in feet/sec) divided by 450240. The

If you use the pounds force defined in order for that formula to be true, then 1 lbf = 0.45339431 kgf (approximately), whereas 1 lb = 0.45359237 kg exactly.  You can do that, since the pound force doesn’t have an "official" definition. Besides, kilograms force, though once acceptable, are not acceptable in the modern metric system, the International System of Units, so it really doesn’t make much difference if you choose to do it that way.  (The British units are no longer supported and updated, so nobody will ever bother to tell us not to use pounds force any more–looks like we are stuck with them.) > figure 450240 contains the conversions from mass to weight and it converts > grains to lbs to get  foot-pounds.

No.  The 450240 (or 450437 if you define the pound force so it has the same relationship to the kilogram force as a pound has to a kilogram)is just the factor needed for the units chosen, with mass in grains, speed in ft/s, and wanting energy in foot-pounds force.  The more general form of the equation is E = m v^2/(2k) and the factor k, which is only equal to one if you use a coherent system, cannot be omitted in this system.  You could also alternatively look at it as converting grains troy (always units of mass–there is no grain force, nor are any of the other units in the troy system of weights units of force–they are all units of mass as well) to pounds mass (avoirdupois) by dividing by 7000, but if you use pounds as units of mass in a coherent system of units the result is then in foot-poundals, which you’d have to change to the pounds force you want to end up with by dividing by 32.16 ft/s^2 or 32.1740 ft/s^2 or whatever you use to define pounds force.  Or instead you could convert the pounds to slugs (a mass unit to mass unit conversion–you are not starting with a measurement of force) by dividing by 32.16 ft/s^2 or 32.1740 ft/s^2 or whatever you use to define pounds force, and then the result will be in foot-pounds force.  That’s three or four different systems of mechanical units you could use in the these calculations, and they all come out the same in the wash.  You could even use an inch-pound-second if you wanted to, but then you’d have to convert the feet into that system and back out again, making the calculations more difficult than in the systems I have discussed. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/

Response:

Hy, My observation is that these arrows have been shot from the same bow, i.e. with the same stored energy. But we observe that the KE released change. This just means that the efficiency of the bow change, with obviously a peak. Anyway, having the most efficiency from your bow is a good thing but the ballistic is something else, in wich I wouldn’t go into. In other word how Weight(Mass) and KE influence the trajectory ?

– Hide quoted text — Show quoted text -> Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting > arrows so as to find the  most suitable combination of trajectory and > penetration for hunting.  As I worked on my Kinetic Energy tables, > some of my results were: > I began to wonder; at what point does the mass of the arrow begin to > impact the arrows velocity that you get a negative effect on Kinetic > Energy?  The only other useful arrow to experiment with was an > aluminum-clad fiberglass fishing arrow with a finished weight of 1305 > My question is this:  Is there any formula that can be used to > determine where that trade-off begins to occur, or can it only be > obtained through experimentation?  Just curious, but would appreciate > any knowledgeable responses. > Chris > I suppose you could write the KE as an equation as K (Kinetic Energy) > K = 0.5 * mass * velocity ^2 > i.e. 0.5(mv^2) > I suppose you could take the differential of this either by mass or > velocity.  Where the derivative = 0 this gives the maximum KE. > Or something like that…  its been a while since I had to do math like > that….  But I think thats the correct approach to take. > The difficulty might be the fact that there are two unknowns, mass & > velocity. > Interesting problem though! > regards > — > Brian

Response:

> Hy, > My observation is that these arrows have been shot from the same bow, i.e. > with the same stored energy. > But we observe that the KE released change. This just means that the > efficiency of the bow change, with > obviously a peak. > Anyway, having the most efficiency from your bow is a good thing but the > ballistic is something else, in wich > I wouldn’t go into. In other word how Weight(Mass) and KE influence the > trajectory ?

I just thought about that yesterday and calculated the best results at about 93% efficiency with a 480 grain arrow ( If I calculated the efficiency correctly).  I am by no means an advanced student of physics, but I certainly have no aversion to understanding the mechanics of KE, Momentum and trajectory, so if you have something that I could learn from, please fire away!  Thanks, Chris

Response:

It’s not a simple question… You’ve got to consider quite a few factors, not just the energy of the shaft, but also it’s momentum, and how much energy it retains downrange… Joe Tapley has a discussion of target penetration on his website: http://homepage.ntlworld.com/joetapley/ under the heading "arrow penetration". It’s written from the point of view of a Target archer, so the idea is a discussion of how to *reduce* target penetration, but if you reverse the argument… *warning* – oversimplification: Fast light arrows decelerate quicker than heavy slow arrows, so you can get better penetration with less KE from heavier shafts. Your results with the fishing arrow aren’t a surprise: KE varies as the square of velocity, so the mass is a lot less important than the speed of the shaft. Momentum on the other hand varies with velocity and mass, so the two are equally important. Double the mass of a shaft and you double it’s momentum and KE. Double the speed of a shaft and you double it’s momentum and quadruple the KE! Just to complicate things: a heavier shaft is usually considered more efficient – it gets more of the stored energy out of the bow… From your figures below: a 15% increase in shaft weight only lowered the shaft speed by about 6%. A 32% increase in shaft weight lowered the shaft speed by 10.5% . In the last instance a massive 270% increase in shaft weight reduced the shaft speed by 55%. If you really want your head to spin, look up some of the small arms ballistics pages on the web. In particular the ones on "kinetic pulse" AKA "killing power" – but remember, again, the people playing with bullets want to *reduce* penetration (so the KE is transferred to the Target) while keeping the KE up. Not what you want with arrows… – Hide quoted text — Show quoted text – > Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting arrows so > as to find the  most suitable combination of trajectory and penetration for > hunting.  As I worked on my Kinetic Energy tables, some of my results were: > I began to wonder; at what point does the mass of the arrow begin to impact > the arrows velocity that you get a negative effect on Kinetic Energy?  The > only other useful arrow to experiment with was an aluminum-clad fiberglass > ft/sec. Yielding only 45.29 ft/lbs. KE. > My question is this:  Is there any formula that can be used to determine > where that trade-off begins to occur, or can it only be obtained through > experimentation?  Just curious, but would appreciate any knowledgeable > responses. > Chris

Response:

Good, easy to understand response, Shadyshark.  Thanks, Chris – Hide quoted text — Show quoted text – > It’s not a simple question… You’ve got to consider quite a few > factors, not just the energy of the shaft, but also it’s momentum, and > how much energy it retains downrange… > Joe Tapley has a discussion of target penetration on his website: > http://homepage.ntlworld.com/joetapley/ > under the heading "arrow penetration". > It’s written from the point of view of a Target archer, so the idea is > a discussion of how to *reduce* target penetration, but if you reverse > the argument… > *warning* – oversimplification: Fast light arrows decelerate quicker > than heavy slow arrows, so you can get better penetration with less KE > from heavier shafts. > Your results with the fishing arrow aren’t a surprise: KE varies as > the square of velocity, so the mass is a lot less important than the > speed of the shaft. > Momentum on the other hand varies with velocity and mass, so the two > are equally important. > Double the mass of a shaft and you double it’s momentum and KE. > Double the speed of a shaft and you double it’s momentum and quadruple > the KE! > Just to complicate things: a heavier shaft is usually considered more > efficient – it gets more of the stored energy out of the bow… > From your figures below: a 15% increase in shaft weight only lowered > the shaft speed by about 6%. A 32% increase in shaft weight lowered > the shaft speed by 10.5% . In the last instance a massive 270% > increase in shaft weight reduced the shaft speed by 55%. > If you really want your head to spin, look up some of the small arms > ballistics pages on the web. In particular the ones on "kinetic pulse" > AKA "killing power" – but remember, again, the people playing with > bullets want to *reduce* penetration (so the KE is transferred to the > Target) while keeping the KE up. Not what you want with arrows… > Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting arrows so > as to find the  most suitable combination of trajectory and penetration for > hunting.  As I worked on my Kinetic Energy tables, some of my results were: > I began to wonder; at what point does the mass of the arrow begin to impact > the arrows velocity that you get a negative effect on Kinetic Energy?  The > only other useful arrow to experiment with was an aluminum-clad fiberglass > ft/sec. Yielding only 45.29 ft/lbs. KE. > My question is this:  Is there any formula that can be used to determine > where that trade-off begins to occur, or can it only be obtained through > experimentation?  Just curious, but would appreciate any knowledgeable > responses. > Chris

Response:

>It’s not a simple question… You’ve got to consider quite a few >factors, not just the energy of the shaft, but also it’s momentum, and >how much energy it retains downrange… >Joe Tapley has a discussion of target penetration on his website: >http://homepage.ntlworld.com/joetapley/ >under the heading "arrow penetration". >It’s written from the point of view of a Target archer, so the idea is >a discussion of how to *reduce* target penetration, but if you reverse >the argument… >*warning* – oversimplification: Fast light arrows decelerate quicker >than heavy slow arrows, so you can get better penetration with less KE >from heavier shafts.

This doesn’t gel. If a light arrow and a heavy arrow are launched from a bow with identical energy-transfer characteristics for both, the lighter arrow will travel faster.   Of course it will decelerate faster when it hits.   Going from 200 fps to zero is "decelerating faster" than going from 100 fps to zero in the same time interval. But if both arrows have the same kinetic energy and cutting properties (blade sharpness, surface area, ruggedness and shape of the broadhead and shaft), why would a lighter arrow penetrate less than a heavier arrow? If a light arrow and a heavier arrow are travelling at the same speed, then surely the heavier arrow will penetrate more, as it carries more kinetic energy.  But it’ll have pretty much the same kinetic energy as the lighter arrow travelling at a suitably faster speed.  So, for the same surface area, shape and texture and the same KE, both arrows will penetrate the same distance. I propose that the issue lies with the entire launch system itself, not with the qualities of the arrow once it has cleared the bow.  For example, given the same materials, a lighter shaft is likely to have a weaker spine than a heavier shaft.  This means that the draw weight of the bow will probably be less that it would be for a heavier shaft. Therefore, the lighter arrow travels slower than it could do, which will decreases its momentum and kinetic energy accordingly. The heavier arrow (of the same materials) may accommodate a higher draw weight.  More momentum and kinetic energy. And then there’s inertia.   I suspect the bow/arrow system becomes somewhat more efficient with heavier arrows, as the greater inertia of the heavier arrow may(?) allow more of the energy stored in the bow to be transferred to the shaft before it clears the string. >Your results with the fishing arrow aren’t a surprise: KE varies as >the square of velocity, so the mass is a lot less important than the >speed of the shaft. >Momentum on the other hand varies with velocity and mass, so the two >are equally important.

Misleading.  Mass and velocity are equally "important", whether you use the KE formula or the momentum formula.  You’ll get a logarithmic curve from the KE formula and a straight line from the momentum formula. >Double the mass of a shaft and you double it’s momentum and KE.

Not quite.  In order to increase the heavier shaft’s momentum and kinetic energy, you’d need to increase the draw weight (energy transfer) of the bow accordingly.  Keep the same draw weight, and as the mass goes up, the velocity goes down accordingly.  Momentum and KE remain the same. >Double the speed of a shaft and you double it’s momentum and quadruple >the KE!

Same as above.   But even still, the KE and momentum figures are just representations of the moving arrow’s properties.  The arrows will behave the same whether you calculate their performance using the KE formula or the momentum formula.  The KE formula produces a logarithmic curve for a range of masses and velocities.  The momentum formula gives a linear plot.  The arrows don’t care one way or t’other. >Just to complicate things: a heavier shaft is usually considered more >efficient – it gets more of the stored energy out of the bow…

Yeah, that’s the inertia in play, right? – Hide quoted text — Show quoted text ->From your figures below: a 15% increase in shaft weight only lowered >the shaft speed by about 6%. A 32% increase in shaft weight lowered >the shaft speed by 10.5% . In the last instance a massive 270% >increase in shaft weight reduced the shaft speed by 55%. >If you really want your head to spin, look up some of the small arms >ballistics pages on the web. In particular the ones on "kinetic pulse" >AKA "killing power" – but remember, again, the people playing with >bullets want to *reduce* penetration (so the KE is transferred to the >Target) while keeping the KE up. Not what you want with arrows… > Hey, All > I apparently have waaaay too much time on my hands today.  I have been > working to find the best set-up for weight/velocity of my Hunting arrows so > as to find the  most suitable combination of trajectory and penetration for > hunting.  As I worked on my Kinetic Energy tables, some of my results were: > I began to wonder; at what point does the mass of the arrow begin to impact > the arrows velocity that you get a negative effect on Kinetic Energy?

At no point.  If you could overlook the characteristics of the launcher (tuning, draw weight) and give it the same efficiency for an entire range of arrow weights, you’d find that as the mass increases, the velocity decreases predictably.  There is no practical (subsonic) "magic point" where the mass alone causes a sudden, dramatic change in the arrow’s kinetic energy. The only significant "forces" acting on the arrow once it has cleared the bow are friction and gravity.   The effect of friction increases as surface area increases, and it’s also affected by the shape of the projectile. Can’t do much about gravity, except perhaps conduct your tests in a low-earth orbit. > The only other useful arrow to experiment with was an aluminum-clad > fiberglass fishing arrow with a finished weight of 1305 grains.  It > My question is this:  Is there any formula that can be used to determine > where that trade-off begins to occur,

The "trade off" occurs at the string and arrow rest. > or can it only be obtained through experimentation?  Just curious, > but would appreciate any knowledgeable responses.

Cheers,

Response:

> If a light arrow and a heavy arrow are launched from a bow with > identical energy-transfer characteristics for both, the lighter arrow > will travel faster.   Of course it will decelerate faster when it > hits. Going from 200 fps to zero is "decelerating faster" than going > from 100 fps to zero in the same time interval. > But if both arrows have the same kinetic energy and cutting properties > (blade sharpness, surface area, ruggedness and shape of the broadhead > and shaft), why would a lighter arrow penetrate less than a heavier > arrow?

Wayl What’s needed here is to understand the difference between momentum and kinetic energy re arrow penetration. Although they are related they are not the same thing anymore than say volume and mass are not they same thing (though related). Suppose you have a target made up of a whole series of glass sheets with air gaps between them and you shoot an arrow at it. The arrow hits the first sheet – what determines whether the arrow breaks through this sheet is the arrow momentum, not it’s kinetic energy. If the arrow has enough momentum it goes through if not it doesn’t. Let’s say it does. In breaking through the sheet the arrow loses energy. It’s KE drops (i.e. its velocity) and hence the arrow momentum drops. The arrow hits the second sheet – enough momentum it breaks through not enough and it doesn’t and so on until the arrow doesn’t have enough momentum to break the sheet and it stops. With this target the penetration is determined solely by the arrow momentum. The arrow KE is irrelevant. If the target instead is a pile of sand then however small the arrow momentum it will keep going, it doesn’t have to ‘cut through’ anything. The arrow just travels along losing energy to friction until the velocity drops to zero. With this target penetration is determined solely by the arrow KE. The arrow momentum is irrelevant. Most targets are bit of both, ‘cutting through’ and friction. What decides the balance between momentum and KE is the nature of the target. > The only significant "forces" acting on the arrow once it has cleared > the bow are friction and gravity.

Frictional forces on an arrow have directly a negligible effect. Its the result of having to move the (mass of) air out of the way to ‘get through’ that generates the drag force.

Response:

- Hide quoted text — Show quoted text -> If a light arrow and a heavy arrow are launched from a bow with > identical energy-transfer characteristics for both, the lighter arrow > will travel faster.   Of course it will decelerate faster when it > hits. Going from 200 fps to zero is "decelerating faster" than going > from 100 fps to zero in the same time interval. > But if both arrows have the same kinetic energy and cutting properties > (blade sharpness, surface area, ruggedness and shape of the broadhead > and shaft), why would a lighter arrow penetrate less than a heavier > arrow? >Wayl >What’s needed here is to understand the difference between momentum and >kinetic energy re arrow penetration. Although they are related they are >not the same thing anymore than say volume and mass are not they same >thing (though related).

Not a good analogy.   Volume exists regardless of mass.  Volume is a measure of space, alone.  Any amount of mass may occupy that volume — or not (there can be a complete vacuum within a volume).  I suspect that you are thinking about one of the physical properties of a gas. Regardless of the mass of the gas, it will "occupy" the available volume of a container. Kinetic energy and momentum are both measures of mass and velocity, nothing more.  They are just calculated differently. >Suppose you have a target made up of a whole series of glass sheets with >air gaps between them and you shoot an arrow at it.

Which pretty much describes the pile of sand you mention later:  glass sheets with air gaps in between.  The primary difference is the size and shape of the grains (sheets) and the volume of the air gaps.  Even the material (silica) is the same for both! >The arrow hits the >first sheet – what determines whether the arrow breaks through this >sheet is the arrow momentum, not it’s kinetic energy.

Wrong. If the arrow has enough kinetic energy to continue moving forward, then it also has enough momentum, or vice versa.  Drop the KE, and you’ll also drop the momentum. Here’s the verbal definition of momentum:  "the property of a moving body that determines the length of time required to bring it to rest when under the action of a constant force or moment". A mass of two kilograms moving along at five meters per second will have a momentum of ten kilogram-meters per second.   This means only that in order to bring the object to a full stop, an opposing force of ten kilogram-meters must be applied for one second. That force may be the instantaneous or cumulative effect of friction. Instantaneous — arrow hits a dense, fixed object.  Cumulative — arrow penetrates into or through a sand pile or a torso. >If the arrow has  enough momentum it goes through if not it doesn’t.

Same with the sand.  If the arrow has enough KE (or momentum) to move the first grain of sand, it gets to try it again on the next grain, and the next, and so on until the kinetic energy is dissipated (transferred to the glass or the grains of sand). >Let’s say it does. In breaking through the sheet the arrow loses energy.

Same with the sand pile.  The arrow loses some energy when it hits the first grain of sand (or hair, skin, muscle, bone and other tissue cells, to bring the analogy closer to the NG theme).  It continues transferring energy to each subsequent particle (cell or grain) it encounters until it comes to rest. Perhaps you’re thinking that the arrow’s act of "breaking" through the glass sheets is fundamentally different from "penetrating" the sand pile (or mass of tissue).  Not really.   Some of the kinetic energy of the arrow is transferred through friction into breaking the molecular bonds of the glass sheet.  In the case of the sand pile, molecular bonds may not be broken, but the arrow’s kinetic energy, through friction, becomes the kinetic energy of the sand particles (they are moved aside). >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops.

Exactly. >The arrow hits the  second sheet – enough momentum it breaks through not >enough and it  doesn’t and so on until the arrow doesn’t have enough >momentum to break  the sheet and it stops. With this target the penetration >is determined solely by the arrow momentum. The arrow KE is irrelevant. >If the target instead is a pile of sand then however small the arrow >momentum it will keep going, it doesn’t have to ‘cut through’ anything.

It depends on what you mean by "cut".  If "cut" means breaking molecular bonds, then there may be an arguable distinction between piercing a solid target and tunneling through a pile of particles.  In the former, the energy of the moving arrow is transferred to the target and it both breaks bonds and moves particles aside.  In the latter, the energy is transferred almost exclusively to moving particles aside; few, if any, chemical bonds are broken in the process. In either case, the arrow doesn’t care; friction makes it give up energy (and lose momentum) to the target, which will use its newly found energy in whatever way it sees fit. >The arrow just travels along losing energy to friction until the >velocity drops to zero. With this target penetration is determined >solely by the arrow KE. The arrow momentum is irrelevant.

Maybe it’ll look like I’m nit-picking words, but I think the distinction is important.  Penetration isn’t "determined solely" by either kinetic energy or momentum.   >Most targets are bit of both, ‘cutting through’ and friction.

"Cutting through" IS friction. >What  decides the balance between momentum and KE is the nature of >the target.

There is no "balance" between kinetic energy and momentum.  They are both expressions of the same three fundamental factors:  mass, distance and time. Take a look at the equations: Momentum —  the product of mass and velocity.   momentum = m * v. Kinetic energy —   one half the product of mass of the particle and the square of its velocity.   KE = 1/2 (m * v^2) Notice the *only* variables in both equations:  mass and velocity. The relationship — the "importance" — of mass and velocity remain the same in both equations.  The kinetic energy equation adds a constant (1/2) and an operation (squaring the velocity), but this only changes the descriptive quality of the number on the left-hand side of the equation.  In the case of the momentum, the number changes in a linear fashion.  In the case of kinetic energy, the number changes in a logarithmic fashion. Consider the stucture of the target.   It changes the momentum and kinetic energy of the projectile  For any given combination of mass and velocity, that projectile will have ONE specific figure for momentum and ONE specific figure for kinetic energy.  As the projectile contacts the target, the energy of the former is transferred to the latter.  In this case, the mass of the projectile remains practically the same, and the energy transfer has the effect only of reducing the projectile’s velocity. If you could measure the change in velocity from the instant before the arrow contacts the target until the time it comes to a full stop, you’d find that the figure for momentum changes in a straight-line fashion (assuming the arrow decelerates at a linear rate), and the KE figure reduces in a logarithmic fashion.   In other words, if you could measure the velocity of the penetrating arrow in regular 1-millisecond intervals, you would see an evenly spaced change in the momentum figure, and you would see a gradually decreasing change in the KE figure.  This is not a sign that KE is "less relevant".   It’s probably a sign that the observer is mis-interpreting the logarithmic change in the KE figure as a linear function. > The only significant "forces" acting on the arrow once it has cleared > the bow are friction and gravity. >Frictional forces on an arrow have directly a negligible effect.

Comparing the frictional forces of air to the frictional forces of the target, yeah, that would be reasonably correct. >Its the result of having to move the (mass of) air out of the way to ‘get >through’ that generates the drag force.

"Moving the mass of air out of the way" IS friction. Perhaps you’re thinking about the transformation of kinetic energy to *thermal* energy — the heating of the arrow and air particles.   This would be one of the effects of friction and a negligible effect on the flight of the arrow.  The other would be the direct transfer of the arrow’s kinetic energy to the kinetic energy of the air particles: they are *moved* out of the way when they contact (friction) the surface of the arrow. Cheers,

Response:

I LOVE it when someone takes up discussing with Joe in a logical and intelligent way – we get to read a VERY good exposition and see thought processes in action from both sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>. Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart enough to make a comfortable decision on some of this stuff so it’s fun to sit in the peanut gallery and try to follow along. Joe and Wayl, thanks!  🙂 hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his point, I also tend to the Wayl arguement about the displacement issue being identical for both materials.   Joe, what am I missing? thanks… – Hide quoted text — Show quoted text ->> If a light arrow and a heavy arrow are launched from a bow with >> identical energy-transfer characteristics for both, the lighter arrow >> will travel faster.   Of course it will decelerate faster when it >> hits. Going from 200 fps to zero is "decelerating faster" than going >> from 100 fps to zero in the same time interval. >> But if both arrows have the same kinetic energy and cutting properties >> (blade sharpness, surface area, ruggedness and shape of the broadhead >> and shaft), why would a lighter arrow penetrate less than a heavier >> arrow? >Wayl >What’s needed here is to understand the difference between momentum and >kinetic energy re arrow penetration. Although they are related they are >not the same thing anymore than say volume and mass are not they same >thing (though related). > Not a good analogy.   Volume exists regardless of mass.  Volume is a > measure of space, alone.  Any amount of mass may occupy that volume — > or not (there can be a complete vacuum within a volume).  I suspect > that you are thinking about one of the physical properties of a gas. > Regardless of the mass of the gas, it will "occupy" the available > volume of a container. > Kinetic energy and momentum are both measures of mass and velocity, > nothing more.  They are just calculated differently. >Suppose you have a target made up of a whole series of glass sheets with >air gaps between them and you shoot an arrow at it. > Which pretty much describes the pile of sand you mention later:  glass > sheets with air gaps in between.  The primary difference is the size > and shape of the grains (sheets) and the volume of the air gaps.  Even > the material (silica) is the same for both! >The arrow hits the >first sheet – what determines whether the arrow breaks through this >sheet is the arrow momentum, not it’s kinetic energy. > Wrong. If the arrow has enough kinetic energy to continue moving > forward, then it also has enough momentum, or vice versa.  Drop the > KE, and you’ll also drop the momentum. > Here’s the verbal definition of momentum:  "the property of a moving > body that determines the length of time required to bring it to rest > when under the action of a constant force or moment". > A mass of two kilograms moving along at five meters per second will > have a momentum of ten kilogram-meters per second. > This means only that in order to bring the object to a full stop, an > opposing force of ten kilogram-meters must be applied for one second. > That force may be the instantaneous or cumulative effect of friction. > Instantaneous — arrow hits a dense, fixed object.  Cumulative — > arrow penetrates into or through a sand pile or a torso. >If the arrow has  enough momentum it goes through if not it doesn’t. > Same with the sand.  If the arrow has enough KE (or momentum) to move > the first grain of sand, it gets to try it again on the next grain, > and the next, and so on until the kinetic energy is dissipated > (transferred to the glass or the grains of sand). >Let’s say it does. In breaking through the sheet the arrow loses energy. > Same with the sand pile.  The arrow loses some energy when it hits the > first grain of sand (or hair, skin, muscle, bone and other tissue > cells, to bring the analogy closer to the NG theme).  It continues > transferring energy to each subsequent particle (cell or grain) it > encounters until it comes to rest. > Perhaps you’re thinking that the arrow’s act of "breaking" through the > glass sheets is fundamentally different from "penetrating" the sand > pile (or mass of tissue).  Not really.   Some of the kinetic energy of > the arrow is transferred through friction into breaking the molecular > bonds of the glass sheet.  In the case of the sand pile, molecular > bonds may not be broken, but the arrow’s kinetic energy, through > friction, becomes the kinetic energy of the sand particles (they are > moved aside). >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops. > Exactly. >The arrow hits the  second sheet – enough momentum it breaks through not >enough and it  doesn’t and so on until the arrow doesn’t have enough >momentum to break  the sheet and it stops. With this target the penetration >is determined solely by the arrow momentum. The arrow KE is irrelevant. >If the target instead is a pile of sand then however small the arrow >momentum it will keep going, it doesn’t have to ‘cut through’ anything. > It depends on what you mean by "cut".  If "cut" means breaking > molecular bonds, then there may be an arguable distinction between > piercing a solid target and tunneling through a pile of particles.  In > the former, the energy of the moving arrow is transferred to the > target and it both breaks bonds and moves particles aside.  In the > latter, the energy is transferred almost exclusively to moving > particles aside; few, if any, chemical bonds are broken in the > process. In either case, the arrow doesn’t care; friction makes it > give up energy (and lose momentum) to the target, which will use its > newly found energy in whatever way it sees fit. >The arrow just travels along losing energy to friction until the >velocity drops to zero. With this target penetration is determined >solely by the arrow KE. The arrow momentum is irrelevant. > Maybe it’ll look like I’m nit-picking words, but I think the > distinction is important.  Penetration isn’t "determined solely" by > either kinetic energy or momentum. >Most targets are bit of both, ‘cutting through’ and friction. > "Cutting through" IS friction. >What  decides the balance between momentum and KE is the nature of >the target. > There is no "balance" between kinetic energy and momentum.  They are > both expressions of the same three fundamental factors:  mass, > distance and time. > Take a look at the equations: > Momentum —  the product of mass and velocity.   momentum = m * v. > Kinetic energy —   one half the product of mass of the particle and > the square of its velocity.   KE = 1/2 (m * v^2) > Notice the *only* variables in both equations:  mass and velocity. > The relationship — the "importance" — of mass and velocity remain > the same in both equations.  The kinetic energy equation adds a > constant (1/2) and an operation (squaring the velocity), but this only > changes the descriptive quality of the number on the left-hand side of > the equation.  In the case of the momentum, the number changes in a > linear fashion.  In the case of kinetic energy, the number changes in > a logarithmic fashion. > Consider the stucture of the target.   It changes the momentum and > kinetic energy of the projectile  For any given combination of mass > and velocity, that projectile will have ONE specific figure for > momentum and ONE specific figure for kinetic energy.  As the > projectile contacts the target, the energy of the former is > transferred to the latter.  In this case, the mass of the projectile > remains practically the same, and the energy transfer has the effect > only of reducing the projectile’s velocity. > If you could measure the change in velocity from the instant before > the arrow contacts the target until the time it comes to a full stop, > you’d find that the figure for momentum changes in a straight-line > fashion (assuming the arrow decelerates at a linear rate), and the KE > figure reduces in a logarithmic fashion. > In other words, if you could measure the velocity of the penetrating > arrow in regular 1-millisecond intervals, you would see an evenly > spaced change in the momentum figure, and you would see a gradually > decreasing change in the KE figure.  This is not a sign that KE is > "less relevant".   It’s probably a sign that the observer is > mis-interpreting the logarithmic change in the KE figure as a linear > function. >> The only significant "forces" acting on the arrow once it has cleared >> the bow are friction and gravity. >Frictional forces on an arrow have directly a negligible effect. > Comparing the frictional forces of air to the frictional forces of the > target, yeah, that would be reasonably correct. >Its the result of having to move the (mass of) air out of the way to ‘get >through’ that generates the drag force. > "Moving the mass of air out of the way" IS friction. > Perhaps you’re thinking about the transformation of kinetic energy to > *thermal* energy — the heating of the arrow and air particles.   This > would be one of the effects of friction and a negligible effect on the > flight of the arrow.  The other would be the direct transfer of the > arrow’s kinetic energy to the kinetic energy of the air particles: > they are *moved* out of the way when they contact (friction) the > surface of the arrow. > Cheers,

– TexARC publicize the sport of Archery! http://www.texasarchery.org Outgoing messages scanned for viruses by Nortons AV 2002

Response:

> I LOVE it when someone takes up discussing with Joe in a logical and intelligent way > – we get to read a VERY good exposition and see thought processes in action from both > sides – without blustering or a lot of finger-poking-in-the-eye stuff<RBG>.

Thanks for the compliment.  It’s fun to argue rational points in a rational manner.  If I’m proven wrong about certain points at issue (which is quite possible, anytime), then I get to learn something new. > Certainly not saying that JOE is wrong, (nor necessarily right,)  just that I’m not smart > enough to make a comfortable decision on some of this stuff so it’s fun to sit in the > peanut gallery and try to follow along.

Unnecessary modesty.  If you’re having fun "watching from the peanut gallery", then you most certainly are smart enough to make a decision, even if it’s the wrong one! :] >Joe and Wayl, thanks!  🙂

My pleasure. >hurm…While I am certain that Joe chose the glass and sand analogy purposely to make his >point, I also tend to the Wayl arguement about the displacement issue being identical for >both materials.   Joe, what am I missing?

Joe will surely give a better explanation for his choice of hypothetical targets, but I’d like to take a guess, and also take the chance to elaborate on my point that kinetic energy and momentum are equally "important for penetration". Being a rigid material, the glass panes will have a measureable, discrete figure for the impact energy required to make it shatter. So, for a certain sheet of glass, we can say that the energy from an arrow that has a momentum of 1000 gram-centimeters per second will shatter the glass.   To get this figure, we can choose a hypothetical arrow of 100 grams travelling at 10 centimeters per second.  100 grams (mass)  x  10 centimeters per second (velocity) = 1000 gram-centimeters per second (momentum). Double the velocity of the arrow to 20 centimeters per second, and you get twice the momentum, 2000 gram-centimeters per second.   If an arrow travelling at 2000 g-cm/sec clears the first sheet of glass and isn’t disturbed by flying shards or any other obstruction, then it will have enough momentum to break the second sheet of similar glass. The calculation follows a nice, linear path.  Twice as much velocity equals twice as much momentum equals sufficient force to break TWO sheets of glass.   But wait!  What happens when you use the basic kinetic energy equation?  (Just for simplicity, I’ll drop the fraction "1/2" in the equation KE = 1/2 m x v^2 — it doesn’t really matter for this discussion.) Double the velocity and you get a quadrupled figure for kinetic energy.   100 grams x 10 cm/sec x 10 cm/sec = 10,000 gm-cm^2/sec^2. 100 grams x 20 cm/sec x 20 cm/sec = 40,000 gm-cm^2/sec^2. This doesn’t mean that kinetic energy is either less or more important than momentum, it just shows that the figures used to represent one label follow a different rule than does the other. The only thing we’ve done is double the velocity, for BOTH equations. We’ve done nothing other than double the velocity of the arrow, and we get two drastically different numbers for momentum and kinetic energy.  But it’s the single change in velocity that produced those two wildly different numbers, not the other way around.     People are accustomed to thinking arithmetically, so the geometric (logarithmic) progression of numbers looks bizarre.   If people were trained to think only geometrically, the linear (arithmetic) progression of numbers would look bizarre. But the arrow behaves the same, regardless of whether you use KE figures or momentum figures to describes its actual performance. Why the difference between KE and momentum?   Momentum is a measure of mechanical force (any action that tends to maintain or alter the position of a body or to distort it).   The momentum equation answsers the question "How much force (work) must I apply to this thing to make it stop moving?" Kinetic energy is a measure of the  energy (the *capacity* for doing work) of motion.  Work is done when energy is transferred from one body to another, moving it or deforming it.  As I’ve said already, BOTH momentum and KE are functions of mass and velocity, nothing more.  For any given arrow mass, the potentional energy of the bow gets transferred into the kinetic energy of the arrow, which is then tranferred into the motion or deformation of the target.  The kinetic energy equation answers the question "How much capacity for work do I have in this moving object?" The questions look quite different, but in a discussion of arrow penetration,  the questions and answers are essentially the same; they’re just expressed in a different dialect. Now, about the sand — can you imagine how terribly messy it would be to calculate the energy needed to move each grain of sand out of the way?  If I were a physics schooteacher, I’d give such a problem to an unruly student for *punishment*. I suspect that’s why Joe chose the glass sheet example. And I’d better shut up, anyway, before this turns in to an unauthorised, somewhat unqualified and certainly unsolicited textbook on Newtonian mechanics. Cheers, – Hide quoted text — Show quoted text ->thanks… > >> If a light arrow and a heavy arrow are launched from a bow with > >> identical energy-transfer characteristics for both, the lighter arrow > >> will travel faster.   Of course it will decelerate faster when it > >> hits. Going from 200 fps to zero is "decelerating faster" than going > >> from 100 fps to zero in the same time interval. > >> But if both arrows have the same kinetic energy and cutting properties > >> (blade sharpness, surface area, ruggedness and shape of the broadhead > >> and shaft), why would a lighter arrow penetrate less than a heavier > >> arrow? > >Wayl > >What’s needed here is to understand the difference between momentum and > >kinetic energy re arrow penetration. Although they are related they are > >not the same thing anymore than say volume and mass are not they same > >thing (though related). > Not a good analogy.   Volume exists regardless of mass.  Volume is a > measure of space, alone.  Any amount of mass may occupy that volume — > or not (there can be a complete vacuum within a volume).  I suspect > that you are thinking about one of the physical properties of a gas. > Regardless of the mass of the gas, it will "occupy" the available > volume of a container. > Kinetic energy and momentum are both measures of mass and velocity, > nothing more.  They are just calculated differently. > >Suppose you have a target made up of a whole series of glass sheets with > >air gaps between them and you shoot an arrow at it. > Which pretty much describes the pile of sand you mention later:  glass > sheets with air gaps in between.  The primary difference is the size > and shape of the grains (sheets) and the volume of the air gaps.  Even > the material (silica) is the same for both! > >The arrow hits the > >first sheet – what determines whether the arrow breaks through this > >sheet is the arrow momentum, not it’s kinetic energy. > Wrong. If the arrow has enough kinetic energy to continue moving > forward, then it also has enough momentum, or vice versa.  Drop the > KE, and you’ll also drop the momentum. > Here’s the verbal definition of momentum:  "the property of a moving > body that determines the length of time required to bring it to rest > when under the action of a constant force or moment". > A mass of two kilograms moving along at five meters per second will > have a momentum of ten kilogram-meters per second. > This means only that in order to bring the object to a full stop, an > opposing force of ten kilogram-meters must be applied for one second. > That force may be the instantaneous or cumulative effect of friction. > Instantaneous — arrow hits a dense, fixed object.  Cumulative — > arrow penetrates into or through a sand pile or a torso. > >If the arrow has  enough momentum it goes through if not it doesn’t. > Same with the sand.  If the arrow has enough KE (or momentum) to move > the first grain of sand, it gets to try it again on the next grain, > and the next, and so on until the kinetic energy is dissipated > (transferred to the glass or the grains of sand). > >Let’s say it does. In breaking through the sheet the arrow loses energy. > Same with the sand pile.  The arrow loses some energy when it hits the > first grain of sand (or hair, skin, muscle, bone and other tissue > cells, to bring the analogy closer to the NG theme).  It continues > transferring energy to each subsequent particle (cell or grain) it > encounters until it comes to rest. > Perhaps you’re thinking that the arrow’s act of "breaking" through the > glass sheets is fundamentally different from "penetrating" the sand > pile (or mass of tissue).  Not really.   Some of the kinetic energy of > the arrow is transferred through friction into breaking the molecular > bonds of the glass sheet.  In the case of the sand pile, molecular > bonds may not be broken, but the arrow’s kinetic energy, through > friction, becomes the kinetic energy of the sand particles (they are > moved aside). > >It’s KE drops (i.e.  its velocity) and hence the arrow momentum drops. > Exactly. > >The arrow hits the  second sheet – enough momentum it breaks through not > >enough and it  doesn’t and so on until the arrow doesn’t have enough > >momentum to break  the sheet and it stops. With this target the penetration > >is determined solely by the arrow momentum. The arrow KE is irrelevant. > >If the target instead is a pile of sand then however small the arrow > >momentum it will keep going, it doesn’t have to ‘cut through’ anything. > It depends on what you mean by "cut".  If "cut" means

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