One of my hobbies is archery which has its own forums such as this where the subjects of KE and Momentum are hotly debated.
One recurring question is the "weight" of the variables of the KE= 1/2MV^2 equation.
Is mass more "significant" or is velocity? Most say it is the velocity because it is squared, but I was told by someone that to determine the more significant contributor to the KE calculus was necessary...integral...brain started smoking......ground to a halt....
Could someone please enlighten me so I can return to my archery discussions with a modicum of intelligence.
Thanks!
QUOTE (500 fps+Jul 28 2009, 03:22 PM)
One of my hobbies is archery ...
Could someone please enlighten me so I can return to my archery discussions with a modicum of intelligence.
Shoot the arrow straight and you will hit the target.
..
Could someone please enlighten me so I can return to my archery discussions with a modicum of intelligence.
Shoot the arrow straight and you will hit the target.
..
This looks like a good (interesting) starting point:-
http://en.wikipedia.org/wiki/Drag_(physics)
http://en.wikipedia.org/wiki/Drag_(physics)
I'm not sure if the question can be answered absolutely without clearly defining what "significant" means. An arrow with no mass doesn't have any energy no matter how fast it's going, and an arrow with no velocity has no energy no matter how massive it is, so obviously both parts are necessary.
One could argue that velocity *must* be more important, because that component is squared, so that if you double the velocity, you quadruple the energy, but the thing is, you then obviously have to impart quadruple the energy to get it to that state. You can't just say, "Gee, I think I'll have better results by doubling the velocity than by doubling the mass," because what you're really saying, then, is that it's better to quadruple the energy you're imparting to the arrow than doubling it. You could just turn it around and say that if you had quadruple the energy to impart to the arrow, it might be better spent by quadrupling the mass rather than merely doubling the velocity. I get the feeling I'm just talking in circles here, but that's kind of my point. Energy is energy. It is what it is. It can't be decomposed into, "The mass carries this percentage of it, and the velocity carries that percentage." If it could, then the formula would have a plus sign in it. (And then the units wouldn't combine correctly, and no one wants to see that.)
That being said, I suppose there's some optimal weight that an arrow could have where a bow would be able to impart the maximum amount of its energy into the arrow, rather than it being dissipated in other ways. If you tried to launch a piece of straw (and forget air resistance for the moment), most of the energy would be dissipated by the bow violently trying to resume its former shape. If you tried to launch a 180 lb. metal rod, the bow would be plenty satisfied to impart its energy into you as much as the rod. Besides, it probably wouldn't go very far anyway. But, in any case, the question here is not "which is more important, mass or velocity," but "at which mass is the bow able to most efficiently impart its stored energy."
Yeah, I know the OP is probably long gone. I just felt like writing something today.
One could argue that velocity *must* be more important, because that component is squared, so that if you double the velocity, you quadruple the energy, but the thing is, you then obviously have to impart quadruple the energy to get it to that state. You can't just say, "Gee, I think I'll have better results by doubling the velocity than by doubling the mass," because what you're really saying, then, is that it's better to quadruple the energy you're imparting to the arrow than doubling it. You could just turn it around and say that if you had quadruple the energy to impart to the arrow, it might be better spent by quadrupling the mass rather than merely doubling the velocity. I get the feeling I'm just talking in circles here, but that's kind of my point. Energy is energy. It is what it is. It can't be decomposed into, "The mass carries this percentage of it, and the velocity carries that percentage." If it could, then the formula would have a plus sign in it. (And then the units wouldn't combine correctly, and no one wants to see that.)
That being said, I suppose there's some optimal weight that an arrow could have where a bow would be able to impart the maximum amount of its energy into the arrow, rather than it being dissipated in other ways. If you tried to launch a piece of straw (and forget air resistance for the moment), most of the energy would be dissipated by the bow violently trying to resume its former shape. If you tried to launch a 180 lb. metal rod, the bow would be plenty satisfied to impart its energy into you as much as the rod. Besides, it probably wouldn't go very far anyway. But, in any case, the question here is not "which is more important, mass or velocity," but "at which mass is the bow able to most efficiently impart its stored energy."
Yeah, I know the OP is probably long gone. I just felt like writing something today.
Nice reply, FSM.
Intuitively, I think velocity is more important if you're trying to penetrate the target more deeply, but mass is more important if you're trying to knock it down.
This might just be related to the surface area of the point, though.
Intuitively, I think velocity is more important if you're trying to penetrate the target more deeply, but mass is more important if you're trying to knock it down.
This might just be related to the surface area of the point, though.
Sure, velocity has to be important if you're talking about deadliness. I'd rather be hit by a freight train going some arbitrarily low speed (say, 1 cm/s) than by a baseball carrying the same kinetic energy. It's not just the surface area, either; I'd say the same thing if you mounted a baseball on a rod on the front of the train. It'd just push me at a nice, comfortable rate until I was going the same speed as the train. (Unless I were backed up to a wall or something, in which case, it'd result in a slow, painful, and gruesome death.)
It's just that the OP's question wasn't about deadliness (which is hard to define exactly anyway), but about the relative contribution of mass and velocity to kinetic energy. Deadliness has more to do with the manner in which the projectile distributes that energy to the target. Someone getting shot while wearing a bulletproof vest has to absorb the same energy as someone not wearing one... or perhaps has to absorb more, in the case that the bullet passes all the way through without it.
It's just that the OP's question wasn't about deadliness (which is hard to define exactly anyway), but about the relative contribution of mass and velocity to kinetic energy. Deadliness has more to do with the manner in which the projectile distributes that energy to the target. Someone getting shot while wearing a bulletproof vest has to absorb the same energy as someone not wearing one... or perhaps has to absorb more, in the case that the bullet passes all the way through without it.
QUOTE (500 fps+Jul 28 2009, 10:22 AM)
One of my hobbies is archery which has its own forums such as this where the subjects of KE and Momentum are hotly debated.
One recurring question is the "weight" of the variables of the KE= 1/2MV^2 equation.
Is mass more "significant" or is velocity? Most say it is the velocity because it is squared, but I was told by someone that to determine the more significant contributor to the KE calculus was necessary...integral...brain started smoking......ground to a halt....
Could someone please enlighten me so I can return to my archery discussions with a modicum of intelligence.
Thanks!
One recurring question is the "weight" of the variables of the KE= 1/2MV^2 equation.
Is mass more "significant" or is velocity? Most say it is the velocity because it is squared, but I was told by someone that to determine the more significant contributor to the KE calculus was necessary...integral...brain started smoking......ground to a halt....
Could someone please enlighten me so I can return to my archery discussions with a modicum of intelligence.
Thanks!
I think the answer is apparent. Whomever told you that calculus was needed to determine the answer to your question was a complete blow hard who was likely just trying to make himself seem more important.
QUOTE (FlyingSpaghettiMonster+)
That being said, I suppose there's some optimal weight that an arrow could have where a bow would be able to impart the maximum amount of its energy into the arrow, rather than it being dissipated in other ways. If you tried to launch a piece of straw (and forget air resistance for the moment), most of the energy would be dissipated by the bow violently trying to resume its former shape. If you tried to launch a 180 lb. metal rod, the bow would be plenty satisfied to impart its energy into you as much as the rod. Besides, it probably wouldn't go very far anyway. But, in any case, the question here is not "which is more important, mass or velocity," but "at which mass is the bow able to most efficiently impart its stored energy."
You make a good point, however, consider this. Different materials spring back into their original shape with differing speeds, as do different thicknesses of a bow. Therefore, choosing a bow made of the right material (choosing one with a stronger draw, and therefore a faster rebound) to use with a random arrow is more likely to get you a good shot than picking just the right arrow for some random bow, as the question of imparting more energy becomes rather moot: You have to pull harder to get a bow with a stronger draw drawn. So long as you don't choose a bow with such a strong draw that it causes you undue strain or limit how far you can draw it, the bow with the stronger draw will inevitably produce better results, which is due to the fact that it shoots the arrow faster.
QUOTE (FlyingSpaghettiMonster+Nov 4 2009, 12:55 AM)
Sure, velocity has to be important if you're talking about deadliness. I'd rather be hit by a freight train going some arbitrarily low speed (say, 1 cm/s) than by a baseball carrying the same kinetic energy. It's not just the surface area, either; I'd say the same thing if you mounted a baseball on a rod on the front of the train. It'd just push me at a nice, comfortable rate until I was going the same speed as the train. (Unless I were backed up to a wall or something, in which case, it'd result in a slow, painful, and gruesome death.)
It's just that the OP's question wasn't about deadliness (which is hard to define exactly anyway), but about the relative contribution of mass and velocity to kinetic energy. Deadliness has more to do with the manner in which the projectile distributes that energy to the target. Someone getting shot while wearing a bulletproof vest has to absorb the same energy as someone not wearing one... or perhaps has to absorb more, in the case that the bullet passes all the way through without it.
With an arrow, though, in contrast to a freight train at any speed, the projectile must be moving sufficiently fast to overcome the tendency to fall, which rules out very slow speeds like 1cm/sec.
I wasn't thinking in terms of deadliness, just the effect on a target (the target in my mind was a cylinder of stiff penetrable material with a bulls eye painted on it - like the ones I've seen on TV archery competitions). I can imagine a very heavy arrow knocking such a target over if it was blunt enough, even if it was going pretty slow. A fast moving yet lighter arrow seems more likely to penetrate, although if it was blunt it might bounce off.
I think a good arrow would be heavy enough to stay its course and not get blown around by wind, but light enough to achieve maximum acceleration and therefore velocity from the bow.
It's just that the OP's question wasn't about deadliness (which is hard to define exactly anyway), but about the relative contribution of mass and velocity to kinetic energy. Deadliness has more to do with the manner in which the projectile distributes that energy to the target. Someone getting shot while wearing a bulletproof vest has to absorb the same energy as someone not wearing one... or perhaps has to absorb more, in the case that the bullet passes all the way through without it.
With an arrow, though, in contrast to a freight train at any speed, the projectile must be moving sufficiently fast to overcome the tendency to fall, which rules out very slow speeds like 1cm/sec.
I wasn't thinking in terms of deadliness, just the effect on a target (the target in my mind was a cylinder of stiff penetrable material with a bulls eye painted on it - like the ones I've seen on TV archery competitions). I can imagine a very heavy arrow knocking such a target over if it was blunt enough, even if it was going pretty slow. A fast moving yet lighter arrow seems more likely to penetrate, although if it was blunt it might bounce off.
I think a good arrow would be heavy enough to stay its course and not get blown around by wind, but light enough to achieve maximum acceleration and therefore velocity from the bow.
QUOTE (light in the tunnel+Nov 3 2009, 09:13 PM)
I wasn't thinking in terms of deadliness, just the effect on a target (the target in my mind was a cylinder of stiff penetrable material with a bulls eye painted on it - like the ones I've seen on TV archery competitions). I can imagine a very heavy arrow knocking such a target over if it was blunt enough, even if it was going pretty slow. A fast moving yet lighter arrow seems more likely to penetrate, although if it was blunt it might bounce off.
The mass of an arrow makes absolutely no difference in determining whether it penetrates or pushes. It's the size of the surface striking the target that does.
An arrow with a blunt tip which strikes a target with the same momentum as an arrow with a sharp tip is more likely to bowl it over, regardless of whether that momentum was achieved mostly from mass or velocity.
The mass of an arrow makes absolutely no difference in determining whether it penetrates or pushes. It's the size of the surface striking the target that does.
An arrow with a blunt tip which strikes a target with the same momentum as an arrow with a sharp tip is more likely to bowl it over, regardless of whether that momentum was achieved mostly from mass or velocity.
It really depends on what you mean by significant. (as FSM pointed out before)
If you are talking about the amount of energy transferred to the target, and you can assume that all energy from the projectile is transferred, (such as in an arrow which does NOT completely penetrate and pass through the target, or a bullet which remains in the target), then a change in velocity has more effect on the change in energy than a change in mass does.
If you are talking about the amount of energy transferred to the target, and you can assume that all energy from the projectile is transferred, (such as in an arrow which does NOT completely penetrate and pass through the target, or a bullet which remains in the target), then a change in velocity has more effect on the change in energy than a change in mass does.
QUOTE (MjolnirPants+Nov 4 2009, 03:32 AM)
The mass of an arrow makes absolutely no difference in determining whether it penetrates or pushes. It's the size of the surface striking the target that does.
An arrow with a blunt tip which strikes a target with the same momentum as an arrow with a sharp tip is more likely to bowl it over, regardless of whether that momentum was achieved mostly from mass or velocity.
You've brought up a good point by bringing momentum into it, but I'm not sure I agree entirely with what you're saying.
Suppose you have two arrows, one with a blunt tip, and one with a sharp tip, and they both have the same mass and velocity, and therefore the same kinetic energy and momentum. I would think the sharp tip would penetrate more deeply, but they both would transfer the same momentum to the target. I think this would still be true whether the two arrows had different masses and velocities, so long as they had the same momentum.
But now it's getting complicated because we're changing too many variables at the same time. Suppose we talked about two identically-shaped arrows with different masses but identical momentums. I would argue that the heavier arrow would penetrate further, because it would have to displace more mass in the target before it would transfer all of its momentum into the target material. This, of course, is making some assumptions about the velocity in both cases being high enough to render the target material's cohesion insignificant, which I'm not sure would be the case for arrows. You might want to check "Impact Depth" on Wikipedia and see whether I'm on the right track.
Looking at it this way, and going back to the original question on kinetic energy, it would seem that if you had the same kinetic energy to spend, you might be better served by increasing the mass at the expense of velocity, but not to the point where the velocity was too low for purposes of aiming or distance or overcoming the internal cohesion of the target. The heavier arrow would penetrate deeper (by virtue of its greater mass), and would have greater knock-down power (because a heavier object has more momentum than a lighter one at the same kinetic energy).
An arrow with a blunt tip which strikes a target with the same momentum as an arrow with a sharp tip is more likely to bowl it over, regardless of whether that momentum was achieved mostly from mass or velocity.
You've brought up a good point by bringing momentum into it, but I'm not sure I agree entirely with what you're saying.
Suppose you have two arrows, one with a blunt tip, and one with a sharp tip, and they both have the same mass and velocity, and therefore the same kinetic energy and momentum. I would think the sharp tip would penetrate more deeply, but they both would transfer the same momentum to the target. I think this would still be true whether the two arrows had different masses and velocities, so long as they had the same momentum.
But now it's getting complicated because we're changing too many variables at the same time. Suppose we talked about two identically-shaped arrows with different masses but identical momentums. I would argue that the heavier arrow would penetrate further, because it would have to displace more mass in the target before it would transfer all of its momentum into the target material. This, of course, is making some assumptions about the velocity in both cases being high enough to render the target material's cohesion insignificant, which I'm not sure would be the case for arrows. You might want to check "Impact Depth" on Wikipedia and see whether I'm on the right track.
Looking at it this way, and going back to the original question on kinetic energy, it would seem that if you had the same kinetic energy to spend, you might be better served by increasing the mass at the expense of velocity, but not to the point where the velocity was too low for purposes of aiming or distance or overcoming the internal cohesion of the target. The heavier arrow would penetrate deeper (by virtue of its greater mass), and would have greater knock-down power (because a heavier object has more momentum than a lighter one at the same kinetic energy).
QUOTE (AlexG+Nov 4 2009,04:16 AM)
If you are talking about the amount of energy transferred to the target, and you can assume that all energy from the projectile is transferred, (such as in an arrow which does NOT completely penetrate and pass through the target, or a bullet which remains in the target), then a change in velocity has more effect on the change in energy than a change in mass does.
That's exactly the kind of response I was trying to avoid. If you had a genie who offered to magically enchant your bow to either magically double the mass of arrows as they left the bow, or magically double the velocity, and you had to choose, then yeah, you'd strongly want to consider taking the free extra velocity for a number of reasons, one of them being that it would quadruple your kinetic energy instead of doubling it. (Though in light of what I said above, I'm kind of leaning towards mass; it's not really the energy that kills, anyway. It's making the target lose blood.) Both of these options cost exactly the same, i.e., one "wish." But without the genie, we're stuck talking about energy. You want to double the energy via increased velocity? Great, you get a 41% increase. You want to do it via increased mass? You get a 100% increase. Having the velocity squared in the formula increases the cost of obtaining it by as much as it increases the benefits of having it.
That's exactly the kind of response I was trying to avoid. If you had a genie who offered to magically enchant your bow to either magically double the mass of arrows as they left the bow, or magically double the velocity, and you had to choose, then yeah, you'd strongly want to consider taking the free extra velocity for a number of reasons, one of them being that it would quadruple your kinetic energy instead of doubling it. (Though in light of what I said above, I'm kind of leaning towards mass; it's not really the energy that kills, anyway. It's making the target lose blood.) Both of these options cost exactly the same, i.e., one "wish." But without the genie, we're stuck talking about energy. You want to double the energy via increased velocity? Great, you get a 41% increase. You want to do it via increased mass? You get a 100% increase. Having the velocity squared in the formula increases the cost of obtaining it by as much as it increases the benefits of having it.
QUOTE (FlyingSpaghettiMonster+Nov 4 2009, 01:34 PM)
You've brought up a good point by bringing momentum into it, but I'm not sure I agree entirely with what you're saying.
Suppose you have two arrows, one with a blunt tip, and one with a sharp tip, and they both have the same mass and velocity, and therefore the same kinetic energy and momentum. I would think the sharp tip would penetrate more deeply, but they both would transfer the same momentum to the target. I think this would still be true whether the two arrows had different masses and velocities, so long as they had the same momentum.
Suppose you have two arrows, one with a blunt tip, and one with a sharp tip, and they both have the same mass and velocity, and therefore the same kinetic energy and momentum. I would think the sharp tip would penetrate more deeply, but they both would transfer the same momentum to the target. I think this would still be true whether the two arrows had different masses and velocities, so long as they had the same momentum.
Not quite. The arrow with the sharper tip would expend more energy overcoming the cohesion of the target to displace some of it's mass (via the act of penetration, sex jokes welcome), and so it would not transfer as much into moving the target backwards. You're right that the same amount of energy gets transfered, but the arrowhead shape helps to determines what that energy does. A blunt arrow would displace some of the mass through the forming of an indentation, but this is much less energy than the sharp tip would expend.
Ah, I don't think you quite caught what I was saying. I still stand by my statement that the amount of momentum transferred to the target (as a whole, not just at the impact point) is the same. If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other. I would also claim that this would be the same if the arrows had identical momentums, but different masses, velocities, energies, and tip shapes, with the added caveat that both arrows' masses are still negligible compared to the target itself (i.e., mass of target plus arrow #1 is approximately equal to mass of target plus arrow #2). The momentum of the system has to be conserved. The energy, too, I suppose, but that's a lot harder to track and account for, but in any case, some of that energy has to go into accelerating the target. How much? Exactly as much as is required to conserve the momentum.
Ah, I don't think you quite caught what I was saying. I still stand by my statement that the amount of momentum transferred to the target (as a whole, not just at the impact point) is the same. If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other. I would also claim that this would be the same if the arrows had identical momentums, but different masses, velocities, energies, and tip shapes, with the added caveat that both arrows' masses are still negligible compared to the target itself (i.e., mass of target plus arrow #1 is approximately equal to mass of target plus arrow #2). The momentum of the system has to be conserved. The energy, too, I suppose, but that's a lot harder to track and account for, but in any case, some of that energy has to go into accelerating the target. How much? Exactly as much as is required to conserve the momentum.
You're right about impact depth ignoring the cohesion, however the concept of impact depth in newtonian dynamics only applies to blunt projectiles. If you consider a target made of the same wood as the arrow, then you can see how cohesion becomes an important factor. Cohesion only becomes unimportant when the arrow is capable of displacing it's own mass' worth of material from the target, or in this case, becomes capable of burying itself slightly deeper than it's own length ('slightly deeper' as opposed to 'as deep' because of the mass of the nock, the metal tip needed to avoid the question of the arrow's cohesion, and the feathers) in a target whose thickness exceeds the arrows length. As you can imagine if you've ever seen archery, this never happens. Indeed, the target's density is usually much less than the arrow's (a bale of hay with a paper target on the front is the most common type), and still the arrows don't tend to bury themselves more than a few inches deep, and this is with a sharp tip.
I won't disagree about the cohesion factor or tip shape, but there are too many variables in flux for me to make general statement about this without opening myself up to a lot of yeah-but's. Shooting at a block of ballistics gel and shooting at a block of wood will obviously have different results. Shooting a sharp arrow will have different results than shooting a blunt one. You say that Newton's approximation only holds for blunt projectiles, and I would agree inasmuch as it applies to trying to predict *how far* the projectile will penetrate, but I would strongly suspect that even for non-blunt projectiles, it will still apply in saying how the *relative* depths might play out. But, then, another problem arises in that you can't exactly just change the masses without necessarily changing other factors. Do you try to hold velocity constant, and thereby change both momentum and kinetic energy? Or do you hold momentum constant, and thereby change velocity and energy? Or hold energy constant and thereby change momentum and velocity? And I would suspect that at arrow-like speeds, changing the velocity would greatly affect how well the arrow would be able to overcome cohesion in the target. I feel comfortable speculating that in *some* target materials, you'll get deeper penetration by increasing the mass, regardless of whether you choose to hold momentum constant or energy, and even if the tip is not blunt. For other materials, I wouldn't be so comfortable. It's really hard to characterize a "general" case.
I said that a heavier object has more momentum than a lighter one at the same kinetic energy, and I stand by it. The math holds. The details of what sorts of bows you're going to need is irrelevant. Just keep trying out different bows and arrows until you find an arrangment where (1/2)*m1*v1^2 = (1/2)*m2*v2^2. When you do, you'll find that if m1>m2, then m1*v1>m2*v2. (In fact, you'll find that m1*v1=sqrt(m1/m2)*m2*v2, and since m1>m2, sqrt(m1/m2)>1.)
Actually, the more I re-read your statement, the more puzzled I am by it. You want to use the weaker bow for the heavier arrow and the stronger one for the faster one. How is this not cheating? What, are you rooting for Team Velocity now? This isn't a football game, it's science!
(Just messing with you.)
FSM
QUOTE
But now it's getting complicated because we're changing too many variables at the same time. Suppose we talked about two identically-shaped arrows with different masses but identical momentums. I would argue that the heavier arrow would penetrate further, because it would have to displace more mass in the target before it would transfer all of its momentum into the target material. This, of course, is making some assumptions about the velocity in both cases being high enough to render the target material's cohesion insignificant, which I'm not sure would be the case for arrows. You might want to check "Impact Depth" on Wikipedia and see whether I'm on the right track.
You're right about impact depth ignoring the cohesion, however the concept of impact depth in newtonian dynamics only applies to blunt projectiles. If you consider a target made of the same wood as the arrow, then you can see how cohesion becomes an important factor. Cohesion only becomes unimportant when the arrow is capable of displacing it's own mass' worth of material from the target, or in this case, becomes capable of burying itself slightly deeper than it's own length ('slightly deeper' as opposed to 'as deep' because of the mass of the nock, the metal tip needed to avoid the question of the arrow's cohesion, and the feathers) in a target whose thickness exceeds the arrows length. As you can imagine if you've ever seen archery, this never happens. Indeed, the target's density is usually much less than the arrow's (a bale of hay with a paper target on the front is the most common type), and still the arrows don't tend to bury themselves more than a few inches deep, and this is with a sharp tip.QUOTE (->
| QUOTE |
| But now it's getting complicated because we're changing too many variables at the same time. Suppose we talked about two identically-shaped arrows with different masses but identical momentums. I would argue that the heavier arrow would penetrate further, because it would have to displace more mass in the target before it would transfer all of its momentum into the target material. This, of course, is making some assumptions about the velocity in both cases being high enough to render the target material's cohesion insignificant, which I'm not sure would be the case for arrows. You might want to check "Impact Depth" on Wikipedia and see whether I'm on the right track. |
You're right about impact depth ignoring the cohesion, however the concept of impact depth in newtonian dynamics only applies to blunt projectiles. If you consider a target made of the same wood as the arrow, then you can see how cohesion becomes an important factor. Cohesion only becomes unimportant when the arrow is capable of displacing it's own mass' worth of material from the target, or in this case, becomes capable of burying itself slightly deeper than it's own length ('slightly deeper' as opposed to 'as deep' because of the mass of the nock, the metal tip needed to avoid the question of the arrow's cohesion, and the feathers) in a target whose thickness exceeds the arrows length. As you can imagine if you've ever seen archery, this never happens. Indeed, the target's density is usually much less than the arrow's (a bale of hay with a paper target on the front is the most common type), and still the arrows don't tend to bury themselves more than a few inches deep, and this is with a sharp tip.
Looking at it this way, and going back to the original question on kinetic energy, it would seem that if you had the same kinetic energy to spend, you might be better served by increasing the mass at the expense of velocity, but not to the point where the velocity was too low for purposes of aiming or distance or overcoming the internal cohesion of the target. The heavier arrow would penetrate deeper (by virtue of its greater mass), and would have greater knock-down power (because a heavier object has more momentum than a lighter one at the same kinetic energy).
Looking at it this way, and going back to the original question on kinetic energy, it would seem that if you had the same kinetic energy to spend, you might be better served by increasing the mass at the expense of velocity, but not to the point where the velocity was too low for purposes of aiming or distance or overcoming the internal cohesion of the target. The heavier arrow would penetrate deeper (by virtue of its greater mass), and would have greater knock-down power (because a heavier object has more momentum than a lighter one at the same kinetic energy).
Not quite. Increasing the mass of an arrow means that more of the potential energy of the bow is required to overcome the arrow's inertia when fired. The end result is that increasing the mass of the arrow decreases it's velocity, while increasing the velocity (via a bow with a stronger pull) will not decrease the mass of the arrow.
QUOTE
Not quite. The arrow with the sharper tip would expend more energy overcoming the cohesion of the target to displace some of it's mass (via the act of penetration, sex jokes welcome), and so it would not transfer as much into moving the target backwards. You're right that the same amount of energy gets transfered, but the arrowhead shape helps to determines what that energy does. A blunt arrow would displace some of the mass through the forming of an indentation, but this is much less energy than the sharp tip would expend.
Ah, I don't think you quite caught what I was saying. I still stand by my statement that the amount of momentum transferred to the target (as a whole, not just at the impact point) is the same. If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other. I would also claim that this would be the same if the arrows had identical momentums, but different masses, velocities, energies, and tip shapes, with the added caveat that both arrows' masses are still negligible compared to the target itself (i.e., mass of target plus arrow #1 is approximately equal to mass of target plus arrow #2). The momentum of the system has to be conserved. The energy, too, I suppose, but that's a lot harder to track and account for, but in any case, some of that energy has to go into accelerating the target. How much? Exactly as much as is required to conserve the momentum.
QUOTE (->
| QUOTE |
| Not quite. The arrow with the sharper tip would expend more energy overcoming the cohesion of the target to displace some of it's mass (via the act of penetration, sex jokes welcome), and so it would not transfer as much into moving the target backwards. You're right that the same amount of energy gets transfered, but the arrowhead shape helps to determines what that energy does. A blunt arrow would displace some of the mass through the forming of an indentation, but this is much less energy than the sharp tip would expend. |
Ah, I don't think you quite caught what I was saying. I still stand by my statement that the amount of momentum transferred to the target (as a whole, not just at the impact point) is the same. If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other. I would also claim that this would be the same if the arrows had identical momentums, but different masses, velocities, energies, and tip shapes, with the added caveat that both arrows' masses are still negligible compared to the target itself (i.e., mass of target plus arrow #1 is approximately equal to mass of target plus arrow #2). The momentum of the system has to be conserved. The energy, too, I suppose, but that's a lot harder to track and account for, but in any case, some of that energy has to go into accelerating the target. How much? Exactly as much as is required to conserve the momentum.
You're right about impact depth ignoring the cohesion, however the concept of impact depth in newtonian dynamics only applies to blunt projectiles. If you consider a target made of the same wood as the arrow, then you can see how cohesion becomes an important factor. Cohesion only becomes unimportant when the arrow is capable of displacing it's own mass' worth of material from the target, or in this case, becomes capable of burying itself slightly deeper than it's own length ('slightly deeper' as opposed to 'as deep' because of the mass of the nock, the metal tip needed to avoid the question of the arrow's cohesion, and the feathers) in a target whose thickness exceeds the arrows length. As you can imagine if you've ever seen archery, this never happens. Indeed, the target's density is usually much less than the arrow's (a bale of hay with a paper target on the front is the most common type), and still the arrows don't tend to bury themselves more than a few inches deep, and this is with a sharp tip.
I won't disagree about the cohesion factor or tip shape, but there are too many variables in flux for me to make general statement about this without opening myself up to a lot of yeah-but's. Shooting at a block of ballistics gel and shooting at a block of wood will obviously have different results. Shooting a sharp arrow will have different results than shooting a blunt one. You say that Newton's approximation only holds for blunt projectiles, and I would agree inasmuch as it applies to trying to predict *how far* the projectile will penetrate, but I would strongly suspect that even for non-blunt projectiles, it will still apply in saying how the *relative* depths might play out. But, then, another problem arises in that you can't exactly just change the masses without necessarily changing other factors. Do you try to hold velocity constant, and thereby change both momentum and kinetic energy? Or do you hold momentum constant, and thereby change velocity and energy? Or hold energy constant and thereby change momentum and velocity? And I would suspect that at arrow-like speeds, changing the velocity would greatly affect how well the arrow would be able to overcome cohesion in the target. I feel comfortable speculating that in *some* target materials, you'll get deeper penetration by increasing the mass, regardless of whether you choose to hold momentum constant or energy, and even if the tip is not blunt. For other materials, I wouldn't be so comfortable. It's really hard to characterize a "general" case.
QUOTE
Not quite. Increasing the mass of an arrow means that more of the potential energy of the bow is required to overcome the arrow's inertia when fired. The end result is that increasing the mass of the arrow decreases it's velocity, while increasing the velocity (via a bow with a stronger pull) will not decrease the mass of the arrow.
I said that a heavier object has more momentum than a lighter one at the same kinetic energy, and I stand by it. The math holds. The details of what sorts of bows you're going to need is irrelevant. Just keep trying out different bows and arrows until you find an arrangment where (1/2)*m1*v1^2 = (1/2)*m2*v2^2. When you do, you'll find that if m1>m2, then m1*v1>m2*v2. (In fact, you'll find that m1*v1=sqrt(m1/m2)*m2*v2, and since m1>m2, sqrt(m1/m2)>1.)
Actually, the more I re-read your statement, the more puzzled I am by it. You want to use the weaker bow for the heavier arrow and the stronger one for the faster one. How is this not cheating? What, are you rooting for Team Velocity now? This isn't a football game, it's science!
(Just messing with you.)
FSM
QUOTE (FlyingSpaghettiMonster+Nov 4 2009, 04:40 PM)
Ah, I don't think you quite caught what I was saying. I still stand by my statement that the amount of momentum transferred to the target (as a whole, not just at the impact point) is the same.
This is true. I'm not arguing that one arrowhead transfers more energy than the other, but that the energy is used for different reactions in the target, based on the shape of the arrowhead.
My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
QUOTE
If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other.
This is what I'm saying isn't right. My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
QUOTE (->
| QUOTE |
| If you had two identical targets sitting on a frictionless plane, and you hit each one with a different arrow with identical mass, velocity, momentum, and energy, but different shaped tips, and assuming both targets absorbed the arrows, rather than them passing through or bouncing off, then both targets would be pushed backwards at identical velocities, albeit one would have an arrow stuck deeper in it than the other. |
This is what I'm saying isn't right.
My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
I won't disagree about the cohesion factor or tip shape, but there are too many variables in flux for me to make general statement about this without opening myself up to a lot of yeah-but's. Shooting at a block of ballistics gel and shooting at a block of wood will obviously have different results. Shooting a sharp arrow will have different results than shooting a blunt one. You say that Newton's approximation only holds for blunt projectiles, and I would agree inasmuch as it applies to trying to predict *how far* the projectile will penetrate, but I would strongly suspect that even for non-blunt projectiles, it will still apply in saying how the *relative* depths might play out. But, then, another problem arises in that you can't exactly just change the masses without necessarily changing other factors. Do you try to hold velocity constant, and thereby change both momentum and kinetic energy? Or do you hold momentum constant, and thereby change velocity and energy? Or hold energy constant and thereby change momentum and velocity? And I would suspect that at arrow-like speeds, changing the velocity would greatly affect how well the arrow would be able to overcome cohesion in the target. I feel comfortable speculating that in *some* target materials, you'll get deeper penetration by increasing the mass, regardless of whether you choose to hold momentum constant or energy, and even if the tip is not blunt. For other materials, I wouldn't be so comfortable. It's really hard to characterize a "general" case.
My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
I won't disagree about the cohesion factor or tip shape, but there are too many variables in flux for me to make general statement about this without opening myself up to a lot of yeah-but's. Shooting at a block of ballistics gel and shooting at a block of wood will obviously have different results. Shooting a sharp arrow will have different results than shooting a blunt one. You say that Newton's approximation only holds for blunt projectiles, and I would agree inasmuch as it applies to trying to predict *how far* the projectile will penetrate, but I would strongly suspect that even for non-blunt projectiles, it will still apply in saying how the *relative* depths might play out. But, then, another problem arises in that you can't exactly just change the masses without necessarily changing other factors. Do you try to hold velocity constant, and thereby change both momentum and kinetic energy? Or do you hold momentum constant, and thereby change velocity and energy? Or hold energy constant and thereby change momentum and velocity? And I would suspect that at arrow-like speeds, changing the velocity would greatly affect how well the arrow would be able to overcome cohesion in the target. I feel comfortable speculating that in *some* target materials, you'll get deeper penetration by increasing the mass, regardless of whether you choose to hold momentum constant or energy, and even if the tip is not blunt. For other materials, I wouldn't be so comfortable. It's really hard to characterize a "general" case.
I think we're getting too hung up on kinetic energy, which is not very meaningful when it comes to penetration, only to determining the energy needed to impart a certain momentum to the arrow. I think we can both agree that it takes more potential energy in the bow to impart more kinetic energy to the arrow, so when it comes to the arrow's performance at impact, it's more meaningful to discuss it's momentum.
p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
I think you're being inconsistent sometimes in distinguishing between momentum and kinetic energy. I mean, I can tell by your arguments that you know there's a difference, but you're not applying it correctly all the time.
One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No.
Another important difference is that the kinetic energy doesn't have to be conserved. It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
I think you're being inconsistent sometimes in distinguishing between momentum and kinetic energy. I mean, I can tell by your arguments that you know there's a difference, but you're not applying it correctly all the time.
One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No.
Another important difference is that the kinetic energy doesn't have to be conserved. It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
This is true, however to maintain the kinetic energy of a heavier arrow vs. a lighter arrow requires the imparting of more potential energy from the bow. If we're assuming the use of one bow with a set amount of potential energy (Ep) which it can impart, then we need to take into account acceleration, because it's momentum is p = mv where v = a(t') and a = F/m. So when we put those together, we get;
p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
I tried to follow your formula, but I noticed a couple of errors. First, momentum should be in units of kg*m/s, but your formula ((Ep/m)(t'))m would result in units of kg*m^2/s (which is neither energy nor momentum). (Or, on second look, I guess I should've just pointed out that you can't just substitute energy for force.) Second, when you try to apply your formula, you're using the same amount of time in both cases, which strongly implies that both arrows will wind up with the same velocity, regardless of mass. (If the bowstring is going to move the same distance in the same period of time, it necessarily has to take the arrow with it at the same speed.) I don't think this is what you intended.
If I were to try to set up this problem myself, I'm not sure what would be the best way to go about it. I was thinking of assuming the bow obeys Hooke's Law. If that were so, I think it'd be pretty easy to show that both arrows would necessarily *have* to have the same kinetic energy, regardless of mass. Since work = force * distance, you could integrate the force (F=k*x) over the distance ( [integral](k*x)*dx ), and get the amount of energy tranferred to the arrow. Neither mass nor time are needed here. Heck, you don't even need to assume Hooke's Law (which definitely wouldn't apply to a compound bow anyway). You could use any expression that gives force as a function of displacement.
QUOTE
I said that a heavier object has more momentum than a lighter one at the same kinetic energy, and I stand by it. The math holds. The details of what sorts of bows you're going to need is irrelevant. Just keep trying out different bows and arrows until you find an arrangment where (1/2)*m1*v1^2 = (1/2)*m2*v2^2. When you do, you'll find that if m1>m2, then m1*v1>m2*v2. (In fact, you'll find that m1*v1=sqrt(m1/m2)*m2*v2, and since m1>m2, sqrt(m1/m2)>1.)
This is true, however to maintain the kinetic energy of a heavier arrow vs. a lighter arrow requires the imparting of more potential energy from the bow. If we're assuming the use of one bow with a set amount of potential energy (Ep) which it can impart, then we need to take into account acceleration, because it's momentum is p = mv where v = a(t') and a = F/m. So when we put those together, we get;p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
QUOTE (->
| QUOTE |
| I said that a heavier object has more momentum than a lighter one at the same kinetic energy, and I stand by it. The math holds. The details of what sorts of bows you're going to need is irrelevant. Just keep trying out different bows and arrows until you find an arrangment where (1/2)*m1*v1^2 = (1/2)*m2*v2^2. When you do, you'll find that if m1>m2, then m1*v1>m2*v2. (In fact, you'll find that m1*v1=sqrt(m1/m2)*m2*v2, and since m1>m2, sqrt(m1/m2)>1.) |
This is true, however to maintain the kinetic energy of a heavier arrow vs. a lighter arrow requires the imparting of more potential energy from the bow. If we're assuming the use of one bow with a set amount of potential energy (Ep) which it can impart, then we need to take into account acceleration, because it's momentum is p = mv where v = a(t') and a = F/m. So when we put those together, we get;
p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
Actually, the more I re-read your statement, the more puzzled I am by it. You want to use the weaker bow for the heavier arrow and the stronger one for the faster one. How is this not cheating? What, are you rooting for Team Velocity now? This isn't a football game, it's science!
(Just messing with you.)
p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
Actually, the more I re-read your statement, the more puzzled I am by it. You want to use the weaker bow for the heavier arrow and the stronger one for the faster one. How is this not cheating? What, are you rooting for Team Velocity now? This isn't a football game, it's science!
(Just messing with you.)
That's not quite what I meant.
What I said was that you can achieve deeper penetration or a further push back by using a bow which imparts more velocity (by having a stronger pull, imparting more force to accelerate the arrow), while simply using a heavier arrow only means that while momentum remains the same, the velocity decreases.
And don't make fun of Team Velocity. We're going to smash Team Mass in the Superposition Bowl this year! (Sorry, I couldn't think of any newtonian terms with the word "super" in it...)
What I said was that you can achieve deeper penetration or a further push back by using a bow which imparts more velocity (by having a stronger pull, imparting more force to accelerate the arrow), while simply using a heavier arrow only means that while momentum remains the same, the velocity decreases.
And don't make fun of Team Velocity. We're going to smash Team Mass in the Superposition Bowl this year! (Sorry, I couldn't think of any newtonian terms with the word "super" in it...)
1) I believe MJPants is right about penetration absorbing energy and reducing momentum-transfer to the target. I'm basing this on the idea of the same toddler jumping into a pool of plastic balls of low viscosity or a pool of tennis balls of higher viscosity. The toddler is more likely to get his wind knocked out jumping into the tennis balls.
2) question: if two blunt arrows, one heavy and one light, hit a target, doesn't the lighter one have a greater propensity to bounce off the target than the heavier one, which would be more prone to knocking the same target over, or at least giving it a good wobble?
This would seem to have something to do with the energy of momentum reverberating back through the arrow that's lighter more so than transferring into the target. Isn't this similar to hitting a billiard ball with a marble or another billiard ball? Hit by a marble, a billiard ball will move a little but the marble will bounce backward, won't it? Whereas a billiard ball hitting another billiard ball will transfer all its momentum and stop. Or am I oversimplifying?
2) question: if two blunt arrows, one heavy and one light, hit a target, doesn't the lighter one have a greater propensity to bounce off the target than the heavier one, which would be more prone to knocking the same target over, or at least giving it a good wobble?
This would seem to have something to do with the energy of momentum reverberating back through the arrow that's lighter more so than transferring into the target. Isn't this similar to hitting a billiard ball with a marble or another billiard ball? Hit by a marble, a billiard ball will move a little but the marble will bounce backward, won't it? Whereas a billiard ball hitting another billiard ball will transfer all its momentum and stop. Or am I oversimplifying?
QUOTE (light in the tunnel+Nov 4 2009, 10:51 PM)
question: if two blunt arrows, one heavy and one light, hit a target, doesn't the lighter one have a greater propensity to bounce off the target than the heavier one, which would be more prone to knocking the same target over, or at least giving it a good wobble?
The mass of the arrow is irrelevant, except for it's contribution to momentum. If you look at the formula I used in my last post, you'll see that the only thing which affects the total momentum of two different arrows is the energy imparted into them by the bow. The factor which determines whether or not the arrow will bounce is the shape of the tip. The factors which determine how far the arrow will bounce is the resistance to movement of the target (it's mass and the friction or cohesion or both involved in it's mounting to the ground). The flexibility of the arrow determines how it will bounce (a born rigid arrow would bounce straight back, while a rubber arrow would bounce wildly) as well as whether the arrow will bounce or shatter (splinter, in the case of a fibrous arrow), as well as contributing to how far it will bounce (as it takes energy to deform the arrow, which is then released in a different manner through shattering/splintering, or in a different direction through it's ensuing wild bounce).
I could probably come up with the formula to show you how much of the arrow's momentum is returned to it, but I don't feel like sitting here for the next hour doing algebra to try and figure it out. You could probably find it online if you're interested. I know that ignoring gravity, rigidity (assuming both the arrow and target to be born rigid) and friction (and any mounting the target may have), the momentum imparted back to the arrow will be 1/2 of the energy expended during the impact.
The larger cue ball has a larger mass yes, but that's not the direct cause of it's ability to push the target ball further, it is it's larger momentum which results from this larger mass, as I'm sure you were imagining the projectile billiard ball travelling at the same velocity as the marble.
If the marble were moving faster so that it's momentum was equal to the momentum of the billiard ball, then the effect on the target billiard ball would be the same (assuming the billiard ball was the same rigidity as the marble). The reason the marble would bounce off the target billiard ball with more speed than the projectile billiard ball is simply because the marble has less mass, and therefore the amount of energy returned to it is enough to make it move further.
The mass of the arrow is irrelevant, except for it's contribution to momentum. If you look at the formula I used in my last post, you'll see that the only thing which affects the total momentum of two different arrows is the energy imparted into them by the bow. The factor which determines whether or not the arrow will bounce is the shape of the tip. The factors which determine how far the arrow will bounce is the resistance to movement of the target (it's mass and the friction or cohesion or both involved in it's mounting to the ground). The flexibility of the arrow determines how it will bounce (a born rigid arrow would bounce straight back, while a rubber arrow would bounce wildly) as well as whether the arrow will bounce or shatter (splinter, in the case of a fibrous arrow), as well as contributing to how far it will bounce (as it takes energy to deform the arrow, which is then released in a different manner through shattering/splintering, or in a different direction through it's ensuing wild bounce).
I could probably come up with the formula to show you how much of the arrow's momentum is returned to it, but I don't feel like sitting here for the next hour doing algebra to try and figure it out. You could probably find it online if you're interested. I know that ignoring gravity, rigidity (assuming both the arrow and target to be born rigid) and friction (and any mounting the target may have), the momentum imparted back to the arrow will be 1/2 of the energy expended during the impact.
QUOTE
This would seem to have something to do with the energy of momentum reverberating back through the arrow that's lighter more so than transferring into the target. Isn't this similar to hitting a billiard ball with a marble or another billiard ball? Hit by a marble, a billiard ball will move a little but the marble will bounce backward, won't it? Whereas a billiard ball hitting another billiard ball will transfer all its momentum and stop. Or am I oversimplifying?
Your analogy is pretty decent, but it's all about the momentum of the projectile ball and the mass of the target ball.The larger cue ball has a larger mass yes, but that's not the direct cause of it's ability to push the target ball further, it is it's larger momentum which results from this larger mass, as I'm sure you were imagining the projectile billiard ball travelling at the same velocity as the marble.
If the marble were moving faster so that it's momentum was equal to the momentum of the billiard ball, then the effect on the target billiard ball would be the same (assuming the billiard ball was the same rigidity as the marble). The reason the marble would bounce off the target billiard ball with more speed than the projectile billiard ball is simply because the marble has less mass, and therefore the amount of energy returned to it is enough to make it move further.
QUOTE
This is what I'm saying isn't right.
My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded.
The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless.
I think you're being inconsistent sometimes in distinguishing between momentum and kinetic energy. I mean, I can tell by your arguments that you know there's a difference, but you're not applying it correctly all the time.
One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No.
Another important difference is that the kinetic energy doesn't have to be conserved. It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
QUOTE (->
| QUOTE |
| This is what I'm saying isn't right. My point is that energy is required for the arrow to embed itself in a target. It must overcome the cohesion of the target, then impart kinetic energy perpendicular to it's direction of travel (by expending some of it's momentum) through the target to move it's mass out of the way to create the hole in which it is embedded. The blunt arrow would not penetrate the target as deeply, and so would not expend as much energy overcoming the cohesion of the target in order to embed itself. Therefore, the target hit by the blunt arrow would be pushed further back. It would be a rather small difference, but a difference nonetheless. |
I think you're being inconsistent sometimes in distinguishing between momentum and kinetic energy. I mean, I can tell by your arguments that you know there's a difference, but you're not applying it correctly all the time.
One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No.
Another important difference is that the kinetic energy doesn't have to be conserved. It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
This is true, however to maintain the kinetic energy of a heavier arrow vs. a lighter arrow requires the imparting of more potential energy from the bow. If we're assuming the use of one bow with a set amount of potential energy (Ep) which it can impart, then we need to take into account acceleration, because it's momentum is p = mv where v = a(t') and a = F/m. So when we put those together, we get;
p = ((Ep/m)(t'))m where
p is momentum, m is mass, t' is time spent accelerating, and Ep is equivalent to F, because we're ignoring efficiency. (in all reality, F < Ep, but since we need not consider the amount of potential energy which is wasted due to inefficient transference into the arrow, that's unimportant here)
So to plug in some numbers, we say that we're imparting 10 units of potential energy over the course of 1 unit of time into the arrow when we fire, and the arrow weighs 2 units, then p = 10 because;
p = ((10/2)(1))2
If we change this so that the mass is then 4 units, then we still get p = 10 because;
p = ((10/4)(1))4
Since we're assuming that we're using one single bow to fire two different arrows, we thus cannot change the amount of potential energy imparted into it (the strength of the draw), or the amount of time spent accelerating (the length of the draw), so changing the mass of the arrow makes no difference in the end to it's overall momentum, and since p = mv and p does not change, as m increases, v must decrease, as evinced by the fact that the acceleration decreases as the mass increases, while the time spent accelerating remains the same.
I tried to follow your formula, but I noticed a couple of errors. First, momentum should be in units of kg*m/s, but your formula ((Ep/m)(t'))m would result in units of kg*m^2/s (which is neither energy nor momentum). (Or, on second look, I guess I should've just pointed out that you can't just substitute energy for force.) Second, when you try to apply your formula, you're using the same amount of time in both cases, which strongly implies that both arrows will wind up with the same velocity, regardless of mass. (If the bowstring is going to move the same distance in the same period of time, it necessarily has to take the arrow with it at the same speed.) I don't think this is what you intended.
If I were to try to set up this problem myself, I'm not sure what would be the best way to go about it. I was thinking of assuming the bow obeys Hooke's Law. If that were so, I think it'd be pretty easy to show that both arrows would necessarily *have* to have the same kinetic energy, regardless of mass. Since work = force * distance, you could integrate the force (F=k*x) over the distance ( [integral](k*x)*dx ), and get the amount of energy tranferred to the arrow. Neither mass nor time are needed here. Heck, you don't even need to assume Hooke's Law (which definitely wouldn't apply to a compound bow anyway). You could use any expression that gives force as a function of displacement.
QUOTE (FlyingSpaghettiMonster+Nov 5 2009, 12:15 AM)
I think you're being inconsistent sometimes in distinguishing between momentum and kinetic energy. I mean, I can tell by your arguments that you know there's a difference, but you're not applying it correctly all the time.
I think you're just not understanding me, which is my fault for saying that kinetic energy is meaningless after the release. I should have said that it's meaningless from the moment of release until the moment of impact. Kinetic energy is stored within the arrow and released upon impact, but the characteristics of the arrow itself are best described in terms of momentum.
Momentum can be changed by imparting energy into the system. In a frame of reference which is stationary wrt to target, the impact releases the kinetic energy of the arrow, which in turn imparts the energy needed to change the momentum of the arrow in order to allow it to push aside the material of the target.
QUOTE
One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No.
Where do you think the energy to displace the material comes from, then? You seem to be suggesting that the energy needed to displace the material comes from somewhere else, but there is only one possible source if momentum is to be conserved. This source varies with the frame of reference (being that momentum is frame dependent), but it always comes from the total momentum being considered. Since we're discussing this from a frame of reference where the arrow moves and the target it stationary, that leaves only one source of energy: The impact created by the arrow's momentum. We could easily discuss the situation from the arrows frame of reference, where it's momentum is 0, but from that frame of reference, the only source of energy is the same impact, but this time created by the momentum of the target. Momentum can be changed by imparting energy into the system. In a frame of reference which is stationary wrt to target, the impact releases the kinetic energy of the arrow, which in turn imparts the energy needed to change the momentum of the arrow in order to allow it to push aside the material of the target.
QUOTE (->
| QUOTE |
| One important difference is that momentum is a vector quantity, and energy is a scalar. If, during the impact, the arrow pushes some of the target material perpendicular to the arrow's path, this does nothing to conserve the momentum in the direction the arrow is travelling. Does it require more energy? Sure. But does it contribute to the net momentum of the system? No. |
Where do you think the energy to displace the material comes from, then? You seem to be suggesting that the energy needed to displace the material comes from somewhere else, but there is only one possible source if momentum is to be conserved. This source varies with the frame of reference (being that momentum is frame dependent), but it always comes from the total momentum being considered. Since we're discussing this from a frame of reference where the arrow moves and the target it stationary, that leaves only one source of energy: The impact created by the arrow's momentum. We could easily discuss the situation from the arrows frame of reference, where it's momentum is 0, but from that frame of reference, the only source of energy is the same impact, but this time created by the momentum of the target.
Momentum can be changed by imparting energy into the system. In a frame of reference which is stationary wrt to target, the impact releases the kinetic energy of the arrow, which in turn imparts the energy needed to change the momentum of the arrow in order to allow it to push aside the material of the target.
Another important difference is that the kinetic energy doesn't have to be conserved.
Momentum can be changed by imparting energy into the system. In a frame of reference which is stationary wrt to target, the impact releases the kinetic energy of the arrow, which in turn imparts the energy needed to change the momentum of the arrow in order to allow it to push aside the material of the target.
Another important difference is that the kinetic energy doesn't have to be conserved.
No, but the total energy does.
QUOTE
It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved.
I think you're misunderstanding the conservation of momentum... Energy is being imparted into the arrow by the bow, which is what changes it's momentum, energy is then being released from the arrow upon impact, which is what changes it's momentum again. QUOTE (->
| QUOTE |
| It can be lost as heat or sound, or converted to potential energy somewhere. It's scalar, so it doesn't even have to preserve direction. You could have an elastic collision between, say, a fast-moving 0.05 kg rubber ball and a stationary 10,000 kg steel ball. There's no problem with the rubber ball carrying most of the energy away in the opposite direction it came from, but there definitely is a problem if the *momentum* of the system isn't preserved. |
I think you're misunderstanding the conservation of momentum... Energy is being imparted into the arrow by the bow, which is what changes it's momentum, energy is then being released from the arrow upon impact, which is what changes it's momentum again.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
So, the target with the embedded arrow better have the same momentum as the arrow had before it hit. How the energy was dissipated is irrelevant.
No, the momentum of the arrow is changed by the force released upon impact. Remember, the arrow wasn't always in flight, it started out stationary, and this is the momentum that needs to be conserved, because this is the momentum we started with.
The acceleration of the arrow which is a = F/m, the velocity of the arrow which is v = a(t') and the momentum of the arrow from the point of view of the archer, which is p = mv. All I did was substitute F/m for a in the velocity formula and substitute a(t') for the v in the momentum formula and combine them. You can do each one separately, and you can specify the actual units used by each formula and you still get the exact same result. Observe:
I'll use the same quantities as specified before, but substituting the specified units for the unspecified ones I used last time, and doing each formula independently of the others.
a = F/m where F is 10 Newtons and m is 2kg, so a = 5m/s^2
v = a(t') where a is 5m/s^2 and t' is 1 second, so v = 5m/s
p = mv where m is 2kg and v is 5m/s so p = 10kg*m/s or 10N*s.
The second one, where the mass doubles is:
a = F/m where F is 10 Newtons and m is 4kg, so a = 2.5m/s^2
v = a(t') where a is 2.5m/s^2 and t' is 1 second, so v = 2.5m/s
p = mv where m is 4kg and v is 2.5m/s so p = 10kg*m/s or 10N*s.
It doesn't matter what the mass of the projectile is, it's momentum depends only on the force imparted to it.
QUOTE
I tried to follow your formula, but I noticed a couple of errors. First, momentum should be in units of kg*m/s, but your formula ((Ep/m)(t'))m would result in units of kg*m^2/s (which is neither energy nor momentum).
No, it doesn't. I used unspecified units simply for the sake of brevity, but that doesn't change the actual units used by each formula I used. My formula is simply the combination of three newtonian formulas: The acceleration of the arrow which is a = F/m, the velocity of the arrow which is v = a(t') and the momentum of the arrow from the point of view of the archer, which is p = mv. All I did was substitute F/m for a in the velocity formula and substitute a(t') for the v in the momentum formula and combine them. You can do each one separately, and you can specify the actual units used by each formula and you still get the exact same result. Observe:
I'll use the same quantities as specified before, but substituting the specified units for the unspecified ones I used last time, and doing each formula independently of the others.
a = F/m where F is 10 Newtons and m is 2kg, so a = 5m/s^2
v = a(t') where a is 5m/s^2 and t' is 1 second, so v = 5m/s
p = mv where m is 2kg and v is 5m/s so p = 10kg*m/s or 10N*s.
The second one, where the mass doubles is:
a = F/m where F is 10 Newtons and m is 4kg, so a = 2.5m/s^2
v = a(t') where a is 2.5m/s^2 and t' is 1 second, so v = 2.5m/s
p = mv where m is 4kg and v is 2.5m/s so p = 10kg*m/s or 10N*s.
It doesn't matter what the mass of the projectile is, it's momentum depends only on the force imparted to it.
QUOTE (->
| QUOTE |
| I tried to follow your formula, but I noticed a couple of errors. First, momentum should be in units of kg*m/s, but your formula ((Ep/m)(t'))m would result in units of kg*m^2/s (which is neither energy nor momentum). |
No, it doesn't. I used unspecified units simply for the sake of brevity, but that doesn't change the actual units used by each formula I used. My formula is simply the combination of three newtonian formulas:
The acceleration of the arrow which is a = F/m, the velocity of the arrow which is v = a(t') and the momentum of the arrow from the point of view of the archer, which is p = mv. All I did was substitute F/m for a in the velocity formula and substitute a(t') for the v in the momentum formula and combine them. You can do each one separately, and you can specify the actual units used by each formula and you still get the exact same result. Observe:
I'll use the same quantities as specified before, but substituting the specified units for the unspecified ones I used last time, and doing each formula independently of the others.
a = F/m where F is 10 Newtons and m is 2kg, so a = 5m/s^2
v = a(t') where a is 5m/s^2 and t' is 1 second, so v = 5m/s
p = mv where m is 2kg and v is 5m/s so p = 10kg*m/s or 10N*s.
The second one, where the mass doubles is:
a = F/m where F is 10 Newtons and m is 4kg, so a = 2.5m/s^2
v = a(t') where a is 2.5m/s^2 and t' is 1 second, so v = 2.5m/s
p = mv where m is 4kg and v is 2.5m/s so p = 10kg*m/s or 10N*s.
It doesn't matter what the mass of the projectile is, it's momentum depends only on the force imparted to it.
(Or, on second look, I guess I should've just pointed out that you can't just substitute energy for force.)
The acceleration of the arrow which is a = F/m, the velocity of the arrow which is v = a(t') and the momentum of the arrow from the point of view of the archer, which is p = mv. All I did was substitute F/m for a in the velocity formula and substitute a(t') for the v in the momentum formula and combine them. You can do each one separately, and you can specify the actual units used by each formula and you still get the exact same result. Observe:
I'll use the same quantities as specified before, but substituting the specified units for the unspecified ones I used last time, and doing each formula independently of the others.
a = F/m where F is 10 Newtons and m is 2kg, so a = 5m/s^2
v = a(t') where a is 5m/s^2 and t' is 1 second, so v = 5m/s
p = mv where m is 2kg and v is 5m/s so p = 10kg*m/s or 10N*s.
The second one, where the mass doubles is:
a = F/m where F is 10 Newtons and m is 4kg, so a = 2.5m/s^2
v = a(t') where a is 2.5m/s^2 and t' is 1 second, so v = 2.5m/s
p = mv where m is 4kg and v is 2.5m/s so p = 10kg*m/s or 10N*s.
It doesn't matter what the mass of the projectile is, it's momentum depends only on the force imparted to it.
(Or, on second look, I guess I should've just pointed out that you can't just substitute energy for force.)
In this case, we can because the efficiency of the system which changes the energy into force is irrelevant. All of that happens before any force is imparted to the arrow. If you want to incorporate it, then you simply quantify the efficiency of the bow and archer and use the resultant formula to derive the force from the energy, then continue on as I've outlined.
Consider what you are saying: You are claiming that imparting the same amount of force into a heavier object produces more momentum than imparting it into a lighter object. If this were true, then the total energy of the system is not conserved.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
QUOTE
Second, when you try to apply your formula, you're using the same amount of time in both cases, which strongly implies that both arrows will wind up with the same velocity, regardless of mass. (If the bowstring is going to move the same distance in the same period of time, it necessarily has to take the arrow with it at the same speed.)
That's the thing, if you look at the formula, the bowstring is not moving the same distance each time. When the mass of the arrow is doubled, it's momentum carries it off the bowstring earlier, because it's inertia slows the bowstring's snap. If the mass doubles, the bowstring's velocity slows by half and the arrow is released halfway through it's return. The bowstring continues to return, naturally, but the key here is that it's no longer imparting any energy to the arrow, merely moving along with it.Consider what you are saying: You are claiming that imparting the same amount of force into a heavier object produces more momentum than imparting it into a lighter object. If this were true, then the total energy of the system is not conserved.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
QUOTE (->
| QUOTE |
| Second, when you try to apply your formula, you're using the same amount of time in both cases, which strongly implies that both arrows will wind up with the same velocity, regardless of mass. (If the bowstring is going to move the same distance in the same period of time, it necessarily has to take the arrow with it at the same speed.) |
That's the thing, if you look at the formula, the bowstring is not moving the same distance each time. When the mass of the arrow is doubled, it's momentum carries it off the bowstring earlier, because it's inertia slows the bowstring's snap. If the mass doubles, the bowstring's velocity slows by half and the arrow is released halfway through it's return. The bowstring continues to return, naturally, but the key here is that it's no longer imparting any energy to the arrow, merely moving along with it.
Consider what you are saying: You are claiming that imparting the same amount of force into a heavier object produces more momentum than imparting it into a lighter object. If this were true, then the total energy of the system is not conserved.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
If I were to try to set up this problem myself, I'm not sure what would be the best way to go about it. I was thinking of assuming the bow obeys Hooke's Law. Neither mass nor time are needed here. Heck, you don't even need to assume Hooke's Law (which definitely wouldn't apply to a compound bow anyway). You could use any expression that gives force as a function of displacement.
Consider what you are saying: You are claiming that imparting the same amount of force into a heavier object produces more momentum than imparting it into a lighter object. If this were true, then the total energy of the system is not conserved.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
If I were to try to set up this problem myself, I'm not sure what would be the best way to go about it. I was thinking of assuming the bow obeys Hooke's Law. Neither mass nor time are needed here. Heck, you don't even need to assume Hooke's Law (which definitely wouldn't apply to a compound bow anyway). You could use any expression that gives force as a function of displacement.
While Hooke's law describes a linear spring, I'm sure the bowstring could be expressed in such terms, although I'm not sure quite how to do it off the top of my head. A compound bow would make things even more difficult.
QUOTE
If that were so, I think it'd be pretty easy to show that both arrows would necessarily *have* to have the same kinetic energy, regardless of mass. Since work = force * distance, you could integrate the force (F=k*x) over the distance ( [integral](k*x)*dx ), and get the amount of energy tranferred to the arrow.
I'm not disagreeing with that. In fact, what you're saying here is the same thing I'm saying: The amount of energy released upon impacting the target depends entirely upon the amount of energy imparted to the arrow from the bow. A bow which imparts more energy to the arrow thus produces a larger release of energy against the target, while the mass of the arrow makes no difference.
What about the fact that two differently weighted arrows will leave the bow with different velocities when shot with the same amount of KE (I hope I'm correctly assuming that)? In that case, the heavier arrow moving slower would arc more on the way to the target than the faster moving one. In that case, doesn't the arrow arcing more lose energy doing so, or is the energy lost going up gained as potential, which is gained back on the way back down? I think I just answered my own question. But isn't any energy lost in the shift of inertia of an object as it changes direction from up to down?
Alright, I don't have enough time to argue and nitpick every single point any more, so I'll just pick one.
Here's at least one place you're confusing units.
No, I'm not claiming that imparting the same *force* into a heavier object produces more momentum in a heavier one than a lighter one. Such a statement is meaningless. You *apply* a force, and you apply it over some distance which would impart energy.
So, if we change your statement to: "You are claiming that imparting the same *energy* into a heavier object produces more momentum than imparting into a lighter object," then yes, this is precisely what I'm saying.
>> So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. <<
Similar thing here. You got the units of momentum correct, but talking about "imparting" forces, as if the arrow now "contains force" which it then applies elsewhere is utter nonsense.
You have to be precise about these things. Force * time = momentum; force * distance = energy.
Suppose we simplify the bow now so that it applies the same force over the entire time it's accelerating the arrow, i.e., F=k, instead of F=kx.
The bow applies its force over some distance, and therefore imparts some exact amount of energy to any arrow, regardless of mass. But now you want to talk about momentum? Okay, well, momentum would be force times time. The heavier arrow, although it's being accelerated over the same distance as the lighter one, is being accelerated for a longer period of time. I mean, if F=ma, then if the force remains constant but the mass increases, doesn't the acceleration decrease? And if so, won't it take longer for the arrow to be pushed out of the bow? So, it takes more time, and therefore carries more momentum.
FSM
QUOTE
Consider what you are saying: You are claiming that imparting the same amount of force into a heavier object produces more momentum than imparting it into a lighter object. If this were true, then the total energy of the system is not conserved.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
The formula for the force generated by the impact is simply a time reversed version of the formula used to determine it's momentum, albeit one where the acceleration and time spent accelerating have changed, where the change in t' is inversely proportional to the change in a. So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. We've just pulled energy out of no-where. Now even if there isn't a 1:1 ratio between the increase in mass and the increase in momentum, this still involves the creation of free energy, which violates conservation of moment. If the 10 Newtons imparted by the arrow produce even 10.00001N*s, then there's still 0.00001 Newtons being created for free.
Here's at least one place you're confusing units.
No, I'm not claiming that imparting the same *force* into a heavier object produces more momentum in a heavier one than a lighter one. Such a statement is meaningless. You *apply* a force, and you apply it over some distance which would impart energy.
So, if we change your statement to: "You are claiming that imparting the same *energy* into a heavier object produces more momentum than imparting into a lighter object," then yes, this is precisely what I'm saying.
>> So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. <<
Similar thing here. You got the units of momentum correct, but talking about "imparting" forces, as if the arrow now "contains force" which it then applies elsewhere is utter nonsense.
You have to be precise about these things. Force * time = momentum; force * distance = energy.
Suppose we simplify the bow now so that it applies the same force over the entire time it's accelerating the arrow, i.e., F=k, instead of F=kx.
The bow applies its force over some distance, and therefore imparts some exact amount of energy to any arrow, regardless of mass. But now you want to talk about momentum? Okay, well, momentum would be force times time. The heavier arrow, although it's being accelerated over the same distance as the lighter one, is being accelerated for a longer period of time. I mean, if F=ma, then if the force remains constant but the mass increases, doesn't the acceleration decrease? And if so, won't it take longer for the arrow to be pushed out of the bow? So, it takes more time, and therefore carries more momentum.
FSM
QUOTE (FlyingSpaghettiMonster+Nov 5 2009, 10:18 AM)
Alright, I don't have enough time to argue and nitpick every single point any more, so I'll just pick one.
Here's at least one place you're confusing units.
No, I'm not claiming that imparting the same *force* into a heavier object produces more momentum in a heavier one than a lighter one. Such a statement is meaningless. You *apply* a force, and you apply it over some distance which would impart energy.
So, if we change your statement to: "You are claiming that imparting the same *energy* into a heavier object produces more momentum than imparting into a lighter object," then yes, this is precisely what I'm saying.
>> So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. <<
Similar thing here. You got the units of momentum correct, but talking about "imparting" forces, as if the arrow now "contains force" which it then applies elsewhere is utter nonsense.
Now you're just arguing semantic. Whether "imparts" or "applies" is the more proper term doesn't do anything to change the validity of my point. There's no confusion of units (as you then go on to point out yourself, which is odd), but me using one word when another, synonymous -in this context- word is the one you're saying I should use. I'll make an effort to be precise about everything if it bothers you that much, but it won't change anything except the vocabulary of my posts.
Which is why it was an assumption on my part as I didn't read the post. I suppose they say "don't judge a book by it's cover" for a reason. Again, my mistake. Sorry.
As for momentum, the heavier arrow would indeed have more momentum but I haven't been able to find anything solid on the relationship between momentum and penetration.
No worries.
No worries.
As for momentum, the heavier arrow would indeed have more momentum but I haven't been able to find anything solid on the relationship between momentum and penetration.
It's hard to say anything that's both absolute and simple concerning the relationship that mass, velocity, momentum, and kinetic energy have to penetration depth. You can't change one and hold the rest constant. If try to raise the momentum by raising the mass, but try to hold kinetic energy constant by lowering the velocity, then it's not really the same experiment any more. The target will be more able to distribute the impact of a slower arrow. How much, I don't know; it seems like something that might require some actual experimentation to see what happens.
Here's at least one place you're confusing units.
No, I'm not claiming that imparting the same *force* into a heavier object produces more momentum in a heavier one than a lighter one. Such a statement is meaningless. You *apply* a force, and you apply it over some distance which would impart energy.
So, if we change your statement to: "You are claiming that imparting the same *energy* into a heavier object produces more momentum than imparting into a lighter object," then yes, this is precisely what I'm saying.
>> So if we impart 10 Newtons into the arrow, and it's doubled mass means it then has a momentum of 20N*s, then the force imparted to the target is 20 Newtons. <<
Similar thing here. You got the units of momentum correct, but talking about "imparting" forces, as if the arrow now "contains force" which it then applies elsewhere is utter nonsense.
Now you're just arguing semantic. Whether "imparts" or "applies" is the more proper term doesn't do anything to change the validity of my point. There's no confusion of units (as you then go on to point out yourself, which is odd), but me using one word when another, synonymous -in this context- word is the one you're saying I should use. I'll make an effort to be precise about everything if it bothers you that much, but it won't change anything except the vocabulary of my posts.
QUOTE
Suppose we simplify the bow now so that it applies the same force over the entire time it's accelerating the arrow, i.e., F=k, instead of F=kx.
QUOTE (->
| QUOTE |
| Suppose we simplify the bow now so that it applies the same force over the entire time it's accelerating the arrow, i.e., F=k, instead of F=kx. You have to be precise about these things. Force * time = momentum; force * distance = energy. Where did you get that first one from? Momentum is velocity * mass, or P = mv. You can't determine momentum without knowing the mass, so force * time is meaningless on it's own. The bow applies its force over some distance, and therefore imparts some exact amount of energy to any arrow, regardless of mass. But now you want to talk about momentum? Okay, well, momentum would be force times time. The heavier arrow, although it's being accelerated over the same distance as the lighter one, is being accelerated for a longer period of time. I mean, if F=ma, then if the force remains constant but the mass increases, doesn't the acceleration decrease? And if so, won't it take longer for the arrow to be pushed out of the bow? So, it takes more time, and therefore carries more momentum. |
You're still saying that a greater mass means greater momentum from the same initial energy, which means it applies a greater force to the target. That requires the creation of free energy.
It just doesn't work that way, you don't get energy for free. If you want to hit the target harder, you must apply more force to the arrow initially, regardless of it's mass.
Consider the following device:
User posted image: User posted image
Ignoring the bulb, this is a device which converts linear motion into electricity. Of course, the achieve linear motion you must use kinetic energy, which is converted from chemical energy by your muscles. Now if what you are saying is true, that the more massive arrow has greater momentum after being acted on by the same level of force, then this is a prototype for a free energy device.
It takes about 0.2N to generate X watts of electrical energy, using this device, as it probably weighs about a fifth of a kilo (I have one here with me). Now, if I were to increase the mass of the magnet inside by twice, then according to what you're saying, I could produce 2X watts of electicity for my 0.2N of force. If I were to quadruple it, then I could produce 4X watts. Make it ten times more massive, and I get 10X watts, a thousand times more massive and I get 1000X watts, all from 0.2N of force. Do you see why this cannot be so? Before you argue that this is invalid, bear in mind that it is by applying a force derived from it's momentum to the electromagnetic field that this device generates electricity. It is essentially no different.
Maybe I'm wrong about when the arrow leaves the string, but I guarantee you that the momentum will remain the same. If the time spent accelerating doubles, the acceleration will be half: There's simply no other way that energy is being conserved in all possible cases. You might argue that the extra energy is being sapped away by gravity, or through conversion to heat by the friction of the air, but then you'd still have the creation of free energy if the arrow were fired in micro gravity and a near perfect vacuum.
It just doesn't work that way, you don't get energy for free. If you want to hit the target harder, you must apply more force to the arrow initially, regardless of it's mass.
Consider the following device:
User posted image: User posted image
Ignoring the bulb, this is a device which converts linear motion into electricity. Of course, the achieve linear motion you must use kinetic energy, which is converted from chemical energy by your muscles. Now if what you are saying is true, that the more massive arrow has greater momentum after being acted on by the same level of force, then this is a prototype for a free energy device.
It takes about 0.2N to generate X watts of electrical energy, using this device, as it probably weighs about a fifth of a kilo (I have one here with me). Now, if I were to increase the mass of the magnet inside by twice, then according to what you're saying, I could produce 2X watts of electicity for my 0.2N of force. If I were to quadruple it, then I could produce 4X watts. Make it ten times more massive, and I get 10X watts, a thousand times more massive and I get 1000X watts, all from 0.2N of force. Do you see why this cannot be so? Before you argue that this is invalid, bear in mind that it is by applying a force derived from it's momentum to the electromagnetic field that this device generates electricity. It is essentially no different.
Maybe I'm wrong about when the arrow leaves the string, but I guarantee you that the momentum will remain the same. If the time spent accelerating doubles, the acceleration will be half: There's simply no other way that energy is being conserved in all possible cases. You might argue that the extra energy is being sapped away by gravity, or through conversion to heat by the friction of the air, but then you'd still have the creation of free energy if the arrow were fired in micro gravity and a near perfect vacuum.
1) Force * time = momentum. Look up "Impulse" on Wikipedia. Impulse and momentum are interchangeable here, because impulse is a change in momentum, and we're starting with no momentum, and thus, the change in momentum *is* the ending momentum.
I'm not sure why you're arguing this. You correctly used N*s as units of momentum, so why would it be hard to see that this is force * time?
Momentum
= force * time
= N * s
= (kg * m / s^2) * s
= kg * (m / s)
= mass * velocity
2) The distinction between "imparting" and "applying" is important, and is not merely semantic, and I pointed it out not to pick on you, but with the hopes that you'll see how you're not being consistent in your terms. You were talking about "imparting" 10 N to an arrow, so that it now has 10 N of force in it. But objects don't contain force; it's nonsensical to speak of it that way. Force is just... well... force. The force has to be applied over some distance and over some time for it to "contain" something, i.e., energy and momentum, respectively. It can then transfer those things by applying a force to something else, again over some distance and over some period of time.
There would be no contradiction, for example, if you applied 1 N of force to a rocket for a long period of time, and then slammed it into an asteroid, generating millions of N of force, because the impact is over a much smaller period of time and distance.
3) About your hand-powered flashlight: it's incorrect to say that a certain amount of force will generate a certain amount of watts. Force is not the same thing as power. It *is* correct to say, however, that a certain amount of force, when applied over a certain distance within a certain period of time, is power. So if you increase the resistance within the device, and still try to apply the same amount of force, you're not going to get an increase in power, because you're going to increase the time it takes to move the... whatever it is that moves, I can't tell from the pic... the same distance as before.
4)
If the time spent accelerating doubles, the acceleration will be quartered.
x = (1/2) * a * t^2
Since the arrows are both being accelerated over the same distance, then:
(1/2) * a1 * t1^2 = (1/2) * a2 * t2^2,
Where a1 and t1 are the acceleration and time of the first arrow, and a2 and t2 are the acceleration and time of the second.
Suppose that t2 = 2 * t1. Then:
(1/2) * a1 * t1^2 = (1/2) * a2 * (2 * t1)^2
==> a1 * t1^2 = 4 * a2 * t1^2
==> a1 = 4 * a2
==> a2 = (1/4) * a1
I'm not sure if this is the part that's hanging you up, but I figured I'd throw it out there and see if it helps make it click for you.
I'm not sure why you're arguing this. You correctly used N*s as units of momentum, so why would it be hard to see that this is force * time?
Momentum
= force * time
= N * s
= (kg * m / s^2) * s
= kg * (m / s)
= mass * velocity
2) The distinction between "imparting" and "applying" is important, and is not merely semantic, and I pointed it out not to pick on you, but with the hopes that you'll see how you're not being consistent in your terms. You were talking about "imparting" 10 N to an arrow, so that it now has 10 N of force in it. But objects don't contain force; it's nonsensical to speak of it that way. Force is just... well... force. The force has to be applied over some distance and over some time for it to "contain" something, i.e., energy and momentum, respectively. It can then transfer those things by applying a force to something else, again over some distance and over some period of time.
There would be no contradiction, for example, if you applied 1 N of force to a rocket for a long period of time, and then slammed it into an asteroid, generating millions of N of force, because the impact is over a much smaller period of time and distance.
3) About your hand-powered flashlight: it's incorrect to say that a certain amount of force will generate a certain amount of watts. Force is not the same thing as power. It *is* correct to say, however, that a certain amount of force, when applied over a certain distance within a certain period of time, is power. So if you increase the resistance within the device, and still try to apply the same amount of force, you're not going to get an increase in power, because you're going to increase the time it takes to move the... whatever it is that moves, I can't tell from the pic... the same distance as before.
4)
QUOTE
If the time spent accelerating doubles, the acceleration will be half
If the time spent accelerating doubles, the acceleration will be quartered.
x = (1/2) * a * t^2
Since the arrows are both being accelerated over the same distance, then:
(1/2) * a1 * t1^2 = (1/2) * a2 * t2^2,
Where a1 and t1 are the acceleration and time of the first arrow, and a2 and t2 are the acceleration and time of the second.
Suppose that t2 = 2 * t1. Then:
(1/2) * a1 * t1^2 = (1/2) * a2 * (2 * t1)^2
==> a1 * t1^2 = 4 * a2 * t1^2
==> a1 = 4 * a2
==> a2 = (1/4) * a1
I'm not sure if this is the part that's hanging you up, but I figured I'd throw it out there and see if it helps make it click for you.
500fps
I also do archery both hunting and target shooting. One thing that I have noticed, through experience and from talking to those who have been at it for longer than me, is that given a constant poundage and draw length you will obtain higher velocities and a flatter trajectory with a lighter arrow. However you get more penetration with a heavier arrow.
Now the equation you mention, KE = 1/2MV^2, is an important one but important to penetration. Another simple equation that I believe to be equally important (if not more so) is F = ma.
However to answer your question simply is to plug in values.
These are the arrows I use (also the place I get'em)
http://www.basspro.com/webapp/wcs/stores/s...0002000_450-2-0
Now typically arrows are cut to fit the bow so we'll just go with 30" to make it easier. At 9.9 grains per inch that is 297 grains plus a 100 grain head bringing us to 397 grains which equates to about 26 grams which is .026 kg.
Now lets say you crono 500fps which is 152.4 mps.
KE = 1/2(.026)*(152.4)^2
KE = 302 Joules
Double the mass
KE = 1/2(.052)*(152.4)^2
KE = 604 Joules
Now double the velocity
KE = 1/2(.026)*(304.8)^2
KE = 1208 Joules
So as you can see when you double the velocity you get 4 times the increase in kinetic energy but when you double the mass you only double the kinetic energy.
But that brings me to the other equation...
F = ma which following the same units above is N = kg * m/s^2
This is where bow weight comes in because that equates to the force applied to the arrow which can be converted from pounds to newtons easily enough, just google it. Draw length basically determines for how long the force is being applied before it is on its own. The longer the draw length with a given weight the longer the acceleration and the higher the velocity is obtained.
So I use an 80 lb bow with a 28" draw. Converted that is a 356 newtons and .71 meters.
We know the arrow to be .026 kg so with the equation we know the acceleration to be
N = kg * m/s^2
N/kg = m/s^2
356/.026 = 13692 m/s^2
Now your thinking wow that's huge, arrows don't travel nearly that fast. Well arrows aren't pushed for a complete second. They are pushed for a fraction of a second. Obviously with such a high acceleration the .71 meters is covered pretty quickly so only a fraction of that potential velocity is obtained. I believe the equation is...
Vf^2 = Vi^2 + 2 * A * d
Where Vf is final velocity, Vi is initial velocity, A is acceleration and d is distance.
We know Vi to be 0, A to be 13692 m/s^2 and d to be .71
Vf^2 = 2 * 13692 * .71
Vf^2 = 19442.64
Vf = 139.4 m/s which is 457.3 feet per second. So that is basically the velocity my own bow would get with an arrow of the specified weight.
What does all this mean? What's the conclusion? As you can see there is a relationship between weight and velocity. The lighter the weight the higher the velocity. As also demonstrated obtaining a higher velocity is better than obtaining a higher mass when it comes to KE. The question is how do you obtain the higher velocity?
Let's start with an arrow that is .052 kg and with my own bow specs come up with the energy..
N = kg * m/s^2
356/.052 = 6846.15 m/s^2
Vf^2 = 2 * 6846.15 * .71
Vf = 89.59 m/s
KE = 1/2(.056) * (89.59)^2
KE = 224.73 joules
Now for the .026 kg arrow
356/.026 = 13692.3
sqrt(2 * 13692.3 * .71) = 139.43
1/2(.026) * (139.43)^2 = 252.72 joules
So not much difference but I get more kinetic energy with the lighter arrow than with the heavier. Even though I decrease the mass to increase the velocity because velocity has more effect it beats out the drop in mass.
Now finally is to cover the same equation N = kg * m/s^2 because it applies to deceleration. For every action there is an equal and opposite reaction and so when the arrow decelerates it does so by a force being applied to it. Now if that force is because of the target then the same force applied to the arrow is also applied to the target. What penetrates the target is pressure. The greater the pressure the greater the penetration. So how is this done with the same arrow and bow? By decreasing the frontal surface area of the arrow. Basically all this means is what everyone already knows, the sharper the object the easier it will penetrate. So when you have an arrow that decelerates at say half the acceleration rate (since it will really depend on what the target is). Then you will get...
N = kg * m/s^2
N = .026 * 6846.15
N = 178 which equals 40 pounds.
Now I don't know what the frontal surface area of arrow tips are but lets say .0025 square inches. That means the arrow will produce 16 000 psi pressure. If you get a sharper tip say .001 square inches then you will produce 40 000 psi of pressure with the same arrow.
Now I'm sorry if this repeats what anyone else may have said but when I saw the majority of the posts I made the assumption that this thread had gotten taken over by bickering. I hope it answers your questions all the same.
I also do archery both hunting and target shooting. One thing that I have noticed, through experience and from talking to those who have been at it for longer than me, is that given a constant poundage and draw length you will obtain higher velocities and a flatter trajectory with a lighter arrow. However you get more penetration with a heavier arrow.
Now the equation you mention, KE = 1/2MV^2, is an important one but important to penetration. Another simple equation that I believe to be equally important (if not more so) is F = ma.
However to answer your question simply is to plug in values.
These are the arrows I use (also the place I get'em)
http://www.basspro.com/webapp/wcs/stores/s...0002000_450-2-0
Now typically arrows are cut to fit the bow so we'll just go with 30" to make it easier. At 9.9 grains per inch that is 297 grains plus a 100 grain head bringing us to 397 grains which equates to about 26 grams which is .026 kg.
Now lets say you crono 500fps which is 152.4 mps.
KE = 1/2(.026)*(152.4)^2
KE = 302 Joules
Double the mass
KE = 1/2(.052)*(152.4)^2
KE = 604 Joules
Now double the velocity
KE = 1/2(.026)*(304.8)^2
KE = 1208 Joules
So as you can see when you double the velocity you get 4 times the increase in kinetic energy but when you double the mass you only double the kinetic energy.
But that brings me to the other equation...
F = ma which following the same units above is N = kg * m/s^2
This is where bow weight comes in because that equates to the force applied to the arrow which can be converted from pounds to newtons easily enough, just google it. Draw length basically determines for how long the force is being applied before it is on its own. The longer the draw length with a given weight the longer the acceleration and the higher the velocity is obtained.
So I use an 80 lb bow with a 28" draw. Converted that is a 356 newtons and .71 meters.
We know the arrow to be .026 kg so with the equation we know the acceleration to be
N = kg * m/s^2
N/kg = m/s^2
356/.026 = 13692 m/s^2
Now your thinking wow that's huge, arrows don't travel nearly that fast. Well arrows aren't pushed for a complete second. They are pushed for a fraction of a second. Obviously with such a high acceleration the .71 meters is covered pretty quickly so only a fraction of that potential velocity is obtained. I believe the equation is...
Vf^2 = Vi^2 + 2 * A * d
Where Vf is final velocity, Vi is initial velocity, A is acceleration and d is distance.
We know Vi to be 0, A to be 13692 m/s^2 and d to be .71
Vf^2 = 2 * 13692 * .71
Vf^2 = 19442.64
Vf = 139.4 m/s which is 457.3 feet per second. So that is basically the velocity my own bow would get with an arrow of the specified weight.
What does all this mean? What's the conclusion? As you can see there is a relationship between weight and velocity. The lighter the weight the higher the velocity. As also demonstrated obtaining a higher velocity is better than obtaining a higher mass when it comes to KE. The question is how do you obtain the higher velocity?
Let's start with an arrow that is .052 kg and with my own bow specs come up with the energy..
N = kg * m/s^2
356/.052 = 6846.15 m/s^2
Vf^2 = 2 * 6846.15 * .71
Vf = 89.59 m/s
KE = 1/2(.056) * (89.59)^2
KE = 224.73 joules
Now for the .026 kg arrow
356/.026 = 13692.3
sqrt(2 * 13692.3 * .71) = 139.43
1/2(.026) * (139.43)^2 = 252.72 joules
So not much difference but I get more kinetic energy with the lighter arrow than with the heavier. Even though I decrease the mass to increase the velocity because velocity has more effect it beats out the drop in mass.
Now finally is to cover the same equation N = kg * m/s^2 because it applies to deceleration. For every action there is an equal and opposite reaction and so when the arrow decelerates it does so by a force being applied to it. Now if that force is because of the target then the same force applied to the arrow is also applied to the target. What penetrates the target is pressure. The greater the pressure the greater the penetration. So how is this done with the same arrow and bow? By decreasing the frontal surface area of the arrow. Basically all this means is what everyone already knows, the sharper the object the easier it will penetrate. So when you have an arrow that decelerates at say half the acceleration rate (since it will really depend on what the target is). Then you will get...
N = kg * m/s^2
N = .026 * 6846.15
N = 178 which equals 40 pounds.
Now I don't know what the frontal surface area of arrow tips are but lets say .0025 square inches. That means the arrow will produce 16 000 psi pressure. If you get a sharper tip say .001 square inches then you will produce 40 000 psi of pressure with the same arrow.
Now I'm sorry if this repeats what anyone else may have said but when I saw the majority of the posts I made the assumption that this thread had gotten taken over by bickering. I hope it answers your questions all the same.
H20--
Nice work. It's a slightly different approach than the one I took, but it comes up with the same results...
...except...
Redo the computation. You should get Vf = 98.59 m/s. Actually, 98.60, if you do the rounding correctly.
After that, the kinetic energy of the two arrows are equal, but the momentum of the heavier arrow is greater.
Which brings me to your comment about bickering. It seems that the majority of the disagreement between MjolnirPants and me concerns exactly that point above. If it's proper for you to come in here and try to cut through the bullcrap and provide a correct understanding of the mechanics involved here, then why is it "bickering" for me to do the same? Should I give up out of politeness, or should I stand my ground?
Nice work. It's a slightly different approach than the one I took, but it comes up with the same results...
...except...
QUOTE
Vf^2 = 2 * 6846.15 * .71
Vf = 89.59 m/s
Vf = 89.59 m/s
Redo the computation. You should get Vf = 98.59 m/s. Actually, 98.60, if you do the rounding correctly.
After that, the kinetic energy of the two arrows are equal, but the momentum of the heavier arrow is greater.
Which brings me to your comment about bickering. It seems that the majority of the disagreement between MjolnirPants and me concerns exactly that point above. If it's proper for you to come in here and try to cut through the bullcrap and provide a correct understanding of the mechanics involved here, then why is it "bickering" for me to do the same? Should I give up out of politeness, or should I stand my ground?
Ah yes, my mistake.
KE = 1/2(.026) * (139.43)^2
KE = 252.72
KE = 1/2(.052) * (98.59)^2
KE = 252.71
If I had rounded off rather than simply drop off the rest then it would've been 252.73 vs 252.72 Really no difference at all. I would have thought a bigger one. I guess the only benefit I see with the smaller arrow is higher velocity so better range and accuracy.
KE = 1/2(.026) * (139.43)^2
KE = 252.72
KE = 1/2(.052) * (98.59)^2
KE = 252.71
If I had rounded off rather than simply drop off the rest then it would've been 252.73 vs 252.72 Really no difference at all. I would have thought a bigger one. I guess the only benefit I see with the smaller arrow is higher velocity so better range and accuracy.
QUOTE
Which brings me to your comment about bickering. It seems that the majority of the disagreement between MjolnirPants and me concerns exactly that point above. If it's proper for you to come in here and try to cut through the bullcrap and provide a correct understanding of the mechanics involved here, then why is it "bickering" for me to do the same? Should I give up out of politeness, or should I stand my ground?
Which is why it was an assumption on my part as I didn't read the post. I suppose they say "don't judge a book by it's cover" for a reason. Again, my mistake. Sorry.
As for momentum, the heavier arrow would indeed have more momentum but I haven't been able to find anything solid on the relationship between momentum and penetration.
QUOTE
Which is why it was an assumption on my part as I didn't read the post. I suppose they say "don't judge a book by it's cover" for a reason. Again, my mistake. Sorry.
No worries.
QUOTE (->
| QUOTE |
| Which is why it was an assumption on my part as I didn't read the post. I suppose they say "don't judge a book by it's cover" for a reason. Again, my mistake. Sorry. |
No worries.
As for momentum, the heavier arrow would indeed have more momentum but I haven't been able to find anything solid on the relationship between momentum and penetration.
It's hard to say anything that's both absolute and simple concerning the relationship that mass, velocity, momentum, and kinetic energy have to penetration depth. You can't change one and hold the rest constant. If try to raise the momentum by raising the mass, but try to hold kinetic energy constant by lowering the velocity, then it's not really the same experiment any more. The target will be more able to distribute the impact of a slower arrow. How much, I don't know; it seems like something that might require some actual experimentation to see what happens.
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