Ok, this post is going to be INSANELY long….sorry 
.
Ah, I hate it when I am wrong….but it did get me thinking more on this topic so in the end I greatly profited from it. As you will see, my comments about the COM are not entirely correct, but you will also see that my basic conclusions were not too far off.
So, I think, Possum, a lot of what you see in practice that might be different from what I try to show below is due to the fact that you are changing all four parameters: length, mass, weight distribution, and very importantly speed at the same time. But this obscures in my opinion some of the underlying principles. So in the following, I am trying to look at a system in which the blade is always moved with the same angular velocity. Obviously, in real life, if you make the blade too long or too heavy, you are will not be able to swing it with the same speed as a smaller or lighter blade, but to take that into account, I would have to know what the limits of the man who swings the blade are. However, if you would be able to measure that, it would be easy enough taking it into account. So for the present we ignore that variable.
Also, I would like you to realize, that, despite the simplifications that am using, this is a fully determined system. Unlike talking about metallurgy, which is still more or less a form of alchemy (ok, before anybody gets on my case for this, take it with a bit of a smile), chopping is based on simple classical mechanics and this is a fully determined system. In other words, if you tell me what the exact limitations of your arm is, I can in principle calculate the exact weight distribution and even a three dimensional model, that will optimize a given chopping situation, without any ifs buts and whens….which, by the way is constantly done for other mechanical systems such as engine parts by people a lot more skill at this than me (its been 12 years since I had to do such calculations). If you find historical blades (swords, and such) that do not conform to the basic principles you have to realize that Newton lived in the 18th century and is laws were not widely employed till the 19th century and anyone living before that would not have had the proper tools to do these calculations, nor did the medieval knights access to 3-D simulation programs. In other words, despite their experience with blades, they were shooting completely in the dark.
But all of these theories rely on the human parameter, which can not be calculated, so in no way should what I said above discourage you from conducting your experiments and finding what works best for you.
What I was really mostly interested in my initial post in thread 1, which originated from the Rogue review, was how weight distribution affects chopping ability. So I really wanted to reduce the entire problem to this parameter. For everything below, I will make the following demands. The angular velocity shall always be the same, the total mass of the blade shall stay the same and the length of the blade is not to be changed. And to restrict things further, the COM shall be the same as well, but I hope you will see that any of the above is easy enough to take into consideration. So you can see that I am taking a very abstract view on this, because that is the only thing that I can bring to the table. You are far more experienced at swinging blades than I am, so all I can do is point out a few principle laws which hopefully allow you to get a more complete picture of what is happening when you modify your blades.
So on with the good stuff ?. Scroll down for the figures!
You are absolutely right in that we should look at the swing an the impact separately. So we look at the swing first. A lot of people seem to have trouble with the moment of inertia, so I figure, we completely ignore that an work only with the kinetic energies involved.
In Fig.1 I am trying to show the system. There are three rotational centers, wrist, elbow and shoulder. Since the weight of your arm and the distances of these centers to each other always stay the same and since all three centers are outside the body of interest, we might as well simplify the problem as I have shown in Fig. 2. It is irrelevant to the problem, whether you “snap out of the wrist” or swing out of the shoulder, all it changes is the velocity and the masses of your arm involved, but non of these things change when you change the weight distribution in the blade. It also does not matter whether you have 3 rotational centers or just one. You can verify that with the same equations I give with the figure. All you have to do is assume a mass and length for your hand, forearm, and upper arm.
So since we got into all of this through mass distribution, it would be boring to assume that the weight is evenly distributed along the knife. Instead we are assuming for the moment a weight distribution with two centers, one towards the handle and one towards the blade. You can also see that the hand is now part of the knife since the closest rotation center (axis) is behind the hand (the wrist). Now hardheart asked why the COM is so important and I said, that a rotation of a body around an external axis can be decomposed into one around the COM and one of the COM. This is what I show with the calculation going with Fig. 2. The total energy Etot is calculated as the sum of the energy of the two masses, which for the moment we assume as equal (makes the math easier, but is not necessary) rotating around A1. If you calculated the sum out, you get an expression which is exactly the same as when you add the energy of the COM around A1 and the rotation of the two masses around the COM (A2).
From this, one thing becomes immediately clear: the further out you space the mass distribution, the larger the kinetic energy at the same angular velocity! In the extreme case, if one mass is placed at A1, and the other at 2R, the kinetic energy is twice as large as the same total mass located at the COM.
But how does this play out in a more realistic model? So completely arbitrarily I made up a numeric example which I thought was reasonable for a knife, but you can change around the numbers however you like (Excel works very well if you want to play around with some sample numbers) if you feel that the examples are not reasonable, it will not change the physics. So in Fig. 3 I made up two knives of equal weight with the same COM but with two different weight distributions. Let’s assume that the knife is 12” long and we have somehow the capabilities of measuring how much weight is in each inch of the knife (for example in oz). The dot represents the rotational center. You can see that the total weight of both knifes is 30 (in oz this is pretty heavy for a knife….oh well
), and the COM is 6.5” from the rotational center. However, while one knife has a weight distribution that is centered around the COM (2.), the other one as a bi-modal weight distribution (1.). Immediately is clear that knife 1.) has a higher kinetic energy than knife 2.), However, since in a knife (unlike in an axe) the bimodality can not be very large, the difference between the two is actually pretty small. We see this even better if we decompose this again into a rotation of the COM and around the COM. You see that the bulk of the energy is from the rotation of the COM (about 80%). You can simplify this model again to the dumbbell rotating around the COM and as you can see in Fig.4 each side of the COM has an average mass of 15 and a kinetic energy of 155.75. From the equation for a single mass rotating around the COM you can determine that the weight distribution of knife 1 is the same as two masses of 15 rotating at a distance r of 3.22, while for knife 2 the same mass would be rotating only a distance of 2.82 from the COM. Hence this distance is a direct indication of the weight distribution of the knife. Since the moment of inertia is something very abstract it can thus be visualized and made much more tangible if this distance is viewed as the moment of inertia with respect to the COM. This is what I meant, when I talked in my first post as the moment of inertia as a distance.
Ok, sofar I have used conveniently a symmetric system so that you can see immediately were the COM is and that it hasn’t moved. Now, let’s see, what happens if you have an uneven bi-modal distribution without shifting the COM or the total mass. It is late and this post is already very long, so I will not do this for a numeric example, but only for the reduced dumbbell system. You have now all the tools to play around with numbers yourself, and hopefully you believe me by now that the dumbbell system is completely representative of this problem. Now, in Fig. 5 I have increased the mass of the handle side and moved it towards the tip such that the position of the COM is unchanged. I calculated the symmetric system below. And here you see something that you might not have expected: Even though you increase the mass and move it forward, the total kinetic energy goes DOWN!!! Now, it is time to show your faith in physics, because there is a reason for it, which you can only see in the decomposed system. It becomes immediately apparent, that the rotation of the COM is the same, because the total mass is still the same and the position of the COM has not changed. However, the energy of the rotation around the COM has decreased, because the distance from the rotational center (the COM) enters quadratic and the mass increased only linearly.
Now, a final word about the COM. With the simple equation of the kinetic energy we can now figure out just as easily what happens if you move the COM forward. Let’s compare two knives, one 10” long, the other 7”. Let’s assume (to make the calculation in our head) that both knives have completely even weight distribution (pretty unrealistic, I agree, but if you have a forward balance the effect is even bigger) and let’s also assume that the two knives have the same mass (again unrealistic, since the 10” knife should certainly be heavier if they are made in the same style). This means that the COM of the 10” knife is at 5 and the COM of the 7” knife is at 3.5. Taking the squares, you get 25/12.25 = 2.04. More than double the kinetic energy! Of course, as I said in the introduction, this assumes that you can actually swing the 10” knife just as fast as the 7” knife, which might not be the case.
Now, only very quickly onto the second part of chopping: the impact. On impact a new rotational center is created, and one only: the point of impact (POI). Because by definition the blade is static at the point of impact (disregarding that the blade bites into the target, because that is a small, linear motion) while it can move at all other points, which is particularly true if you loosen your grip right before impact as you usually would do. However, this is not a necessary assumption. Necessary assumptions though are, that you do not strike through the target and that the target does not move itself (hard target). Now, instead of the kinetic energy, we have to consider a momentum balance now. If the momentum on one side of the target is greater than on the other, the blade will start rotating. To prevent the knife from flying in the direction of the residual momentum, the hand has to stop the rotation, which is felt as shock. Let’s call the point at which the momentum is balanced the momentum balance point (MBP), which is, I believe, what Cliff called the “dynamic balance point”. We can calculate the MBP for our sample knife 1 and 2 which I did in Fig.6 and from its very definition you can see that a strongly bi-modal weight distribution will shift this point forward as will moving the COM forward or lengthening the blade, but now the distance enters the equation only linearly. Knife 1 has the MBP very close to 9, while knife 2 has the MBP between 7 and 8. Two things immediately apparent though, without making a single calculation: The MBP will always be forward of the COM and it will never be exactly at the tip or beyond (because there would be nothing to create a balance anymore).
There is one more very important realization to make: It is completely irrelevant to this problem where the MBP is. As long as you hit exactly the MBP, there is no excess momentum and all the kinetic energy is transferred into the target (bouncing back from the target we ignore for the moment because that is also independent of the MBP). Since energy must always be conserved, you can not generate a more powerful strike just by shifting the MBP, if the kinetic energy that you created in the swing is the same! So contrary to what Cliff said or what you think you observe, the power of the swing is not connected to the MBP and couldn’t care less whether you swing out of the wrist or out of the shoulder (even though I am pretty sure that Cliff was not talking about a stationary target, see below). If you argue against that, you are violating energy conservation. HOWEVER, the velocity of the POI is directly dependent on the MBP (assuming that you deliver a strike with the POI matching the MBP as you should for maxium energy transfer). That means that you get a direct benefit if you move the MBP forward if your target is not stationary. If your target requires a certain velocity to prevent a bending away (like a thin branch) a blade with a forward MBP might be successful were one with a rearward MBP might not, even though they might carry the same energy, the slower blade just doesn’t get to transfer all of it.
With the momentum calculation you can also calculate the excess momentum and get an idea of how much shock is transferred in case of a mis-strike. I haven’t done the calculation, but I am pretty sure that the strongly bi-modal distribution is more forgiving.
Lastly, since a strongly bi-modal distribution both increase kinetic energy and moves the MBP forward and hence increases POI velocity, you usually do get a more powerful swing with a forward MBP.
You can now assume any weight distribution and shift around COMs as you like or keep them in a certain place as you like and see how it affect the kinetic energy and the MBP. The only thing missing is how to map out the weight distribution of an actual blade. I haven’t quite figured that out, but I am working on it.
.Ah, I hate it when I am wrong….but it did get me thinking more on this topic so in the end I greatly profited from it. As you will see, my comments about the COM are not entirely correct, but you will also see that my basic conclusions were not too far off.
So, I think, Possum, a lot of what you see in practice that might be different from what I try to show below is due to the fact that you are changing all four parameters: length, mass, weight distribution, and very importantly speed at the same time. But this obscures in my opinion some of the underlying principles. So in the following, I am trying to look at a system in which the blade is always moved with the same angular velocity. Obviously, in real life, if you make the blade too long or too heavy, you are will not be able to swing it with the same speed as a smaller or lighter blade, but to take that into account, I would have to know what the limits of the man who swings the blade are. However, if you would be able to measure that, it would be easy enough taking it into account. So for the present we ignore that variable.
Also, I would like you to realize, that, despite the simplifications that am using, this is a fully determined system. Unlike talking about metallurgy, which is still more or less a form of alchemy (ok, before anybody gets on my case for this, take it with a bit of a smile), chopping is based on simple classical mechanics and this is a fully determined system. In other words, if you tell me what the exact limitations of your arm is, I can in principle calculate the exact weight distribution and even a three dimensional model, that will optimize a given chopping situation, without any ifs buts and whens….which, by the way is constantly done for other mechanical systems such as engine parts by people a lot more skill at this than me (its been 12 years since I had to do such calculations). If you find historical blades (swords, and such) that do not conform to the basic principles you have to realize that Newton lived in the 18th century and is laws were not widely employed till the 19th century and anyone living before that would not have had the proper tools to do these calculations, nor did the medieval knights access to 3-D simulation programs. In other words, despite their experience with blades, they were shooting completely in the dark.
But all of these theories rely on the human parameter, which can not be calculated, so in no way should what I said above discourage you from conducting your experiments and finding what works best for you.
What I was really mostly interested in my initial post in thread 1, which originated from the Rogue review, was how weight distribution affects chopping ability. So I really wanted to reduce the entire problem to this parameter. For everything below, I will make the following demands. The angular velocity shall always be the same, the total mass of the blade shall stay the same and the length of the blade is not to be changed. And to restrict things further, the COM shall be the same as well, but I hope you will see that any of the above is easy enough to take into consideration. So you can see that I am taking a very abstract view on this, because that is the only thing that I can bring to the table. You are far more experienced at swinging blades than I am, so all I can do is point out a few principle laws which hopefully allow you to get a more complete picture of what is happening when you modify your blades.
So on with the good stuff ?. Scroll down for the figures!
You are absolutely right in that we should look at the swing an the impact separately. So we look at the swing first. A lot of people seem to have trouble with the moment of inertia, so I figure, we completely ignore that an work only with the kinetic energies involved.
In Fig.1 I am trying to show the system. There are three rotational centers, wrist, elbow and shoulder. Since the weight of your arm and the distances of these centers to each other always stay the same and since all three centers are outside the body of interest, we might as well simplify the problem as I have shown in Fig. 2. It is irrelevant to the problem, whether you “snap out of the wrist” or swing out of the shoulder, all it changes is the velocity and the masses of your arm involved, but non of these things change when you change the weight distribution in the blade. It also does not matter whether you have 3 rotational centers or just one. You can verify that with the same equations I give with the figure. All you have to do is assume a mass and length for your hand, forearm, and upper arm.
So since we got into all of this through mass distribution, it would be boring to assume that the weight is evenly distributed along the knife. Instead we are assuming for the moment a weight distribution with two centers, one towards the handle and one towards the blade. You can also see that the hand is now part of the knife since the closest rotation center (axis) is behind the hand (the wrist). Now hardheart asked why the COM is so important and I said, that a rotation of a body around an external axis can be decomposed into one around the COM and one of the COM. This is what I show with the calculation going with Fig. 2. The total energy Etot is calculated as the sum of the energy of the two masses, which for the moment we assume as equal (makes the math easier, but is not necessary) rotating around A1. If you calculated the sum out, you get an expression which is exactly the same as when you add the energy of the COM around A1 and the rotation of the two masses around the COM (A2).
From this, one thing becomes immediately clear: the further out you space the mass distribution, the larger the kinetic energy at the same angular velocity! In the extreme case, if one mass is placed at A1, and the other at 2R, the kinetic energy is twice as large as the same total mass located at the COM.
But how does this play out in a more realistic model? So completely arbitrarily I made up a numeric example which I thought was reasonable for a knife, but you can change around the numbers however you like (Excel works very well if you want to play around with some sample numbers) if you feel that the examples are not reasonable, it will not change the physics. So in Fig. 3 I made up two knives of equal weight with the same COM but with two different weight distributions. Let’s assume that the knife is 12” long and we have somehow the capabilities of measuring how much weight is in each inch of the knife (for example in oz). The dot represents the rotational center. You can see that the total weight of both knifes is 30 (in oz this is pretty heavy for a knife….oh well
), and the COM is 6.5” from the rotational center. However, while one knife has a weight distribution that is centered around the COM (2.), the other one as a bi-modal weight distribution (1.). Immediately is clear that knife 1.) has a higher kinetic energy than knife 2.), However, since in a knife (unlike in an axe) the bimodality can not be very large, the difference between the two is actually pretty small. We see this even better if we decompose this again into a rotation of the COM and around the COM. You see that the bulk of the energy is from the rotation of the COM (about 80%). You can simplify this model again to the dumbbell rotating around the COM and as you can see in Fig.4 each side of the COM has an average mass of 15 and a kinetic energy of 155.75. From the equation for a single mass rotating around the COM you can determine that the weight distribution of knife 1 is the same as two masses of 15 rotating at a distance r of 3.22, while for knife 2 the same mass would be rotating only a distance of 2.82 from the COM. Hence this distance is a direct indication of the weight distribution of the knife. Since the moment of inertia is something very abstract it can thus be visualized and made much more tangible if this distance is viewed as the moment of inertia with respect to the COM. This is what I meant, when I talked in my first post as the moment of inertia as a distance.Ok, sofar I have used conveniently a symmetric system so that you can see immediately were the COM is and that it hasn’t moved. Now, let’s see, what happens if you have an uneven bi-modal distribution without shifting the COM or the total mass. It is late and this post is already very long, so I will not do this for a numeric example, but only for the reduced dumbbell system. You have now all the tools to play around with numbers yourself, and hopefully you believe me by now that the dumbbell system is completely representative of this problem. Now, in Fig. 5 I have increased the mass of the handle side and moved it towards the tip such that the position of the COM is unchanged. I calculated the symmetric system below. And here you see something that you might not have expected: Even though you increase the mass and move it forward, the total kinetic energy goes DOWN!!! Now, it is time to show your faith in physics, because there is a reason for it, which you can only see in the decomposed system. It becomes immediately apparent, that the rotation of the COM is the same, because the total mass is still the same and the position of the COM has not changed. However, the energy of the rotation around the COM has decreased, because the distance from the rotational center (the COM) enters quadratic and the mass increased only linearly.
Now, a final word about the COM. With the simple equation of the kinetic energy we can now figure out just as easily what happens if you move the COM forward. Let’s compare two knives, one 10” long, the other 7”. Let’s assume (to make the calculation in our head) that both knives have completely even weight distribution (pretty unrealistic, I agree, but if you have a forward balance the effect is even bigger) and let’s also assume that the two knives have the same mass (again unrealistic, since the 10” knife should certainly be heavier if they are made in the same style). This means that the COM of the 10” knife is at 5 and the COM of the 7” knife is at 3.5. Taking the squares, you get 25/12.25 = 2.04. More than double the kinetic energy! Of course, as I said in the introduction, this assumes that you can actually swing the 10” knife just as fast as the 7” knife, which might not be the case.
Now, only very quickly onto the second part of chopping: the impact. On impact a new rotational center is created, and one only: the point of impact (POI). Because by definition the blade is static at the point of impact (disregarding that the blade bites into the target, because that is a small, linear motion) while it can move at all other points, which is particularly true if you loosen your grip right before impact as you usually would do. However, this is not a necessary assumption. Necessary assumptions though are, that you do not strike through the target and that the target does not move itself (hard target). Now, instead of the kinetic energy, we have to consider a momentum balance now. If the momentum on one side of the target is greater than on the other, the blade will start rotating. To prevent the knife from flying in the direction of the residual momentum, the hand has to stop the rotation, which is felt as shock. Let’s call the point at which the momentum is balanced the momentum balance point (MBP), which is, I believe, what Cliff called the “dynamic balance point”. We can calculate the MBP for our sample knife 1 and 2 which I did in Fig.6 and from its very definition you can see that a strongly bi-modal weight distribution will shift this point forward as will moving the COM forward or lengthening the blade, but now the distance enters the equation only linearly. Knife 1 has the MBP very close to 9, while knife 2 has the MBP between 7 and 8. Two things immediately apparent though, without making a single calculation: The MBP will always be forward of the COM and it will never be exactly at the tip or beyond (because there would be nothing to create a balance anymore).
There is one more very important realization to make: It is completely irrelevant to this problem where the MBP is. As long as you hit exactly the MBP, there is no excess momentum and all the kinetic energy is transferred into the target (bouncing back from the target we ignore for the moment because that is also independent of the MBP). Since energy must always be conserved, you can not generate a more powerful strike just by shifting the MBP, if the kinetic energy that you created in the swing is the same! So contrary to what Cliff said or what you think you observe, the power of the swing is not connected to the MBP and couldn’t care less whether you swing out of the wrist or out of the shoulder (even though I am pretty sure that Cliff was not talking about a stationary target, see below). If you argue against that, you are violating energy conservation. HOWEVER, the velocity of the POI is directly dependent on the MBP (assuming that you deliver a strike with the POI matching the MBP as you should for maxium energy transfer). That means that you get a direct benefit if you move the MBP forward if your target is not stationary. If your target requires a certain velocity to prevent a bending away (like a thin branch) a blade with a forward MBP might be successful were one with a rearward MBP might not, even though they might carry the same energy, the slower blade just doesn’t get to transfer all of it.
With the momentum calculation you can also calculate the excess momentum and get an idea of how much shock is transferred in case of a mis-strike. I haven’t done the calculation, but I am pretty sure that the strongly bi-modal distribution is more forgiving.
Lastly, since a strongly bi-modal distribution both increase kinetic energy and moves the MBP forward and hence increases POI velocity, you usually do get a more powerful swing with a forward MBP.
You can now assume any weight distribution and shift around COMs as you like or keep them in a certain place as you like and see how it affect the kinetic energy and the MBP. The only thing missing is how to map out the weight distribution of an actual blade. I haven’t quite figured that out, but I am working on it
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and read the whole thing but a xanax (3 Mg) and ambien (30 Mg) cocktail sounds like a better choice. I'll save the (10 Mg) Lortabs for later.
Likewise, if we could accellerate hatchets the same as knives, there would be basically no need for chopping knives. I think we'd see the same real-world problems with your dumbbell shaped knife.