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Post by Deleted on Nov 21, 2007 11:32:43 GMT
How would the impact pressure change if you assume 0.15 second for deceleration? Decreasing the time it takes for deceleration (negative acceleration) to occur from 0.5 s to 0.15 s increases the force to 73 lbs. The pressure of that force distributed over a 1 inch length of a blade with a edge thickness of 0.02 inches is 3650 psi. I guess it all really depends on how wide the pumpkin is that you are cutting. Keep in mind there are a lot of assumptions in all these calculations.
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Post by Deleted on Nov 21, 2007 14:15:07 GMT
Mike, is it possible for to get an accurate acceleration number with your equipment? Can we get a starting velocity, an ending velocity and the time between? I am not comfortable with the way we have obtained the V2 figure.
Mark, I don't fully understand the mass number you are using in the calculation of .0466. My mass number of 1 lb was obtained by putting the tip of a sword on a scale and holding it horizontally by the pommol.
I know you said there are a lot of assumptions. I want to try and isolate out the assumptions and see what we are left with and how we can use that info. I don't want to consider edge thickness, cause that is hard for us to accurately measure. I will be happy if you get a good Force calculation from actual measures.
As far as the katana, we would have to think about that some more. In a slice the katana is moving slower, 48 mph by an expert according to the Mythbusters. We would have to add to that the additional effect of downward force from the arms that one normally makes during a slice. That effectively adds to the mass. I did not consider this before because it is much less with a tip cut. It seems to add the equivalent of an extra 3 lbs of force to the tip based on a scale test I did. In a slice, pressing down with the body might add 15 to 30 lbs of extra downward force depending on your form. In the Mythbusters vid above, that guy is really pressing down with his body. If we decide to use the katana as we do the longsword and make a tip cut using the leverage of the handle, we will have to consider that the tip of the katana has much less mass.
note: So far I have edited this post 20 times re-thinking about all this. I guess its ok to edit as long as no one posts a new response. I just though I would clarify this incase anyone has noticed.
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Post by ShooterMike on Nov 22, 2007 2:53:16 GMT
Mike, is it possible for to get an accurate acceleration number with your equipment? Can we get a starting velocity, an ending velocity and the time between? I am not comfortable with the way we have obtained the V2 figure. Sorry Bill. All it does is start and stop a clock across a fixed distance. Then it uses the figures to calculate a constant speed.
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Post by Deleted on Nov 22, 2007 4:37:23 GMT
So I guess we will have to settle for momentum computations. I suspect the difference would not have been that much if we had accurate measures of V2 and time.
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Post by Deleted on Nov 23, 2007 0:59:55 GMT
Guys,
Just saw this thread, interesting! I'm a mathematician, and frequently teach mechanical dynamics. I centainly don't want to sound definitive on the issue here (It's actually quite a difficult problem!) or step on any toes, but I'll be happy to add a bit if anybody is still interested.
r
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Post by Deleted on Nov 23, 2007 3:57:22 GMT
Absolutely Rjs100 . I never thought my way to do it was best. It is the just the only way I can do it given my knowledge. It will be interesting to see how big the variation is.
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Post by Deleted on Nov 24, 2007 1:11:04 GMT
Ok, so what is the power we can get? The tip speed of 270 f/s is probably close, but the mass moving that fast is no where near 1.5 lbs. Force is not really a relevant way to measure what is going on here. (Think about the 21 lb. figure, certainly you could push on the sword with more than that force with the blade-even a relatively small linear portion of the blade-resting on the stationary target, like a pumpkin or thick leather coat, and not get any penetration at all,.) Additionally, the calculations Mark provided use average acceleration and hence the average force over the entire contact time.
The more important measure, I think, is the total kinetic energy deposited on the target.We could measure the CG (center of mass) speed and then the rotation rate of the blade about that to compute the total kinetic energy transferable as the linear kinetic (1/2mv^2) plus the rotational kinetic (1/2Iw^2) where I is the polar moment and w is the angular velocity.)Measurements would be difficult to effectuate. (high speed cameras etc.)
If what you are really after however is knowing how much damage the long sword could do, (Say crashing into plate armor or a helmet etc.) I think we can very accurately measure the energy transfer in this case.
If you set up a ballistic pendulum, that is, a known mass on the end of a pivoting rod that you secure to a frame, you then just strike the mass as hard as you can with the sword (the mass being something dense and heavy so the sword does not bounce back, giving total energy transfer) and observe the angle to which you get the weight to swing (affix a protractor with a pointer to the pivot end and the maximum swing angle will be recorded by the sliding pointer.) This angle can be translated directly into energy by a simple formula. If the angle you measure is theta say, and the length of the pivoting rod is L then the total energy transfered to the target would be mgL(1-cos(theta)) where m is the mass the sword struck and g is the gravitational acceleration. (I've neglected the mass of the rod, but if the mass of the bob is a lot more than that of the rod as it should be, then this will be pretty accurate) You can then compare the total energy of the sword with the energy one gets from various firearms etc., as those figures are readily found in the literature.
Interestingly, in a clean cut very little energy is actually transfered to the target, the little that is is an artifact of the blade having to displace the material as it passes (one sees this effect in ballistic gelatin rippling away from the impact site as an approximately orthogonal to the edge movement compression wave propagates from the contact front.) One generally attempts to minimize energy transfer, ideally reduce it as close to zero as possible evidenced by the perfect cut leaving the target undisturbed other than being cut through! This is where the curvature of the blade (virtual or actual) is significant. The principal of "draw" cutting really attempts to alter the angle of incidence of the blade which in effect increases the sharpness of the edge. This is NOT due to a smaller linear contact patch from the curved blade. Think of the cross section of a blade being just a wedge, perhaps if it is sharp, the wedge will subtend say 20 degrees or so. By drawing the blade across the target (or having a lot of curvature built in) we present a cross section that is no longer perpendicular to the edge, but rather at a substantial angle toward the hilt for a standard "draw". The effect is to decrease the wedge angle seen by the material substantially, thus lowering the pressure and causing a cut. In fact, the perceived wedge angle is the product of the original wedge angle times the cosine of this draw angle. This is why we slice tomatoes rather than chop them. Most blades would have to be too sharp to be durable to cut well without drawing on some materials. (There is a secondary issue here dealing with the shear response of the material to an edge sliding parallel to it, but it is much less important.) So "drawing" either by actually drawing or curving the blade, or doing both, will give the effect of having a blade much sharper than the material it's made from could support with any durability.
Sorry for being so long winded and pedantic!
r
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Post by Deleted on Nov 24, 2007 15:49:55 GMT
Ok so lets try to figure out the Kenetic energy in pounds. Lets assume we are not cutting through but instead hitting solid armor. Working with the formula k= (1/2mv^2)
tip of blade is 1 lb velocity is 180 mph
If we square v we get 32,400. Multiply by mass 1 and take half of that we get 16,200 units of something. How do we convert kenetic energy into pounds of force going into the target.
I can't work the other formula because I don't have a cos chart or calculator handy that can calculate that or thata. Could you work the calculation and see what we get.
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Post by Deleted on Nov 24, 2007 18:15:30 GMT
Tsafa,
We need the miles per hour in feet per second, for that multiply by 5280ft/mile times 1hr/3600sec. That is multiply by 1.467. So whatever mph number you have multiply by 1.467 to convert to feet per second, then square etc. The mass must be in slugs to make the calculation come out correctly in lb-ft of energy, so divide the number of lbs you have moving by 32.2.
For the example of 1 lb moving at 180mph we get 1082 lb-feet of energy deposited on target. That is close to the muzzle energy from an M16!
I am skeptical that you actually have 1 lb moving at that velocity, as I mentioned before to calculate the KE accurately we need how fast the CG is moving plus the angular velocity of the mass to get the correct KE. Again as stated before, these measurements are problematic. I think it best to employ a ballistic pendulum to get really accurate results.
r
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Post by Deleted on Nov 25, 2007 1:50:08 GMT
Thanks for helping me out.
The 1 lb measure was taken by putting the sword tip of my Gen 2 Lucern on the scale and holding it by the pommol horizontally. If I hold it under the cross-guard the tip weight is 8 oz. I recalculate the formula to equal 541.36 lb-feet. Does this sound like a more reasonable number?
I will work on setting up a ballistic pendulum too.
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Post by Deleted on Nov 25, 2007 5:33:36 GMT
Tsafa,
I actually think that you might be able to generate that kind of energy (that is 1000 ft-lbs or so) with a long sword!
I just don't think that there is that much mass concentrated near the tip of the sword. I suppose we could figure out how much mass there was in the tip by dipping it into water to find out how much volume it displaces and then calculate it's weight from the density of steel (specific weight of steel is .285lb/cubic inch) but of course that presumes you know how much of the sword is moving at the assumed speed etc. Best to go the pendulum route.
When you put the sword horizontal and measure the tip weight at 1 lb, you are really getting the weight of the sword times the distance from the handle end to the POB, divided by the length of the sword (assuming you balanced the very end of the handle only). This is not the weight of the tip itself. (Think of a perfectly uniform bar, balanced at one end and the other end supported by the scale, then the scale would read half the weight of the bar, which is certainly not the weight of the tip alone)
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Post by Deleted on Nov 25, 2007 10:37:50 GMT
Thank you RJS100. This project has shed light on a lot of ideas.
ShooterMike, I recall you sawed the tip off your Gen2 12 century to shorten it. Could you put that on a scale and tell us how much it weighs. Perhaps you know some other people that have broken swords tips off and can give use a few weight measures.
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Post by Deleted on Nov 25, 2007 17:08:34 GMT
You're welcome. It will be interesting to see the data you get.
r
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