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Comments
Stormtroper you are my hero for the day.
Huh, I'd thought we could treat the hilt as being part of the blade. On second thought, that's not a very accurate assumption. This is going to make things more complicated.
We'll still need to know what the cross-sections of the blade and hilt look like (please tell me it's one of the upper ones from here) and their measurements, however, including knowing where the sharp point starts (that's the end part where the blades converge, right?). I think we can treat the hilt as a cylinder, so that part should be easy.
Also, we still need to find a way to know the lineal and angular velocities of the sword. Knowing the velocity of any two points of the sword is enough. Also, more or less where on the blade is the impact point?
Also, why does this guy have a violin on his head?
^ Just doing my job.
Diamond, lenticular and hexagonal are the only "real" types of blade cross-section -- all others are variations on them, at least for double-edged swords. Which one we want to use depends on the particular sword type we want to discuss, but your average 15th century knightly longsword would definitely have a diamond cross-section.
The hilt can probably be treated as a 30cm (approx. 12 inch) cylinder. Now, the angle at which the blade tapers is a tricky thing because there's so much variance. Some swords don't taper at all until the last few inches, Some taper from almost all the way from the bottom to the top, ending in a fine point. With that in mind, you can really pick and choose, although if we're going with the theme of "15th century knightly longsword", we should lean on the stronger side for blade taper.
Not sure about specific velocities. I do know about impact, though; the ideal place to hit is two or three inches below the point.
I am not quite sure why that man has a violin on his head.
Just popping in to brag about how I personally know the pony physics presentation guy.
Ahh, my brain hurts. I tried to find the load given arbitrary dimensions and got hugeass equations that I really don't want to do. Maybe if there was an online calculator that could do multiple integrals and singularity functions.
Also, I got the joke right off, too.