Scion of the Mists Posted April 5, 2019 Posted April 5, 2019 15 hours ago, Clever Username said: The books say that a shardblade isn't that heavy but when it changes into a shield it is the same weight. Now, shields are usually heavier than swords but in shard form they are equal. When Kaladin changes Syl to a knife I don't think Brandon said anything about weight but with the sword shield ratio we can assume the knife was the same weight as the sword. This would mean that you wouldn't need a lot of leverage with a 30 foot long sword as is weighs the same as a normal sword. But this point is moot anyway since there is a limit on how large a shardblade can be. (I actually haven't read the books in a long time so if you could come up with some text that would be great) Do we know that the Syl-blade and Syl-shield weighed the same? I can't imagine that the knife and the sword weighed the same. Also, yeah, it's the torque that's a problem, not the weight. Torque is dependent on where the center of mass is. Imagine picking up a really heavy bookshelf. Now imagine leaning it over so it's horizontal, sticking out away from you.
Lopen Armblessed he/him Posted April 5, 2019 Author Posted April 5, 2019 *face palms* I completely forgot about that. Even if it was the same wieght as a normal sized shardblade it would apply more torque so you would need a lot of leverage. My mistake.
Gasper he/him Posted April 6, 2019 Posted April 6, 2019 On 4/5/2019 at 10:29 AM, Clever Username said: *face palms* I completely forgot about that. Even if it was the same wieght as a normal sized shardblade it would apply more torque so you would need a lot of leverage. My mistake. Just balance it, apply a equal moment to each end with the hand being the point of rotation. A very long and very thin blade with a short and heavy pommel.
Scion of the Mists Posted April 8, 2019 Posted April 8, 2019 On 4/6/2019 at 2:45 PM, Gasper said: Just balance it, apply a equal moment to each end with the hand being the point of rotation. A very long and very thin blade with a short and heavy pommel. Let's say the sword is about 2kg, with a length of 10m, and the grip is 0.2m long. So you have a 2kg weight 5m away on one side and an Xkg weight 0.1m on the other. To make them balance, you'd need a 100kg (220lb) pommel. Not very feasible.
Gasper he/him Posted April 8, 2019 Posted April 8, 2019 (edited) 1 hour ago, Scion of the Mists said: Let's say the sword is about 2kg, with a length of 10m, and the grip is 0.2m long. So you have a 2kg weight 5m away on one side and an Xkg weight 0.1m on the other. To make them balance, you'd need a 100kg (220lb) pommel. Not very feasible. I was thinking something about 2-3 meters long, not something as long as a school bus, also, you have to factor in the force exerted by the hand on the sword grip, the pommel, and that the weight of the blade is going to be at the center of mass of the blade, not at the tip. The weight of the blade is actually distributed across the length, but lets make this easy. I had a drawing but I cant get it to work. B is the balance point, O is pommel, H is hand, M is center of force on the blade. B->M=1.5 m B->O=.5 m B->H=.2 m M=2N H= -3N O=? Moment = 0, (2*1.5)+(-3*.2)+(O*.5)=0 O=.6 N Edited April 8, 2019 by Gasper
Scion of the Mists Posted April 8, 2019 Posted April 8, 2019 2 hours ago, Gasper said: I was thinking something about 2-3 meters long, not something as long as a school bus, also, you have to factor in the force exerted by the hand on the sword grip, the pommel, and that the weight of the blade is going to be at the center of mass of the blade, not at the tip. The weight of the blade is actually distributed across the length, but lets make this easy. I had a drawing but I cant get it to work. B is the balance point, O is pommel, H is hand, M is center of force on the blade. B->M=1.5 m B->O=.5 m B->H=.2 m M=2N H= -3N O=? Moment = 0, (2*1.5)+(-3*.2)+(O*.5)=0 O=.6 N I was referring to the specified problem of a 30ft blade, where the center of mass would be around 15ft away from you (~5m).
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