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MLP has ruined physics class.
Okay, so I just started my second semester of my sophomore year in college. Right now I'm taking a physics class, and we've been going over the basics from high school, starting with "Force = Mass x Acceleration."
This, of course, is abbreviated "F = MA." Now I can't stop seeing ponies in the problems we do.
Comments
I should probably get my basic physics equations down pat. It'd be interesting to be able to measure the precise kinetic force of my sword strikes.
If a sword of 1.5kg (approx. 3lbs) can move .9 meters (approx. 3 feet) in 0.2 of a second, what would be the force of the kinetic impact?
Wouldn't you have to know the length and weight distribution of the sword to answer that?
^More like the force at the point of impact, to find that we need to treat the sword as a rod and find the moment of inertia. We already have the time, so we could presumably find the angular velocity and then use that to find the torque.
The .9 meters/3 feet is the length of the blade whipping out. I guess that could technically be 1.2 meters (4 feet) with arms at extension, though.
As for weight distribution, all close combat weapons are designed to deliver the maximum amount of kinetic force within their design specs. A mace or war hammer will deliver more for its comparative weight due to the nature of the surface that makes contact as compared to a sword, but that requires an additional measurement as well.
>OP's about ponies
>Alex derails it into swords in the second post
I don't know who am i supposed to hate
halp
What does F = MA have to do with ponies?
Or...more specifically to do with ponies than it does with everything else, I guess?
Friendship = Magic.
All I can read is Fullmetal Alchemist
Funny, I just read it as Newton's Secon-
Oh goddammit Mass Effect.
I thought this was going to be about the physics class where they were supposed to analyze Hollywood Physics and some kid brought in the ponies.
EDIT: found it
Also relevant:
I read that as "Fuck = My Awesome" and I was wondering what the hell that meant.
Wow, I'm surprised Anonus is the only other person who didn't immediately see this.
I haven't even seen the show, and this was my first thought.
I saw it. Just didn't post because I thought it was obvious >.>
I think the reason I don't see it is because I see the "ma" as two things, rather than one. Also because I don't particularly care for MLP.
Wild Mass Guessing: In Mid-Childa, Newton's Second Law is written ma = F.
It took me a short while to see it, admittedly.
I didn't get it at first either.
Due to work-related "injuries", the only thing I can possibly see in F=ma is force equals mass times acceleration. I simply can't unsee Newton's second law. Don't worry, if you plan on pursuing physics, you'll become just like me.
It's even worse with greek letters. What do you think when you see ψ? If your answer is wave function, you know what I'm talking about.
All this sword mechanics made me regret not having been able to take the relevant elective class, I just spent a while reading about this.
Anyhow, what y'all seem to be talking about is not a force, but an impulse. I have no idea how to go about finding a theoretical force vs. time curve, but I think we can find a load P equivalent to the maximum force (correct terminology might be lost in translation, and might not be easy to see without diagrams):
Assuming the sword's center of mass is within the force's line of action (otherwise we'll need to know what INUH said) and assuming it's not rotating (sword strikes do that, don't they? We'll need to know the sword's angular velocity otherwise), the sword's kinetic energy is:
U = 1/2 m v^2
This is equivalent to the strain energy the load will do on the sword, which we can find as a function of said load.
Assuming the sword's inertia acts uniformly distributed along the whole sword (because I have no idea what it'd actually look like. This assumption holds better if it's a large area what is being hit), we have a distributed force equal to:
w = P / L
Assuming its center of mass is in the middle, we can calculate the strain energy. Take the sword as if it were two fixed beams of half its length. First, the normal strain energy (the integral is from 0 to L / 2, E is its elasticity/Young's module (200 GPa for all steels... or was it building steels?) and I its moment of inertia of area,for which we need the sword's profile dimensions to know):
Moment: M = 1/2 w x^2
Normal strain energy: Uσ = int M^2 / (2 E I) dx = int (1/2 w x^2)^2 / (2 E I) dx = w^2 / (8 E I) int x^4 dx = w^2 L^5 / (1280 EI)
Substituting w and adding both beam's normal strain energies, we have:
Uσ = P^2 L^3 / (640 E I)
Now, finding the beams' shear strain energies (G is the rigidity module, 77 GPa for all steels I know of, Q the first moment of inertia, and t the width):
Shear force: V = w x
Shear stress: τ = V Q / I t
Shear strain energy: Uτ = int τ^2 / (2 G) dv = int (V Q / I t)^2 / (2 G) dv
Doing anything else requires knowing the sword's profile dimensions, so let's leave it at that. Remember that the total shear strain energy is two times that.
The total strain energy is:
U = Uσ + Uτ
Ignoring Uτ
P^2 L^3 / (640 E I) = 1/2 m v^2
=> P = √(320 m v^2 E I / L^3)
IJBM is forcing me to learn stuff.
what
Physics
fucking mathematicians
^^^^, ^^^, ^^, ^ That reminds me I haven't finished this thing on Puella Magis and thermodynamics.
This, of course, is abbreviated "F = MA." Now I can't stop seeing ponies in the problems we do.
Really it should be F = ma.
To be fair, what hasn't MLP ruined at this point?
^^ Pompeii?
^ They used to wonder what volc ash could be...
...then they took an eruption to the city.
Now THAT joke is ruining everything.
On swords:
Firstly, have a shitty graphic.
(a)====||_(b1)__(b2)__________________>
What I'm representing (horribly) here is a longsword, which is about 30cm (approx. 1 foot) of hilt and about 90cm (approx. 3 feet) of blade length.
(a) represents the pommel, which is one of the balance points of the weapon.
(b1) represents the minimum ideal balance point of the blade itself, essentially the heaviest point. This can be as little as 5cm (approx. 2 inches) from the hilt.
(b2) represents the maximum ideal balance point, no more than about 25cm (approx. 10 inches) up from the hilt.
Swords usually have a balance point much closer to b1 than b2, and only ever have b1 or b2. a is usually lighter than b.
The true ideal balance point varies from sword to sword and person to person, but is generally within that range.
With one's hand or hands on the hilt, the sword is swung with the hand(s) between a and b. It's a fulcrum action, rotating around those points of balance. The upper portion of the blade is therefore set in motion by proxy of the work done between the balance points. Taking a sword as two fixed beams is therefore accurate, but not entirely specific enough. The energy provided by the movement of the human hand and the natural fulcrum motion of the balance points powers the end, striking portion of the blade.
One-handed swords and two-handed swords are slightly different in this manner because a two-handed sword is used by pulling the pommel while pushing just below the crossguard. The natural counterbalance of the pommel is, in this case, less important than the pulling force of its wielder. That said, exactly the same concept applies -- moving two points of balance around one another so that the sword "self generates" force to contribute to the technique. A second hand is better than a heavier pommel, though, increasing acceleration, speed and force.
If you look at pictures of historical medieval swords, you might note one particularly interesting difference. One-handed swords tend to have "wheel" pommels -- essentially thick discs. Longswords, on the other hand, tend to have oval, pear or other shapes more well-suited to being held in the human hand. Earlier longswords do have wheel pommels, but they're primitive two-handed swords, primarily for this very reason. Wheel pommel longswords fall out of favour some time in the 14th century and never re-emerge. Single-handed swords, on the other hand, continue to have wheel pommels until the cut-and-thrust styles of the Renaissance gain popularity.
One that note, the Renaissance eludes me in some ways when it comes to swordsmanship. Some things don't make much sense. Why were longswords abandoned around 1550 in favour of less sturdy, more demanding forms? Why were wheel pommels abandoned in favour of heavier, nearly unpommelled swords? As much as we see the Renaissance as a time of advancement in all things, I can't help but feel it was the beginning of the end for swordsmanship. For what my observations are worth, I'd say the period from the late Middle Ages to the early Renaissance is the pinnacle of fencing in Europe.