Reply To: Bows and archery

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#2439
Thaeris
Participant

Well, to start, if you just want to make things easy, here is the old discussion:

http://codex.masterplanfoundation.com/viewtopic.php?f=5&t=641&sid=975141cdcf0f0761b26c2a1c81ee6545&start=30

…Without making things easy, one of the first things you can make note of about a bolt – almost any bolt, regardless of the style of crossbow – is that they are shorter than arrows in general. Some may state that this potentially makes them less aerodynamically efficient, which is a statement I’d need to analyze further, but it also makes them stronger, so long as the diameter of the missile meets or exceeds that of a comparative arrow. My first argument in this regard is going to be in regards to something called the Slenderness Ratio, which you can read about here:

https://en.wikipedia.org/wiki/Slenderness_ratio

Regarding actual numbers for strength, the article does mention the Modulus of Elasticity, which would be covered in a separate set of equations. We’re not there, so there is no need to worry about that just yet.

So, here we are so far: Other articles I refreshed my mind with used the Greek Lambda as the variable to denote the Slenderness Ratio. So we’ll go with that. Your equation so far is thus:

[Lambda] = l / k

Refer back to the article, and it tells you how to calculate k:

k = (I / A)^0.5,

…Or the square root of I / A. Note that I’ve re-written the article’s formula because I’m actually solving for k here, and not just writing an equation a certain way for convenience.

A is the cross-section area of the bolt shaft. Yes, I do believe some of those buggers taper, which would make computations a bit more complex, but we will worry about that later of the conversation ever gets to that place (I actually do need a refresher on some more complex calcs like that, to be honest). Area of a circle is pretty easy, of course:

A = [Pi] * [r^2]

…Where Pi ~ 3.14 and r is the radius (half the diameter) of the bolt shaft.

I is the Second Moment of Area of the shape in question, which is of course the circular cross-section of the bolt. Use Ix or Iy from the circle formula on this page here:

https://en.wikipedia.org/wiki/List_of_second_moments_of_area

In general, you use the second moment of area to evaluate how strong a structure is in relation to its cross-section in relation to a given direction. A circle is the same in all directions, and so that is why Ix and Iy are the same.

…So, you calculate all this stuff, kind of going in reverse-order from reading at this point. Eventually you can calculate that [Lambda] = l / k noted from earlier, etc., etc.

And what’s the point? Bolts will have a lower slenderness ratio than arrows, which makes them tougher given that both missiles are made of the same materials.

AND THEN, the final point is this: Why didn’t Europeans just make Chinese-style crossbows with wide, bow-like prods with long power-strokes? They could have. Maybe they wanted the bolts short and stocky explicitly for the purpose of dealing with increasingly heavy armor found on European battlefields…