Realistic Stronger Box and Crossbow

[AlexanderHowl]

Since bows and crossbows have their own ST ratings, representing the strength of the materials that they are made from, bows and crossbows with higher ST should be costlier and heavier than bows and crossbows with lower ST. In order to represent this realistically, I propose that the cost and weight bows and crossbows should be multiplied by (Actual ST/Minimum ST) squared.

For example, a composite bow possesses a minimum ST 10, costs $900, and weighs 4 lbs. A ST 15 composite bow would cost $2,025 and weigh 9 lbs while a ST 20 composite bow would cost $3,600 and weigh 16 lbs. Each would represent the cost and weight of the stronger materials required to make the bow.

What do you think? Would you want to increase the cost and weight of stronger bows to make them more realistic or would it be needlessly complicated?

[Turhan's Bey Company]

Have you read "The Deadly Spring"?

[DouglasCole]

Someone else will probably jump in, but the stiffness of a beam varies as the cube of it's thickness. So if you double the thickness, it's 8x the pull to first order.

And of course it's not quite that simple due to tillering, and changes in width if you like, etc.

Even so: Doubling the ST score of the bow, quadrupling its draw weight, is only cube root of 4 times thicker: about 60%.

Given that bows tend to be fairly light (a ST 16 yew longbow from the aforementioned Pyramid 3/33 is about 1.9 lbs), going to a ST 32 bow is about 3 lbs. So it's probably a bit more calculation than you need.

[Fred Brackin]

Originally Posted by AlexanderHowl

Since bows and crossbows have their own ST ratings, representing the strength of the materials that they are made from, ?

To my knowledge this statement is not accurate. To make a stronger bow you make it thicker rather than from different materials.

[AlexanderHowl]

Yes, the increase in the strength of the materials would mean thicker materials, since thickness is related to strength. I was unaware of the cube relationship for a beam though. A ST 11 longbow possesses a center thickness of 5/8", so a ST 22 longbow would then presumably possess a center thickness of 1", as it would represent a quadrupling of the draw weight, and it would only increase the cost and weight by 1.6x. The formula would then be the cube root (square [Actual ST/Minimum ST]).

[Anthony]

Originally Posted by DouglasCole

Someone else will probably jump in, but the stiffness of a beam varies as the cube of it's thickness. So if you double the thickness, it's 8x the pull to first order.

And of course it's not quite that simple due to tillering, and changes in width if you like, etc.

Even so: Doubling the ST score of the bow, quadrupling its draw weight, is only cube root of 4 times thicker: about 60%.

Incorrect. While you are correct that stiffness for a beam of a given aspect ratio varies with the cube of thickness, what that's missing is that doing so also halves the maximum safe curvature rate, and thus if you wanted the same draw length you'd need to increase the length of the bow.

In general, elastic potential energy is equal to 0.5 * strain^2 * elastic modulus * volume, and you store maximum energy in an elastic material when strain is equal to yield strain (less some safety margin), thus giving maximum energy storage of 0.5 * yield strain^2 * elastic modulus * volume (alternately, as yield strain = yield strength / elastic modulus, you can write this as 0.5 * yield strength^2 / elastic modulus). Because not all parts of a bow staff are under equal strain, the actual energy that can be stored is 1/3 of this limit for a rectangular cross-section, 1/6 for a circular cross-section.

As all of those terms are constants except volume, this tells us energy storage for a well-constructed bow is linear in total volume of elastic material, and thus linear in mass (I was writing a response to TDS at one point, never got finished with it, but my basic take is that TDS has the wrong amount of math -- it should either assume a competent bowyer and be much simpler, or it should involve calculus).

[Rupert]

Note that while the bow gets bigger and heavier with increased draw, the rest of a crossbow may not, because the stock is probably over-strength anyway.

[Varyon]

Originally Posted by Anthony

Incorrect. While you are correct that stiffness for a beam of a given aspect ratio varies with the cube of thickness, what that's missing is that doing so also halves the maximum safe curvature rate, and thus if you wanted the same draw length you'd need to increase the length of the bow.

Does a typical bow design use the maximum (given some margin) safe draw length possible? If so, getting the same draw length would require increased length of the bow, but if not there's certainly room to play around there.

[DouglasCole]

Originally Posted by Varyon

Does a typical bow design use the maximum (given some margin) safe draw length possible? If so, getting the same draw length would require increased length of the bow, but if not there's certainly room to play around there.

It's going to try to come close, because you want as little moving bits of the bow as possible (which is why long) and you want them to be as light as possible (so properly tillered/tapered).

[Flyndaran]

Originally Posted by Anthony

...

As all of those terms are constants except volume, this tells us energy storage for a well-constructed bow is linear in total volume of elastic material, and thus linear in mass (I was writing a response to TDS at one point, never got finished with it, but my basic take is that TDS has the wrong amount of math -- it should either assume a competent bowyer and be much simpler, or it should involve calculus).

So that would have little real effect for the realistic human strength range. But it would cause a big disagreement with TDS for my low grade but realistic giant humanoids?
And I suppose those odd hobbits and their tiny "weapons"?