TL3+1 Mechanical Artillery

[Icelander]

Originally Posted by Ze'Manel Cunha

Kinetic Energy is 0.5 * mass * velocity squared

So your velocity change by reducing the weight of throwing arm will be:

Velocity change = square root (2 * KE * Y/X) - square root (2 * KE * 1/X)

Or it's just easier when you know what X and Y are, so lets call it an even 100 pounds, 38.6kg arm plus 6.8kg shot for X, and a 30% weight reduction in the arm to 75 pounds, 34.1kg for Y.

KEx = 0.5 * 45.4 * (110)squared = 274,670 joules

KEy = 274,670 joules = 0.5 * 34.1 * v(squared) -> v = 127 m/s

So by reducing arm weight by 30%, (total arm & missile weight reduction of 25%), you get a speed increase of 15%.

Doesn't it make any difference that the 15 lb. rock missile is at the end of the lever? And that the lever might not be equally thick everywhere?

I mean, you're only accelerating the end of it to the maximum speed, surely? The other end hardly moves.

[Ze'Manel Cunha]

Originally Posted by Icelander

Doesn't it make any difference that the 15 lb. rock missile is at the end of the lever? And that the lever might not be equally thick everywhere?

I mean, you're only accelerating the end of it to the maximum speed, surely? The other end hardly moves.

That's the ke at the point of release, it's easier to measure than the forces involved, otherwise you start getting into design decisions and whatnot...

[Anthony]

Originally Posted by Icelander

Doesn't it make any difference that the 15 lb. rock missile is at the end of the lever? And that the lever might not be equally thick everywhere?

Yes. The efficiency (based solely on arm weight) will be (moment of inertia of rock) / (moment of inertia of rock + moment of inertia of arm).

The moment of inertia of an object at the end of an arm is mass*length^2. The moment of inertia of a fixed width arm is mass/3*length^2. The moment of inertia of various tapering shapes generally requires calculus, though experimentation should be able to optimize.

[Icelander]

Originally Posted by Anthony

Yes. The efficiency (based solely on arm weight) will be (moment of inertia of rock) / (moment of inertia of rock + moment of inertia of arm).

The moment of inertia of an object at the end of an arm is mass*length^2. The moment of inertia of a fixed width arm is mass/3*length^2. The moment of inertia of various tapering shapes generally requires calculus, though experimentation should be able to optimize.

How much do you reckon the arm will weigh?

Do you have an opinion on the 'strength' of various materials when compared to wood here? Is there a better material for this job at TL4 than wood?

[Anthony]

Originally Posted by Icelander

How much do you reckon the arm will weigh?

Ugh. That would require a bunch of math. I will say that your best bet is an arm that is variable in length, probably wood with rope at the end. I would also note that this device will not be meaningfully smaller than a trebuchet of equivalent power, and a modern floating-arm trebuchet (which really needs good quality wheels and axles) is about as efficient as you're going to get out of mechanical artillery.

Originally Posted by Icelander

Do you have an opinion on the 'strength' of various materials when compared to wood here? Is there a better material for this job at TL4 than wood?

Probably iron or brass axles (you need smooth, fairly thick, round rods; not sure if iron can do that at TL 4), wooden structure and arms. You'll need late TL 5-6 high strength steels to really do better than wood.

[teviet]

Originally Posted by Icelander

Good, thanks. Would the same proportion hold true for other ranges between 550-700 or would there be a major difference?

It won't be the same proportion. For short ranges speeds are low and drag is minimal, so the two projectiles will have more similar performance. For longer ranges the difference in drag forces will dominate. This is something that requires a numerical integration.

If the granite sphere has a maximum range of 550m, that means launching at 91m/s at 42 degrees. Firing the lead ball at the same speed at 45 degrees will carry it 830m. That's only a factor 1.5 increase rather than 1.7 earlier.

For further comparison, at ranges less than 100m there is less than a 10% difference between stone and lead balls. At the other end, to throw the granite sphere 1500m requires a launch at 240m/s at 35 degrees; a lead ball thrown at that speed at 45 degrees will go over 5000m.

Originally Posted by Icelander

Now, how much of an increase in penetration are we looking at here?

The projectile, at any range beyond point blank, will be travelling slightly faster. Also, the impact surface is much smaller.

Does anyone know how to figure the damage increase?

Maybe one of the gun-fu gamers would know? GURPS has an interesting hybrid system for dealing with penetration and wounding. A larger projectile should either do less damage with a larger wounding modifier, or do more damage but with a DR multiplier. GURPS seems to allow either or both of these mechanics to be used in any situation; I'm not sure which one would be appropriate. In either case, the ratio of wounding to penetration should scale as the diameter of the projectile, I think.

TeV

[Anthony]

Originally Posted by Icelander

Does anyone know how to figure the damage increase?

Answer unclear, ask again later? The formula for guns seems to have wound up being something like sqrt( ke/diameter ), meaning it will scale with the 1/6 power of projectile density. That works out to about +26% for lead, +19% for iron, -25% for wood (density 0.5).

[Icelander]

Originally Posted by Anthony

Ugh. That would require a bunch of math. I will say that your best bet is an arm that is variable in length, probably wood with rope at the end. I would also note that this device will not be meaningfully smaller than a trebuchet of equivalent power, and a modern floating-arm trebuchet (which really needs good quality wheels and axles) is about as efficient as you're going to get out of mechanical artillery.

TL2 (late) onagers have a short rope at the end on which the sling pouch swings. According to a reenactor this adds about +1/3 to the launch energy.

Originally Posted by Anthony

Probably iron or brass axles (you need smooth, fairly thick, round rods; not sure if iron can do that at TL 4), wooden structure and arms. You'll need late TL 5-6 high strength steels to really do better than wood.

Venture a guess about how much of an improvement TL4 axles are over TL2 ones here?

[Icelander]

If we imagine that TL3+1 stone throwers were used to shoot fantasy equivalents to artillery shells, i.e. spheres with magical or alchemical explosive or incendiary substances use magical fuses, what is the practical limit on how much of the shell weight needs to be devoted to the structural integrity of the projectile and how much can be explosive?

Could TL4 technology make a 5 lbs. hollow metal sphere with very thin walls that could contain 10 lbs. of explosive and still tolerate the stresses of being flung from a monakon that could range 800 yards or so?

[Ze'Manel Cunha]

Originally Posted by Icelander

If we imagine that TL3+1 stone throwers were used to shoot fantasy equivalents to artillery shells, i.e. spheres with magical or alchemical explosive or incendiary substances use magical fuses, what is the practical limit on how much of the shell weight needs to be devoted to the structural integrity of the projectile and how much can be explosive?

Could TL4 technology make a 5 lbs. hollow metal sphere with very thin walls that could contain 10 lbs. of explosive and still tolerate the stresses of being flung from a monakon that could range 800 yards or so?

Probably better off with ceramics.