Saturday, October 17, 2009


I’m learning as I go.

Let’s throw out some of what’s been said up to now and let’s simplify some of the numbers where we can. Still using newtons for force, let’s designate a long bow as equal to 1 newton, a ballista as equal to 20 newtons and a catapult equal to 100 newtons – but don’t worry, I’m about to add a few wrinkles to those numbers.

Before we begin, however, I want to make this perfectly clear: I really do not care if this post meets the requirements of physics – my goal is not to pass a university examination, it is to produce a fairly believable, accessible set of rules for game play. In order to do so, I will have to bend the laws of momentum, with the caveat that in some instances the missile would actually have more momentum than what is assessed and in other cases less. We’ll just say that in the long run, it evens out.

For the calculations that follow, I’m going to simply ignore penetration. I can’t be bothered to take the math course necessary to do the calculations ... but you are welcome to take the general concept described below and complicate it however you desire. The principles should work for whatever cranks your brain fluid.

All right, the ballista. Let’s say that any target that can be hit within the period of 1 second will be hit with the full power of the weapon, that being 20 newtons as I said. And let’s incorporate Chris’ point from two posts ago and note that within 2 seconds, the momentum will be less – one half. If the initial velocity is 65.48 m/s, then at the end of 2 seconds we can describe that velocity as equal to 32.74 m/s. But what about the time between 1 and 2 seconds?

We can’t be accurate in any case unless we calculate for every possible distance, so let’s say for simplicity’s sake that for any period greater than 1 but less than 2, the velocity will be divided by 1.5 seconds. This changes our missile’s velocity at 1.2 seconds and at 1.8 seconds to 43.65 m/s. Accurate? No. But as I say, that’s not my concern.

Reducing the momentum also reduces the force of the ballista’s bolt, to the same degree: 20/1.5 = 14 newtons (it’s 13.33 actually, and a repeating fraction, but let’s not quibble; we want an even number).

Following this forward, and rounding it to the nearest whole number, the bolt’s force between 2 and 3 seconds would be 20/2.5 = 8 newtons, and between 3 and 4 seconds would be 20/3.5 = 6 newtons. These are fairly easy numbers to work with.

Given that 1 newton = 1 long bow, then we can define 1 newton as 1d6 damage (average 3.5). To get a wider average, we can define 2 newtons as 1d12 damage (average 6.5). Therefore, 20 newtons would be 10d12 damage. 14 newtons could be 7d12 damage, 8 newtons 4d12 damage and 6 newtons 3d12 damage. Simple? You bet.

So let’s fire a ballista at an opponent 60 hexes away. I define hexes as 5’ in diameter, so that’s 300 feet.

In the first second, the ballista bolt travels 215 feet. Within 2 seconds, the total is 358 feet (remember, the projectile is slowing). Thus, our target, if hit, suffers 7-84 damage. Isn’t that easy?

It is, so let’s throw in a wrench or two.

(By the way, before I forget, you can’t use the slowing momentum described above to determine the maximum range of the weapon. There are variables, like wind resistance, I’m not taking into account. I told you, I’m not a physicist. Sometimes my creativity fails to recognize my limitations. Aristotle would understand)

I said yesterday that it would be wildly difficult to hit a moving individual with a siege weapon – I would think something equivalent to hitting AC -10. But I think we can do something about that (it is at this point, definitely, that I throw penetration considerations out the window - shoot me).

The bolt fired by a ballista weighs, as I said, 0.325 kg, or slightly less than a pound. This would be a springy shaft about 1.5 times as long and four times as thick as an ordinary arrow. I would like to treat it as something that could potentially bounce, and even splinter. This would enable us to treat the bolt within the rules of a ‘grenade-type’ missile - meaning that it would be okay to miss, as long as you got close.

The rules under grenade-like missiles reads (bottom right hand corner of p. 64, DMG), “If the ‘to hit’ die roll indicates a miss, roll 1d6 and 1d8. The d6 indicates the distance in feet the missile was off target ... the d8 indicates the direction in which the distance in feet of the miss is measured ...”

Given that we’re talking a greater range of fire than a hand-held missile, I think we can stipulate that the d6 is the distance in hexes, not feet. But okay, we have something we can work with here.

Let us presuppose that our artillerist, Jeremy, needs a 23 on a d20 to hit Grunk in the front row of his armed force, and that he fully expects to miss. However, Jeremy isn’t concerned with that. Once he’s missed, he has a 1 in 8 chance of dropping his shot right in front of Grunk ... and that given the momentum of the missile, it’s going to bounce or splinter with a low trajectory, still causing damage.

How much damage? Let’s take make an ad hoc assumption on that, and say that the momentum of the missile is reduced by 75% ... dividing the number of newtons by 4. If Jeremy fires at Grunk at a distance of 60 hexes, the number of newtons would then be reduced from 14 to 4 (remember, we’re rounding to whole numbers). Grunk would still suffer 2d12 damage.

But we can do better than that.

When the missile splinters, it need not continue in a straight path. The largest and most deadly piece might spin off in a modified direction ... in D&D this is usually expressed as within a 60-degree angle. Consider the following picture.

Grunk is depicted here as within the 60 degree shadow of the missile, but not on the central trajectory. Suppose that, for every hex removed from that trajectory, we divide the force again – in this case, Grunk is two places removed, and the number of newtons is again divided by 4. Now Grunk need suffer only 1d6 damage ... but at least the ballista is not wasted. Note that under grenade missiles, if you are within a certain range of the missile, damage is automatic - you do not need ‘to roll’ to hit again

If Grunk had a couple of pals who were also within the shrapnel arc, the DM might randomly determine who gets hit by the missile ... or potentially, if more than one individual is hit. It might be suggested that a ballistae missile can only reasonably hit one person ... but a catapult stone will create additional fragments from the ground where it hits – splitting trees and throwing up gravel, depending on the location. More than one person might be hit therefore.

It occurs to me that Jeremy, our artillerist, would be wise to aim his shot in front of Grunk, rather than at Grunk, and thereby reducing the chance of the missile going long. But not too far in front ... there is a sweet spot, where the likelihood is best for some damage, if not maximum. Plus, there’s always a chance that the missed missile will land right in Grunk’s hex. Smoosh!

Get enough siege weapons firing into a massed crowd using these rules, you could do some serious damage. It would be particularly effect on board ship. It was working out shipboard combats that started me thinking along the lines of siege weapons in the first place. That, and mass combat rules – another bugbear of the game, eternally unsolved, eternally needed.

Well, please feel free to cut my math to pieces. It could do with a good chopping. I’m always learning, after all. I’m sure I made lots of mistakes ... I’ve only read it over once.

In the meantime, I shall turn my mind to the question of damaging fortifications – along with destroying ships, other siege weapons and so on. Until then.

No comments: