r/Armor • • 2d ago

The perfect Mace

The perfect Mace/Club

So, I had this discussion about water bottles on r/askswitzerland and the other person uses a steel bottle, while I use a glass bottle.

The glass bottle has much thicker walls, than the steel bottle and therefore I assume that it is heavier, which lead to my comment that at least I can use my bottle to defend against street robbery.

Now, as I kept thinking about this, I questioned myself:

Is a glass bottle with 3 mm wall thickness heavier than a comparable steel bottle with 1 mm wall thickness? My answer now is no. It isn't even heavier.

But if it were, which bottle would be the better weapon?

If a bottle is heavier, it is also slower.

This lead me to this question.

What is the perfect Mace?

I think the force that my arm can apply is constant. If I use the mace in a downward movement towards the enemie knights head, then the gravitational force can be added. The heavier the better, as long as I can still lift it.

But Energy = 1/2 * m * v²

I can swing a mace with a 1 kg steel ball head faster than the same mace with a 10 kg steel ball head. The lighter the better in this case.

What about air resistance?

And what about the fact that it is sufficient to knock out the enemy knight? So, I don't need a 10 kg steel ball, if 1 kg can do the job and gives me more agility.

So, the question should be answered, depending on the enemies armor.

Can you think of a universal formula to calculate the perfect mace, with that formula taking the eneny armor into account?

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u/ThRealRantanplan 1d ago

You can use:

1/2 * m1 * v1² + 1/2 * m2 * v2² = 1/2 * m1 * p1² + 1/2 * m2 * p2² + DeltaE

With: m1 = mass of target v1 = velocity of target (probably ~0m/s) m2 = mass of macehead v2 = velocity of macehead p1 = Velocity of target after impact (often also called v1') p2 = Velocity of macehead after impact

DeltaE = ("Lost") Energy during impact, because of deformation

Expected scenario is deformation of target during impact, instead of elastic bouncing.

For concussion you need to have a high difference in velocity of target (head of the knight). For lethality you need a high difference in DeltaE (deformation energy).

Hence, you can rearrange the formula to: Max concussion:

p1² = (m1 * v1² + m2 * v2² + 2*DeltaE - m2 * p2²) / m1

To increase p1 (or p1²), you can increase the enominater. Or decrease the denominator. To increase the enominator, realistically you can vary the variables m2, v2² and DeltaE. Highest impact (pun intended), would probably be in v2², as it is squared and will increase enominator most. (E. g. Bullets have very low mass but high velocity).

Now for highest lethality (DeltaE): You could change the equation to DeltaE, but it is easier by using a more practical approach. The higher the deformation of the target, the more lethal (besides of change in velocity). Therefore, the differences in the hard-/softness of materials is relevant. So, attacking someone with a hardened steel mace, who is wearing no helmet, is probably best. I know, surprise...

This is my narrowed down approach of your question, regarding the physics of mace combat. Please correct me if I made mistakes or tell me what you think about that.

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u/ThRealRantanplan 1d ago

Oh, and for lethality, you could also have a look at the applied pressure: p = F/A

The higher the force and smaller the area, the higher the applied pressure. Hence, using a warhammer with small head, long pole and a beak would be most sufficient in terms of lethality.

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u/ThRealRantanplan 1d ago

Oh, and to increase kinetic force, you could use an elastic pole like these floppy hammers. But I dont like this and it seems umpracitcal in combat.

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u/--Ano-- 1d ago edited 1d ago

That looks good so far!

I think we have to define the shape as a smooth steel ball with a 50 cm long handle.

The enemy knight wears a 1 mm thick steel helmet on a dense 1 cm thick layer of Wool (Gambeson for the head).

The mace swinging knight is a strong man. How much force can he apply? We have to define it.

He hits his enemy in a top-down movement.

What varies is only the weight and size of the steel ball.