So, BattleTech has a lot of stuff happening in it and there have been occasional disputes how much punishment a BattleMech can actually take were it to ever be confronted with modern weaponry. So I've taken the liberty of calculating the effectiveness of the armor using the BattleTech Machine Gun (for which I will be using BTMG as a shorthand) as reference and comparing it to the M2 Browning heavy machine gun, as for this I will assuming that the BTMG uses 12.5 mm anti-materiel ammunition.
First off, let's take the basic stats of the BT Machine Gun we'll be needing:
- 2 damage per turn
- 200 units of ammo per ton
- Each unit consists of x amount of rounds expended over 10 seconds
First, we have to specify the type of ammunition used for our calculation. For this I have decided to go with the M903 SLAP (Saboted Light Armor Penetrator) rounds as specified here. I chose them for being specifically designed to be used against light armored vehicles, attack helicopters and their high muzzle-velocity. The relevant data from the spreadsheet is as follows:
- Total weight: 94.99 g
- Projectile weight: 23.33 g
- Projectile velocity: 1,219.2 m/s
First we will need to determine how many rounds can be carried in one ton. For this we can simply convert 1 t into 1,000,000 g.
1,000,000 g / 94.99 g = 10,527.42
For the ease of the following calculations (and a speck of realism) I will round that number down to the next largest multiple of 200, where the lost weight can be attributed to ammunition belt and feeding mechanisms, which provides us with 10,400 rounds of ammunition.
From this we can determine the rate of fire by congratulating how many rounds are expended as part of one unit of ammunition over 10 seconds.
10,400 rounds / 200 units = 52 rounds/unit
Multiplying that value by 6 we get an effective rate of fire of 312 rounds per minute. This is a comparatively low rate of fire to the 450-600 rpm of an M2HB manned by infantry or mounted on armored vehicles, but this can have some practical reasons. Machine guns operating at a high rate of fire cause the barrel to become very hot which can lead to deformation. Mounted on a BattleMech, it is difficult to exchange the barrels quickly without the use of vulnerable mechanical equipment which through damage or inaccuracy could render the entire weapon ineffective or useless. A lower rate of fire significantly increases barrel life during sustained fire and may have been a deliberate design choice by manufacturers.
Moving on, we will need the kinetic energy delivered by a single round. With the documented weight and velocity of a M903 SLAP projectile we can calculate the potential energy in Joules.
Joules are calculated in weight in kg times velocity in m/s squared.
(94.99 g / 1000) * (1,219.2 m/s)² = 34,678.85 J
Knowing the energy of one projectile, we can calculate the energy of a full burst over 10 seconds.
34,678.85 J * 52 = 1,803,300.2 J
This is a good start, but we have nothing to compare it against yet. First, we will have to calculate the effective armor strength. In BT, armor is distributed in half ton packages consisting of 8 points each. From this we can determine the weight of one point of armor.
500 kg / 8 points = 62.5 kg/point
Since we now know the weight of each point and the energy of a BTMG burst, we can calculate how much energy it takes to penetrate a single kg of armor.
1,803,300.2 J / (62.5 kg * 2) = 14,426.4 J/kg
Using this knowledge we can figure out how much armor a different projectile could effectively penetrate, like a tank shell.
The Rheinmetall L55 of the Leopard 2 is used used on most current NATO main battle tanks under various names such as the M256 mounted on the M1A2 Abrahams. As projectile I will be using the DM53 120mm KE Projectile as documented here with the following specifications:
- Total Weight: 21.4kg
- Projectile weight: 8.35 kg (with sabot)
- Velocity: 1,750 m/s
Unfortunately it is very difficult to tell the weight of the sabot from a projectile, so for the sake of discussion, I will use the same weight ratio of the projectile to the total weight as the M903 SLAP. This will likely be rather inaccurate, but perhaps more truthful than including the entire weight of the sabot for calculating energy.
21.4 kg / (94.99 g / 23.33 g) = 5.26 kg
Using that weight, we can now determine the energy of the projectile. Using the same formula as above.
5.26 kg * (1,750 m/s)² = 16,108,750 J
From here we can see how much armor this projectile can penetrate through.
16,108,750 J / 14,426.4 J/kg = 1116.62 kg
And to compare in a more easily understood manner:
1116.62 kg / 62.5 kg/point = 17.87 points
For rate of fire, I couldn't seem to find documented estimates, so I'll use the 10 rounds per minute which have been common with previous-generation cannons. Skipping that calculation, 10 rounds per minute equate to 1,67 rounds per 10 seconds.
17.87 points * 1.67 rounds/turn = 29.84 points/turn
In other words, an L55 would equate to an AC30 with about 28 units of ammo per ton, but everything considered this is a quite formidable result for mechanics which were cooked up in the 80s.
I'd really love to know how exactly you can get a single M2HB to weight almost as much as a small car though.
Edited by SethAbercromby, 06 August 2016 - 05:52 AM.