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TheDark October 25th, 2017 10:11 PM

Pre-Gunpowder Artillery
 
Has anyone done a design sequence for pre-gunpowder artillery, such as catapult or ballista? I know WTH has the bow/crossbow design sequence, but it doesn't scale up to artillery sizes.

atpollard October 26th, 2017 08:19 AM

Now there is an interesting problem!

So which are you interested in?
Stored tension like a giant crossbow or a counterbalance lever like a catapult?
They will have very different design physics operating.

The Catapult should scale larger.

After release, ballistics are ballistics. A cannon ball, a musket ball and a stone bow do not care about how it was accelerated. Mass and velocity are all that matter. They will determine range and damage.

atpollard October 26th, 2017 03:35 PM

Hey Robject, I told you I was a gearhead!

So using data from Vitruvius, historical recreations and FF&S I have compiled some rules for creating Roman Artillery for Traveller. Vitruvius design calculations, recreations and historic records indicate the torsion weapons hurled stones or spears at about 50 m/s to a range of about 300 meters. I have avoided historical names since they changed significantly over time, but the weapons remained fairly similar from several centuries BC through most of the middle ages.

The torsion (twisted rope) engines with two arms were the most accurate and had greater range and accuracy than a TL 3 Musket, so I have focused on them.

Roman Torsion Stone Throwing Artillery Creation Guidelines

I. Ammunition (Iron or Stone ball)
1. Select the weight of the Ammo to be thrown (Wa) = 0.3 minimum to 163 maximum = Kilograms
  • For Bolts: Length of Bolt = 1.169 x (Wa)^0.33 = Meters
  • For Iron Balls: Diameter of Ball = (Wa/4000)^0.33 = Meters
  • For Stone Balls: Diameter of Ball = (Wa/1500)^0.33 = Meters
2. Cost of ammo = Wa x 5 = Credits

II. Vitruvius Torsion Engine
3. Energy (E) = Wa x 1250 = Joules
4. Damage (D) = (E^0.5) / 15 = number of D6
5. Penetration (P): Calculate Penetration based on (E) using the table for Rifle Bullets in FF&S.
  • At Short Range: If E<2000, then Pen = 'nil';
  • If 2000<E<3000, then Pen = D
  • If 3000<E, then Pen = D/2
6. Required Strength (S) = E/60
7. Required Crew = S/7 = number of men required to achieve a reload rate of "5".
  • Pulleys and levers can be used to trade mechanical advantage for time so a half crew has a reload rate of "10",
  • a third of a crew has a reload rate of "15",
  • a quarter of a crew has a reload rate of "20"
  • a fifth of a crew has a reload rate of "25".

III. TL 1 Composite Torsion Engine
8. Weight of Engine and Carriage (We) = [EDIT] Wa x 500 = Kilograms
9. Cost of Engine = We x 40 = Credits
10. Span of arms = Length of shaft = [EDIT] = 2 x (Wa)^0.33 = Meters
  • Span and Length are provided for reference only to help visualize the weapon.
11. Special Note: For a TL 2 Steel Torsion Engine, the weight (We) remains unchanged from the TL 1 version, but the Cost and Span/Length are halved.

IV. Short Range
12. Muzzle Velocity (MV) = (2x E/Wa)^0.5 = m/s [typically 50 m/s]
13. Short Range (SR) = (E / Wa)^0.5 = Meters [typically 35 meters]

All of the Roman Torsion Engines are designed with a velocity of 50 m/s and short range of just over 35 meters with the selected ammo, however they are all capable of firing smaller stones at greater ranges and velocities. One quarter the design weight seems to be a practical lower limit for the projectile (the forces on the weapon can damage the frame from too great an impact.)


DESIGN EXAMPLE
So just for fun, let's build a Vitruvian Torsion Stone Thrower ... to hurl a 10 kg ball (equal to a 22 lb cannon ball):

1. (Wa) = 10 kg
  • For Bolts: Length of Bolt = 1.169 x (10)^0.33 = 2.5 meters
  • For Iron Balls: Diameter of Ball = (10/4000)^0.33 = 0.138 meters = 138 mm
  • For Stone Balls: Diameter of Ball = (10/1500)^0.33 = 0.191 meters = 191 mm
2. Cost of ammo = 10 x 5 = Cr 50
3. (E) = 10 x 1250 = 12,500 Joules
4. (D) = (12,500^0.5) / 15 = 7.45 D6
5. (P) value = 2 - 4 - 6
At Short Range: Pen = 7.45/2 = 3.7
6. (S) = 12,500/60 = 208
7. Crew = 208/7 = 30 men @ reload rate "5"; 15 men @ reload rate "10"; 10 men @ reload rate "15", 8 men @ reload rate of "20" and 6 men @ reload rate "25".
8. (We) = [EDIT] 10 x 500 = 5000 Kilograms
9. Cost of Engine = [EDIT] 5000 x 40 = Cr 200,000
10. Span of arms = Length of shaft = [EDIT] 2 x (10)^0.33 = 4.3 meters
11. TL 2 Steel: Cost = Cr 20,000 & Span/Length = 12.5 meters

10 KG BALL (22 pounder)
(MV) = (2x 12,500/10)^0.5 = 50 m/s
(SR) = (12,500 / 10)^0.5 = 35 meters
Extreme Range = 8x SR = 280 meters

9 KG BALL (20 pounder)
(MV) = (2x 12,500/9)^0.5 = 53 m/s
(SR) = (12,500 / 9)^0.5 = 37 meters
Extreme Range = 8x SR = 296 meters

8 KG BALL (18 pounder)
(MV) = (2x 12,500/8)^0.5 = 56 m/s
(SR) = (12,500 / 8)^0.5 = 40 meters
Extreme Range = 8x SR = 320 meters

7 KG BALL (15 pounder)
(MV) = (2x 12,500/7)^0.5 = 60 m/s
(SR) = (12,500 / 7)^0.5 = 42 meters
Extreme Range = 8x SR = 336 meters

6 KG BALL (13 pounder)
(MV) = (2x 12,500/6)^0.5 = 65 m/s
(SR) = (12,500 / 6)^0.5 = 46 meters
Extreme Range = 8x SR = 368 meters

5 KG BALL (11 pounder)
(MV) = (2x 12,500/5)^0.5 = 71 m/s
(SR) = (12,500 / 5)^0.5 = 50 meters
Extreme Range = 8x SR = 400 meters

4 KG BALL (9 pounder)
(MV) = (2x 12,500/4)^0.5 = 79 m/s
(SR) = (12,500 / 4)^0.5 = 56 meters
Extreme Range = 8x SR = 448 meters

3 KG BALL (7 pounder)
(MV) = (2x 12,500/3)^0.5 = 91 m/s
(SR) = (12,500 / 3)^0.5 = 65 meters
Extreme Range = 8x SR = 520 meters

Condottiere October 26th, 2017 03:59 PM

According to Reddit, trebuchets are superior.

atpollard October 26th, 2017 04:22 PM

Quote:

Originally Posted by Condottiere (Post 576053)
According to Reddit, trebuchets are superior.

For range, I read trebuchets were FAR superior. For accuracy, I read that the two-armed torsion engines exceeded even 18th century muskets and some were used by Roman snipers to shoot commanders off enemy walls.

So for modeling with FF&S, the Torsion 'Ballista' seemed closer to the Heavy Crossbow and the Trebuchets seemed closer to FF&S Mortars and Howitzers.

YMMV.

What can you find on Trebuchets to give us some real historic data points to calibrate FF&S constants against?

Condottiere October 26th, 2017 04:44 PM

Stand off range becomes important during artillery duels.

And since for the larger artillery pieces, on land they tend to be used against non moving objects, id est walls and towers, aiming can be adjusted.

https://youtu.be/rAdC2K8-E4U

Take Edward Longshanks' War wolf, who at one one point refused the surrender of one castle in order to demonstrate it's power.

"Aye didnae hae it set up fir nuthin'."

pendragonman October 26th, 2017 06:54 PM

For steps that determine number of dice of damage you should probably indicate the rounding convention you prefer employed. Also, an armature span of 25 meters seems a bit extreme don't you think?

atpollard October 26th, 2017 07:17 PM

Quote:

Originally Posted by pendragonman (Post 576065)
For steps that determine number of dice of damage you should probably indicate the rounding convention you prefer employed. Also, an armature span of 25 meters seems a bit extreme don't you think?

What is the span for a 12,500 joule Composite crossbow (per WTH)?

What was the span for any real world "ballista" capable of firing a 10 kg projectile hundreds of meters?

Find data that suggests it is wrong and I will gladly change the constants.

TheDark October 26th, 2017 07:57 PM

Quote:

Originally Posted by atpollard (Post 576048)
Hey Robject, I told you I was a gearhead!

So using data from Vitruvius, historical recreations and FF&S I have compiled some rules for creating Roman Artillery for Traveller. Vitruvius design calculations, recreations and historic records indicate the torsion weapons hurled stones or spears at about 50 m/s to a range of about 300 meters. I have avoided historical names since they changed significantly over time, but the weapons remained fairly similar from several centuries BC through most of the middle ages.

Thanks, this is a great starting point and everything seems to make sense.

If you're open to a few revisions, what got me thinking about this was reading Osprey's Greek & Roman Artillery and trying to work out some of the math on my own, and I compared a couple numbers to what was mentioned in that book.

The arm length seems very large - 25 meters for a 10 kg thrower. A Greek 1-talent projector had a span of slightly over 5 meters for a projectile massing 26.2 kilograms. Vitruvius gives a moderately complicated formula for the washer diameter for stone throwers - in daktyls of 19.3mm each, the washer diameter is 1.1 times the cube root of 100 times the stone's mass in minas of 436.6 grams. The arm length should be 7 times the washer diameter. Using the 10kg projector:
10 kilograms is 22.9 minas
22.9 * 100 = 2290
2290^(1/3) = 13.1809
13.1809*1.1 = 14.5 (14.49899, I'm rounding because it's so small a difference)
Washer diameter = 279.85 mm, arm length = 1,958.95 mm, or 1.96 meters per arm, or 3.92 meters for total arm length, rather than 25 meters. Off the top of my head, I'm not sure how to simplify that formula to make it easier to work with. For a bolt-thrower, the washer diameter is bolt length divided by 9, so this machine (if refitted as a bolt-thrower) could fire a 2.5 meter bolt. This means the math for bolt-throwers is easier, since arm length is bolt length*.78 (7/9 = .777 repeating).

For weight, Douglas Campbell's work has suggested a ratio of closer to 500:1 than 100:1 for launcher:ammunition.

Going off into speculation, arrow-throwers were much smaller than stone-throwers. A machine capable of firing a 4 foot long arrow was the same size as one capable of firing a 1 kilogram stone. Ammunition to machine size is closer to 200:1 instead of the stone-thrower's 500:1 (possibly because of better aerodynamics on the bolt?).

For Roman machines specifically, archeological finds in Rhodes, Pergamon, Tel Dor, and Carthage have turned up stones of the following masses (in kilograms): 1.3, 2.1, 3.5, 4.4, 5.2, 6.6, 7.9, 8.7, 9.6, 10.9, 13.1, 16.4, 17.5, 21.8, 26.2, 28.4, 30.6, 34.9, 39.3, 43.7, 52.4, 65.5, and 78.6. The 26.2 kilogram (1 talent) seems to have been most common in Roman service.

TheDark October 26th, 2017 08:03 PM

Quote:

Originally Posted by atpollard (Post 576067)
What is the span for a 12,500 joule Composite crossbow (per WTH)?

178.5 meters. Composite is 70 joules per meter. TL 7 Composite Steel would still be 62.5 meters. WTH's numbers are either off or don't scale up at all past personal weapons.


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