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Norfolk nChance

How much Broadside Momentum is delivered...?

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Question:

How much Momentum does a three-deck broadside from a L’Ocean deliver at 100m (109yds)?

Momentum = Mass x Velocity    p=mv

 

This is the gun setup

Lower Deck         16x         36pds    =576 mass

Middle Deck       18x         24pds    =432 mass

Upper Deck        18x         12pds    =216 mass

                                                Total mass = 1,224 stationary

I know the muzzle velocity of a 36pd cannon is 450m/s that’s roughly 1,000mph.

But sort of stuck from here... I was thinking using the same velocity speed... but needs degradation

1,224pds x 1,000mph = 1,224,000pds /2,500 tons =

490 tons momentum is delivered at the muzzle? Is this anywhere close because it seems rather a big punch?

 

 

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Interesting question. Drag is a comlicated subjet, even for relatively simple geometries like spheres.
First off: You jumbled your unit conversions a bit there, momentum is not calculated in tons. When calculating stuff like this it's bes to stick with SI units until calculation is finished and then convert to sth you are more familiar with.

Ordnance mass taken from Wikipedia for French 36pders and British 24s and 12s:

First deck total mass:  m1 = 16*17.6 kg = 281,6 kg

2nd Deck:                       m2 = 18*15,2 kg = 273,6 kg

3rd Deck:                        m3 = 18*5,44 kg = 97,22 kg

Let  muzzle velocity be 450 m/s for lack of a more accuarate value for all guns. Makes the total Impulse P = m*v = (m1+m2+m3)*v = 293,616 kg*m/s or about 294 kN*s. So one could argue the overall force of the broadside applied to a plane right behind the muzzles would be 294 kN which would be the same force of weight that a mass of about 30 tons would create by lying on said surface.

As for the drag (reduction in velocity), that is hard to approximate. As it is dependent on surface area and speed of the object, density and viscosity of the medium (air counts as a fluid here) it travels though and some other stuff. Can't give you a perfect number here (have no time for that :D).

But I'd guesstimate around a 100 yards distance will already lose you at least 20% of velocity, resulting in 20% loss of impulse as well.

If you want to read up on Drag approximation:
https://www.sciencedirect.com/science/article/pii/S2214914717301459

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Does momentum really get delivered in this manner?  because the balls are penetrating at great speed (and passing completely through in many cases), wouldn't a lot of their energy fail to be transferred?

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Thanks Tom,

that's great and highlighting my screw up as well. My 500 tons momentum seemed way too high, why the question.

The 30 tons without any degradation may seem light-ish… But using 24 or 20 tons at 100 yards makes sense. Ships very rarely were completely blown out of the water. The Victory delivery on the Bucentaure must have been powerful and for the French crew horrific

https://en.wikipedia.org/wiki/French_ship_Bucentaure_(1803)

 

Norfolk

 

 

 

https://www.convert-me.com/en/convert/force/kilonewton.html?u=kilonewton&v=294

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Applying high school physics to a complex problem, hmmm.

That formula works only under 'perfect' conditions, no friction, no air resistance, spherical objects, single point of contact etc. It also describes the potential to displace another object, you know like plastic, elastic collisions.

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It looks like mass in motion to me (the ship is moving laterally), and if it hits you in the face and keeps going....

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I asked the question, because my “General” assumption looked wrong. The momentum looked too large, this from me using fingers n toes first before the TI-84 comes out. Doing a little story, I wanted to give the reader a frame of reference to the momentum or force delivered...

“The HMS Indefatigable took the full force of 500 tons the L’Ocean three decker broadside delivered or one and a half 747s.... hmmm...” something looks wrong... No physics needed at this stage...

30 tons or 20 tons  gives a much more accurate reference if not an exact one...

Then again, I too will never be a physicist just a simple humble Investment Banker of the worse kind. 

 

 

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3 minutes ago, Norfolk nChance said:

I asked the question, because my “General” assumption looked wrong. The momentum looked too large, this from me using fingers n toes first before the TI-84 comes out. Doing a little story, I wanted to give the reader a frame of reference to the momentum or force delivered...

 

“The HMS Indefatigable took the full force of 500 tons the L’Ocean three decker broadside delivered or one and a half 747s.... hmmm...” something looks wrong... No physics needed at this stage...

 

 

30 tons or 20 tons  gives a much more accurate reference if not an exact one...

Then again, I too will never be a physicist just a simple humble Investment Banker of the worse kind. 

 

 

 

One of the naval fiction books (I think it was Alan Lewrie) I read compares a broadside from a Gun ketch (below the Rates) as equivalent to a company of Field Artillery.  The broadside of a SOL probably cant be compared to land artillery.  I doubt that Napoleon and Wellington combined (at Waterloo) had the weight of shot that the Victory could deliver.

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2 hours ago, Angus MacDuff said:

Check out the momentum when Iowa lets it loose. 

iowa.png

To be fair...   The ship DOES NOT move laterally...  that is an effect of the shock wave from the muzzle bouncing back against the hull.

The gun has a sophisticated recoil mechanism, like most modern artillery systems.

In the age we're concerned with, that was handled by wheels, lines and tackles....

Edited by Vernon Merrill
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Just now, Vernon Merrill said:

To be fair...   The ship DOES NOT move laterally...  that is an effect of the shock wave from the muzzle bouncing back against the hull.

The gun has a sophisticated recoil mechanism, like most modern artillery systems.

One of the Iowas did a shoot at Vieques in the 80's and while I wasn't there my buddy was and had binoculars on it for the full broadside.  He said it did in fact move sideways.  I can not speak to his level of exaggeration unfortunately.  Look right forward at the bow wave. 

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1 hour ago, Vernon Merrill said:

In the age we're concerned with, that was handled by wheels, lines and tackles....

...and most important, rolling fire :) 

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6 hours ago, Angus MacDuff said:

One of the Iowas did a shoot at Vieques in the 80's and while I wasn't there my buddy was and had binoculars on it for the full broadside.  He said it did in fact move sideways.  I can not speak to his level of exaggeration unfortunately.  Look right forward at the bow wave. 

If you jump the planet earth moves. He is couldn't have been lying. It is impossible the the ship didn't react to the shooting. The heaviest oil tanker also moves to the side if you fire a 9mm pistol from the side of it. 

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The ship weighs too much, the forces exerted by the water add to this effect. Same pic, there is very little movement of the ship itself. No waves from lateral movement, just a blast depression.

WNUS_16-50_mk7_Iowa_pic.jpg

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I read the article that had this pic and it said something about the blast displacing the water below the muzzles.  This water displacement creates a force against the ship's hull.  Over to you, m'sieu scientifique...

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If we take a 24" oak target, drag from naca data and a closed form flatfire approximation for ease of calculation: (input data in customary units, all internal calculations in SI to avoid conversion errors and mishandling)

Using single shot and full charges:
36 livre (38.88lb) 1450fps (and same powder constant used for the two other ordnance - note this is a lower constant than that which matches given RN velocities for greater windage, with similar output velocity, your value)
24 livre (25.9lb) 1430fps (estimated from ordnance dimensions)
12 livre (12.96lb) 1340fps (ditto)

At 100m, the residual velocity would be (roughly)
1352fps, 1320fps and 1216fps respectively.

So the "raw" impulse of the whole broadside, stopped, would be 240,560 Nm
But (using the Didion/Bashforth coefficients for penetration) only around
690fps for the 36livre, 750 fps for the 24 livre and 893fps for the 12 livre are needed to perforate the 24" 'target'. Exit velocity can be estimated (probably with some error, but close enough for general comparison) by leaving a residual penetration equal to the maximum expected less than the target thickness and solving for velocity.

I find these as 850fps, 757fps and 524fps respectively and only a portion of the original impulse is delivered to the target on the first structure. In this case ~102,000 Nm.

note that reducing velocity to the minimum for penetration and 'just falling out' of the rear of the target results in somewhat more impulse (133,400 Nm in this case) than only punching neat holes with excess velocity, despite much lower firing impulse...



 

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8 hours ago, Angus MacDuff said:

I read the article that had this pic and it said something about the blast displacing the water below the muzzles.  This water displacement creates a force against the ship's hull.  Over to you, m'sieu scientifique...

I forgot to mention that the guns themselves also are designed to absorb energy like you arm does firing a gun. 

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On 10/3/2018 at 6:20 AM, Norfolk nChance said:

How much Momentum does a three-deck broadside from a L’Ocean deliver at 100m (109yds)?

Question about in-game or in reality ?

 

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