Norfolk nChance Posted October 3, 2018 Posted October 3, 2018 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? 1
2 Tom Farseer Posted October 3, 2018 Posted October 3, 2018 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 3
1 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 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?
0 Norfolk nChance Posted October 3, 2018 Author Posted October 3, 2018 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
0 Raf Van Boom Posted October 3, 2018 Posted October 3, 2018 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.
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 Check out the momentum when Iowa lets it loose.
0 Raf Van Boom Posted October 3, 2018 Posted October 3, 2018 That's not called momentum in physics, that's force, you know, F = ma sort of thing.
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 It looks like mass in motion to me (the ship is moving laterally), and if it hits you in the face and keeps going....
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 I just blow shit up. If there is a physicist in the way, that's his look-out.
0 Norfolk nChance Posted October 3, 2018 Author Posted October 3, 2018 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. 1
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 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. 1
0 Vernon Merrill Posted October 3, 2018 Posted October 3, 2018 (edited) 2 hours ago, Angus MacDuff said: Check out the momentum when Iowa lets it loose. 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 October 3, 2018 by Vernon Merrill 2
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 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.
0 vazco Posted October 3, 2018 Posted October 3, 2018 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 2
0 HachiRoku Posted October 3, 2018 Posted October 3, 2018 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. 1
0 Raf Van Boom Posted October 3, 2018 Posted October 3, 2018 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. 1
0 Angus MacDuff Posted October 3, 2018 Posted October 3, 2018 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...
0 Raf Van Boom Posted October 3, 2018 Posted October 3, 2018 This? http://www.navweaps.com/index_tech/tech-022.php
0 HachiRoku Posted October 4, 2018 Posted October 4, 2018 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.
0 Hethwill, the Red Duke Posted October 4, 2018 Posted October 4, 2018 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 ?
Question
Norfolk nChance
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?
21 answers to this question
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