Falaise Posted October 21, 2020 Share Posted October 21, 2020 Well this game is driving me crazy !!! after having pushed me to be interested in botany, obliged to use photoshop now I am doing mathematics, a subject that comes out of useful bases interests me as much as the reproduction of the tick on the eastern slopes of the Alps !!! So here is my problem: I have 550 shells of 105mm I am using a shooting zone, the zone I want to hit is a circle 350m in diameter then [(175 x 175) x ft] / 550 = 174.8 m² one shell every 13.2 m does anyone know the number of shells per x m² for with a good average useful destruction ??? 0 Quote Link to comment Share on other sites More sharing options...
Erwin Posted October 21, 2020 Share Posted October 21, 2020 This may not exactly answer your question. However, I once did tests of artillery barrages on an urban/town and found that a longer barrage (ie HARASS) was more effective than short HEAVY barrage in causing casualties. 1 Quote Link to comment Share on other sites More sharing options...
Glubokii Boy Posted October 21, 2020 Share Posted October 21, 2020 Are you perhaps not overcomplicating things ? Finding a good duration and intensity will be good enough...no? This looks like a good time to do some hotseat QB testing... Pick the types of maps(terrain) you would like to test...plink down some defenders on it...and blow it to pieces ... Testing different settings in the artillery interface....and then watching the results. That ought to give you the best answers.. 1 Quote Link to comment Share on other sites More sharing options...
Freyberg Posted October 21, 2020 Share Posted October 21, 2020 Assuming you have about eight gun firing, 'Heavy, Maximum' would look really cool. If the target is in strong buildings, you'll cause serious damage and a lot of casualties; if the target is in anything but strong buildings, you'll utterly obliterate it. 1 Quote Link to comment Share on other sites More sharing options...
Falaise Posted October 21, 2020 Author Share Posted October 21, 2020 (edited) already good leads thank you However, given the Gunner and his obsessions with statistics there must be tables in the RL they could give a reference to the game with some adaptation Edited October 21, 2020 by Falaise 0 Quote Link to comment Share on other sites More sharing options...
Falaise Posted October 21, 2020 Author Share Posted October 21, 2020 Just now, Falaise said: already good leads thank you However, given the Gunner and his obsessions with statistics there must be tables in the RL they could give a reference to the game with some adaptation 0 Quote Link to comment Share on other sites More sharing options...
Erwin Posted October 21, 2020 Share Posted October 21, 2020 FYI: Artillery info and tables that I found: MODERN ARTILLERY EFFECTS.docx US MODERN ARTY DATA.docx CMBN US ARTY DATA.docx 1 Quote Link to comment Share on other sites More sharing options...
Zveroboy1 Posted October 22, 2020 Share Posted October 22, 2020 This is a modern Russian artillery nomogram : Example 3 ( the dashed green line) : Determine how many 100mm cannon and the quantity of rounds needed to destroy dug-in personnel and weapons in a 7.2 hectare target area in a 10 minute artillery strike. Begin at the "duration of fire" axis and find 10 minutes. From that point, move vertically to the 100mm line in "type of fire". Mark that point. Next, drop down to the "area of destruction" axis and find 7.2 hectare. Move horizontally to the 100mm line in "covered and concealed personnel and weapons". Move vertically and determine where the second point on the "type of fire" line and this line intersect. They intersect at 36 on the "quantity of artillery pieces (mortars)". The green line crossed the "quantity of rounds" axis at 1800. Thus the mission will require 36 100mm cannons and 1800 rounds. The Russian Way of War, Lester W. Grau, Charles K. Bartles. Foreign Military Studies Office, Fort Leavenworth, KS. 2 Quote Link to comment Share on other sites More sharing options...
StieliAlpha Posted October 22, 2020 Share Posted October 22, 2020 14 hours ago, Falaise said: Well this game is driving me crazy !!! after having pushed me to be interested in botany, obliged to use photoshop now I am doing mathematics, a subject that comes out of useful bases interests me as much as the reproduction of the tick on the eastern slopes of the Alps !!! So here is my problem: I have 550 shells of 105mm I am using a shooting zone, the zone I want to hit is a circle 350m in diameter then [(175 x 175) x ft] / 550 = 174.8 m² one shell every 13.2 m does anyone know the number of shells per x m² for with a good average useful destruction ??? Eh, your calculation is wrong. (Sorry, old engineers habit. „First of all, check if the numbers match.“) The area is „D squared x Pi / 4“ or 350x350x3.14/4= 96‘162 m2. With 550 shells, this comes to 0.005 shells/m2. But a square meter is a very small area. Quite a long time ago, we discussed the blast radius of artillery shells in a thread (I think, it was naval gun related). That discussion may give some further info. 0 Quote Link to comment Share on other sites More sharing options...
Erwin Posted October 22, 2020 Share Posted October 22, 2020 There is always a disjoint between what effects there are in the game and what effects are in RL. While the chart above is fascinating, it may not have as much relevance to in-game testing as info in the three files I uploaded, which are from in-game testing. 0 Quote Link to comment Share on other sites More sharing options...
Falaise Posted October 22, 2020 Author Share Posted October 22, 2020 8 hours ago, StieliAlpha said: Eh, your calculation is wrong. (Sorry, old engineers habit. „First of all, check if the numbers match.“) The area is „D squared x Pi / 4“ or 350x350x3.14/4= 96‘162 m2. With 550 shells, this comes to 0.005 shells/m2. But a square meter is a very small area. Quite a long time ago, we discussed the blast radius of artillery shells in a thread (I think, it was naval gun related). That discussion may give some further info. like i said math isnt my forte but i checked i got the same numbers my formula is R²x pi or (175 x 175) x 3.14 = 96162.5 / 550 = 174.4 17 hours ago, Zveroboy1 said: This is a modern Russian artillery nomogram : Example 3 ( the dashed green line) : Determine how many 100mm cannon and the quantity of rounds needed to destroy dug-in personnel and weapons in a 7.2 hectare target area in a 10 minute artillery strike. Begin at the "duration of fire" axis and find 10 minutes. From that point, move vertically to the 100mm line in "type of fire". Mark that point. Next, drop down to the "area of destruction" axis and find 7.2 hectare. Move horizontally to the 100mm line in "covered and concealed personnel and weapons". Move vertically and determine where the second point on the "type of fire" line and this line intersect. They intersect at 36 on the "quantity of artillery pieces (mortars)". The green line crossed the "quantity of rounds" axis at 1800. Thus the mission will require 36 100mm cannons and 1800 rounds. The Russian Way of War, Lester W. Grau, Charles K. Bartles. Foreign Military Studies Office, Fort Leavenworth, KS. as Erwin says, this graphic is fascinating by combining the set it will give me a reference thanks beaucoup 0 Quote Link to comment Share on other sites More sharing options...
StieliAlpha Posted October 22, 2020 Share Posted October 22, 2020 1 hour ago, Falaise said: like i said math isnt my forte but i checked i got the same numbers my formula is R²x pi or (175 x 175) x 3.14 = 96162.5 / 550 = 174.4 All good. Please refer to my PM. 0 Quote Link to comment Share on other sites More sharing options...
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