Author Topic: Road To Forgotten Hope 2.4: Part 2  (Read 15902 times)

Offline Oberst

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #165 on: 14-07-2011, 15:07:32 »
Can someone explain again what the "angled damage modifier" does. According to the news I just thought it will instantly drop the damage to zero, if you hit a tank in an angle between 60 to 90 degrees. But a few pages ago someone said the angle will also modify your damage, so you might survive shots, which penetrate and would have instakill you under a lower angle, but now does lower damge because of the angle.   ???

Offline Miklas

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #166 on: 14-07-2011, 15:07:37 »
The way I understood it was that if a shell hits the armor at 90 degrees no damage is dealt. If the shell hits at 60 degrees 100% damage is dealt. If the shell hits between 60 and 90 degrees the damage dealt will drop accordingly the steeper the angle is.
I.e. A 80 degree shot would do very little damage while a 65 degree shot would do a lot of damage.

Offline Kelmola

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #167 on: 14-07-2011, 16:07:35 »
Can someone explain again what the "angled damage modifier" does. According to the news I just thought it will instantly drop the damage to zero, if you hit a tank in an angle between 60 to 90 degrees. But a few pages ago someone said the angle will also modify your damage, so you might survive shots, which penetrate and would have instakill you under a lower angle, but now does lower damge because of the angle.   ???
See this diagram.

Let T be the thickness of armour and T1 the effective thickness. Let A be the angle at which the shell hits the armour. A1 or 90-A is the angle we're interested in, as it helps us to calculate T1. Using basic trigonometry, sin A1 = T / T1 <=> T1 = T / sin A1. For example, a shell hitting a 60mm armour at 75 degree angle, we get T1 = 60 / sin (90-75) = 232mm.

Actually, if the damage modifier is relative to the effective thickness, sin (90-angle) is all we need to know since that is the ratio; in this case, sin 15 =  0,259, meaning that 25,9% of damage is applied to the target.

Offline McCloskey

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #168 on: 14-07-2011, 16:07:53 »
here I thought that if you actually hit something under the 90° angle, you will have the best chance of penetrating (do the most damage).. and if it's more than 90 (91-180°) or less (89-0°) then the shot would more likely deflect (in this case the damage would drop).
The way I understood it was that if a shell hits the armor at 90 degrees no damage is dealt. If the shell hits at 60 degrees 100% damage is dealt. If the shell hits between 60 and 90 degrees the damage dealt will drop accordingly the steeper the angle is.
I.e. A 80 degree shot would do very little damage while a 65 degree shot would do a lot of damage.

Offline Kelmola

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #169 on: 14-07-2011, 16:07:58 »
Whoops. Got confused and assumed that the degrees were "deviation from straight angle" and not absolute. Well my illustration and formula still works, only that A1 is of course the angle used without having to calculate it, so 15 degrees is 15 degrees. And sin 15 = 25,9%

Offline Cory the Otter

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #170 on: 14-07-2011, 16:07:06 »
And I'll be 2 days late for the premiere!  :'(

Offline Kwiot

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Re: Road To Forgotten Hope 2.4: Part 2
« Reply #171 on: 14-07-2011, 18:07:30 »
Well, I'll be 2 weeks late... Going tomorrow morning to Italy...  ::)

I have 2 questions- is Matilda still haxed tank? And was the driving Marder, Panzer IV, M10, Sherman improved? No more rolling over?