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MBT (first edition)» Forums » Rules

Subject: Level, Rising & Falling shots versus geometry rss

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Frank Clarke
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Just out of interest:

A terrain "height" is 25 feet.
At 1 hex the angle is 4.36 degrees, at 4 hexes 1.09 degree. 1 degree doesn't make that much difference really, hardly worth the extra column space on the card.

Of course, it is a fun rule.

Panzer has "Deck hits" with the same geometry problems, at 25 foot up you are hitting the deck at 4.36 degrees at 1 hex, which is 329mm of metal to traverse for 25mm of deck.

Also, Soviet tanks can depress their guns down by 5 degrees, so 17.3.1 weapon depression/elevation doesn't make much difference.

Firing at deck armor from 4.36 degrees with HEAT isn't very effective of you look at a HEAT round:

The impact switch won't touch the hull at 4 degrees off horizontal.

So you have rules increasing complexity and reducing realism.

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groggy froggy
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I tried to dig "deeper" into this. Using trigonomerty and the parabola of the munition trajectory.

Munition, initial velocity, range, resulting impact angle:
APFSDS, 1600 m/s, 400 m, 0.04 deg
APS/HVAP, 1000 m/s, 400 m, 0.11 deg
HEAT, 600 m/s, 400 m, 0.3 deg
HESH, 400 m/s, 400 m, 0.7 deg

And increasing range to 1600 m (100 m per hex) gives
0.2 deg
0.5 deg
1.3 deg
2.8 deg

Maybe shooting at "normal" ranges and heights? 1 elevation up (8 meters) shooting at 4 hexes range, gives 1,15 deg. Or 2 levels up at 16 hex range, gives 0.6 deg.

And now I have ignored the aerodynamical drag, that would increase the impact angle, but not dramatically at these "short" ranges.

So, "maximum" shot impact angle: HESH round at 1600 m at elevation 2 = 2.8+0.6 = 3.4 degrees + contribution from drag. In any case less than 10 degrees.

So, the above analysis by Frank is still valid. Armour values would not be significantly changed. But the possibility for effective Hull Down might be slightly impacted for different tanks / doctrines.


A battle where tank gun depression actually was critical:
It was a crucial moment when Israeli reservists rushed to defend the bridge at the northern tip of the Sea of Galilee over Jordan.

Impact angle formula is
s = v^2/g * sin(2*beta)
s = distance, m
v = initial velocity, m/s^2
g = 9.81 m/s^2
beta = impact angle

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