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KV2 first shot accuracy test result


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forgive me fo going back to this, also i've commited the cardinal sin of not thoroughly reading the posts after mine but it's late & i need my bed.

i was asking about the possibility of dropping shells onto the top armour of tanks from distance. Panzer76 said that this would need a high arc weapon.

now, my physics are a long time ago but for any projectile it seems to me that:

as the Y movement is proportional to the (initial vertical moment of the shell)/g which is a constant resulting in a sin function. and the X movement is proportional to the initial kinetic energy of the shell (.5* MV(sq)) / air resistance which results in (rate of change of shell speed) / distance - which is an intergral, then for point V1 the shell would have given XY cords and for V2 the shell would have X changing by an intergral & Y changing by a contant - resulting in a steepening of the shell fall angle.

the difference between a high velocity & low velocity shell is the flight path and travel time, both of which sum to produce more accurate shot. however, with a lower velocity shell, there seems to me the possibility of using the steeper fall angle for your own means.

now i KNOW someone will actually look this up. and when you do, i'll have a look. maybe get some metric on how bad my memory is really getting.

i've had CMBB for about 3 weeks now & i'm enjoying this forum nearly as much as the game.

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Originally posted by Other Means:

forgive me fo going back to this, also i've commited the cardinal sin of not thoroughly reading the posts after mine but it's late & i need my bed.

i was asking about the possibility of dropping shells onto the top armour of tanks from distance. Panzer76 said that this would need a high arc weapon.

now, my physics are a long time ago but for any projectile it seems to me that:

as the Y movement is proportional to the (initial vertical moment of the shell)/g which is a constant resulting in a sin function. and the X movement is proportional to the initial kinetic energy of the shell (.5* MV(sq)) / air resistance which results in (rate of change of shell speed) / distance - which is an intergral, then for point V1 the shell would have given XY cords and for V2 the shell would have X changing by an intergral & Y changing by a contant - resulting in a steepening of the shell fall angle.

the difference between a high velocity & low velocity shell is the flight path and travel time, both of which sum to produce more accurate shot. however, with a lower velocity shell, there seems to me the possibility of using the steeper fall angle for your own means.

now i KNOW someone will actually look this up. and when you do, i'll have a look. maybe get some metric on how bad my memory is really getting.

i've had CMBB for about 3 weeks now & i'm enjoying this forum nearly as much as the game.

LOL....... yes i think the proportional term may be suspect...related may be more accurate but YES the shell will dip at end of flight........not that much at ranges argued.....resistance is not a liner proportional to speed hence intergral.....a rather complex interplay of of several varibles shot weight /energy/ diameter.......

vertical velocity is constantly modified by g a constant but even this movement is effected by air resistance ..... the X component is not entirly seperate in that the original energy of the shot is not applied along that axis solely.........

As was I previously stated the flight path would be a blunted arc.

the steeping of flight needed to be effective against deck armour requires a higher trajectory than DF at ranges discribed....

this blunting will be more extreme at greater ranges.......

instictive intuition makes me believe that such steeping will not be a factor at 600m scenario debated......

we need a function for air resistance to argue about this. and in the case of 152mm the shot weight may carry better against air resistance.

hunch

Boris

London

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Originally posted by Other Means:

i must say, i'm constantly impressed by the quality of this forum.

so, not to try to hijack the thread but: if a tank is hull down to you, presenting a very small silhouette, are you better off shooting from long range (discounting optics for a sec) thus dropping the shell down onto them, reducing glance angle & going for the top armour? i know you drastically reduce your chance of hitting but could there be a situation where you would have no chance of penetrating the front armour but may get the top from longer range?

well yes .....as long as long range the projectile fall will have enough energy to penertrate the deck.............depends what you mean by LONG.

Boris

london

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Originally posted by mididoctors:

well yes .....as long as long range the projectile fall will have enough energy to penertrate the deck.............depends what you mean by LONG.

Boris

london

True enough, however, the deck is very thin. I imagine some of the larger HE rounds will penetrate just from the explosion (like the 152mm.)

[ January 29, 2003, 01:12 AM: Message edited by: StellarRat ]

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so, talking about the Y component of the shot flight, if we postit a shell flight over flat ground (it gets a bit complicated after that - ha) then the Y component of the flight in the upswing would seem to me to equal the Y component in the downswing. given a constant shell mass the effect of air resistance on the Y component could be discounted.

i also think that over 600m the total blunting effect would be negligable, especially given the strength of the other factors. we'd have ro get to the last 3rd of the shell flight for it to be noticable.

<hr width=80% color=black>

i wrote my last post about 2.30 am GMT (Liverpool UK) i'd wanted to get to bed before 12 but...i've just bought CMBB.

[edited to shorten the HR]

[ January 29, 2003, 09:39 AM: Message edited by: Other Means ]

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Originally posted by Other Means:

[QB] so, talking about the Y component of the shot flight, if we postit a shell flight over flat ground (it gets a bit complicated after that - ha) then the Y component of the flight in the upswing would seem to me to equal the Y component in the downswing. given a constant shell mass the effect of air resistance on the Y component could be discounted.

Tounge firmly in cheek

you would think so but resistance due to air friction of original Y componet is higher due to Y componet V(y)......at apex V(y)=0.....due air resustance intergarl and g.........physics of freefall is misleading as terminal issues will come into play.....IE no extra external input of kinetic energy in Y axis exsits to breakthru natural terminal freefall V(y) max..unlike situation at muzzle where Chem energy of expanding gas overcomes any limitations....IE accn occurs only in barrel....

If you fire a high Vel shot straight up into the Atmosphere.....in a vacum it would come down with the same SPEED as it went up but in air it comes down slower.....this is counter-intuitive........where did the energy go? Surley all that potental energy would be turned back into V(y)equal to that at the muzzle......no some is lost as heat and sound........This is consistently lost thruout shot flight in both axis......

however I am being rather anal (your not kidding) cos while

skydivers reach terminal velocity fairly quickly

the noticable degradation of acc due to gravity on a 152mm shell I imagine is quite low..especialy in as V(y) is very small compared with V(x)......for all practical purposes you are right........

do not sweat it.........as for plunging fire in AFV combat hmmmmmmmm....

well it is certainly a factor in naval big gun combat where tactics revolve around movement thru plunging fire range zones....these weapons are hurling metal the weight of a small cars over immense distances....could it be scaled down .....?

more importantly was it ever tacticaly employed?

Boris

London

signing off

[ January 30, 2003, 11:12 AM: Message edited by: mididoctors ]

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