Real LAV25 Sight Reticle

[***LeGeNDK1LLER***]

Originally Posted by Hotrod525

Here you go whit the sight... as you can see, HE and AP do not had the same graduation. Anyway, here you go. If you need the whole picture just PM me.

O.T. sorry
jesus christ you are using 1 of the best tanks in the world probably the best with the leclerc!!
soooooo jealous, you can and i can't!

i will wait for christmas maybe..

[Hotrod525]

Originally Posted by octo-crab

EDIT: Ahhh you tried to trick me, you didn't say you were an army crewman when I originally wrote this post so I thought you were referring to the PR LAV-25.

Well in my sig' you can read : Me on a L2A4+C... and its not a piece of equipement you can be close at as civilian, [ well not yet ] even L1C2 are not that much accessible to public..

Anyway, sorry for confusing but yes, i'm an armor guy'

And LegendKiller, Abrams M1A2 TUSK and Leopard 2A6M-C had proven themself way much more safer than Leclerc. [ I dont even known if Leclerc ever seen combat... will do some research .. ]

[[R-CON]Burton]

cant help but feel there must be something iffy to do with opsec when posting stuff like that?

[[R-DEV]Ninja2dan]

Originally Posted by Burton

cant help but feel there must be something iffy to do with milsec when posting stuff like that?

Images of a reticle are in no way against OPSEC in this case. There is nothing that could be gained from enemy knowledge of the device, therefore there is no threat to security.

In fact, MCWP 3-14.1 [Light Armored Vehicle-25 (LAV-25) Gunnery and Employment], which is one of the USMC field manuals for the LAV-25, is approved for public distribution and available for review and download on the internet.


Besides, if some kind of terrorist or enemy soldier was to get their hands on a LAV-25, I think we'd have more serious security concerns than them knowing what the reticle looks like.

[Farks]

Originally Posted by [R-DEV]Ninja2dan

Images of a reticle are in no way against OPSEC in this case. There is nothing that could be gained from enemy knowledge of the device, therefore there is no threat to security.

In fact, MCWP 3-14.1 [Light Armored Vehicle-25 (LAV-25) Gunnery and Employment], which is one of the USMC field manuals for the LAV-25, is approved for public distribution and available for review and download on the internet.


Besides, if some kind of terrorist or enemy soldier was to get their hands on a LAV-25, I think we'd have more serious security concerns than them knowing what the reticle looks like.

"- Hey Mohammed, do you know how to aim one of those things?
- No problem, I played Project Reality online for two years! I got very good K/D with this!"

[McBumLuv]

Originally Posted by Th3Exiled

Actually It would be around 5m. If we were to say that acceleration is exactly -9.8 ms^-2, a constant value, which would be equivalant to a=-9.8t^0=dv/dt, then when we integrate to get the velocity we get: v=-9.8t + c. c would be substituted for u (v=-9.8t + u) however initial vertical motion will be considered 0 (hence u=0). Integrate ds/dt=-9.8t to get s=(-9.8/2)t^2.
when t= 1, s=-4.9*1=-4.9
Alternatively, we could have just used the displacement formula for constant acceleration; s=ut+at^2. This approach would have taken less time, but it is much more restrictive(and I felt like integrating).

Exiled.

Yea goddammit I was using the wrong kinematics equation, I should have been using d=v1*t*1/2at^2.

[Th3Exiled]

Originally Posted by Hotrod525

Here you go whit the sight... as you can see, HE and AP do not had the same graduation. Anyway, here you go. If you need the whole picture just PM me.

Exile i think that it is OBVIOUS that you're math are wrong if you look at the sight, at 2400M, the drop is HUGE, following you're calculating, the drop would be only 14M... 14M at 2400 in a sight, is WAY smaller.

Math is correct, your interpretation is wrong. Don't forget that the (horizontal) velocity isn't constant (due to drag) therefore time, t, can not be calculated by t=s/v. However, if we assumed that horizontal velocity is constant:

t=2400/1100
S=ut+1/2at^2 (ut obmitted, initial vertical motion assumed 0 for simplicity [otherwise it would take longer and would require using the quadratic equation])
Sy=-9.8/2(2400/1100)^2 (acceleration, g, is negative)
Sy=-4.9*4.76
Sy=-23.32m

Which is different to your 14m which would require t=1.69, since:
Sy= ut+1/2 at^2
-14=0-9.8/2 t^2
-14=-4.9 t^2
14/4.9=t^2
t^2=2.86
t=1.69

Also since it covers 2400m in 1.69 seconds, if we assume constant velocity:
t=s/v
1.69=2400/v
v=2400/1.69
v=1420.12 ms^-1
However, that is larger than the stated initial velocity of 1100ms^-1 and since it would be exposed to air resistance, the initial velocity, u, would have to be > 1420 ms^-1.

If you wanted to get even more technical, besides accounting for air resistance, you could also take into account the elevation of the barrel needed to hit the target. This would in effect make the initial horizontal velocity smaller thus taking more time to reach the target and being a larger drop on the round. Though unless we had some data such as how long it takes for a round to reach a target at various distances, it would be impossible to incorporate drag, even with the data you still need to use quite a bit of calculus.

Exiled.

[illidur]

Originally Posted by Th3Exiled

Math is correct, your interpretation is wrong. Don't forget that the (horizontal) velocity isn't constant (due to drag) therefore time, t, can not be calculated by t=s/v. However, if we assumed that horizontal velocity is constant:

t=2400/1100
S=ut+1/2at^2 (ut obmitted, initial vertical motion assumed 0 for simplicity [otherwise it would take longer and would require using the quadratic equation])
Sy=-9.8/2(2400/1100)^2 (acceleration, g, is negative)
Sy=-4.9*4.76
Sy=-23.32m

Which is different to your 14m which would require t=1.69, since:
Sy= ut+1/2 at^2
-14=0-9.8/2 t^2
-14=-4.9 t^2
14/4.9=t^2
t^2=2.86
t=1.69

Also since it covers 2400m in 1.69 seconds, if we assume constant velocity:
t=s/v
1.69=2400/v
v=2400/1.69
v=1420.12 ms^-1
However, that is larger than the stated initial velocity of 1100ms^-1 and since it would be exposed to air resistance, the initial velocity, u, would have to be > 1420 ms^-1.

If you wanted to get even more technical, besides accounting for air resistance, you could also take into account the elevation of the barrel needed to hit the target. This would in effect make the initial horizontal velocity smaller thus taking more time to reach the target and being a larger drop on the round. Though unless we had some data such as how long it takes for a round to reach a target at various distances, it would be impossible to incorporate drag, even with the data you still need to use quite a bit of calculus.

Exiled.

what does this have to do with the suggestion? it would be sweet to have thermal on lav though

illidur'd

[gazzthompson]

read the quote at the top of his post, then read the thread.

[[R-CON]nedlands1]

From FM 3-22.1 (BRADLEY GUNNERY) (23.08 megabytes of PDF) regarding the M242 25mm cannon (see part 2-13),

The time of flight is as follows:

M791 APDS-T Rounds

Range, Time (s)
1000m, 0.8
1500m, 1.2
2000m, 1.7
2500m, 2.2


M792 HEI-T Rounds

Range, Time (s)
1000m, 1.2
1500m, 2.2
2000m, 3.6
2500m, 5.3

At 2400m, the time the round has been in the air is 2.1 seconds (from spline interpolation) for M791 APDS-T rounds and 4.9 seconds for M792 HEI-T rounds.

Using the constant acceleration equations which were kindly provided by Exiled and assuming the gun is fired at the horizontal and there are no other external effects on the projectile (eg drag, wind etc), then the drop is roughly 22m and 118m respectively at 2400m.

Back on topic. Stadiametric sights would be cool but BF2 has some weird issues with scaling things for various aspect ratios which would make it difficult to implement effectively.