Real LAV25 Sight Reticle

[sylent/shooter]

Hmmm....... looking forward to seeing this. Possible will join the Canadian Army (armoured) after i grad highschool

[PLODDITHANLEY]

The Canadian one is easy to understand and nice to finally here from someone who knows and see the bullet drop adjustments, noticed the HE round goes much slower than the AP round.

[Hotrod525]

actualy there is a bigger drop on HE due the fact HE is heavier than AP.

[McBumLuv]

Originally Posted by Hotrod525

actualy there is a bigger drop on HE due the fact HE is heavier than AP.

Unless that affects the bullet's friction against the air or slows it down somehow, I don't see how it affects drop, seeing as gravity's constant and independent of the lateral movement.

[Th3Exiled]

Originally Posted by Hotrod525

actualy there is a bigger drop on HE due the fact HE is heavier than AP.

Please tell me you didn't say that..... Gravitational acceleration is independent of the mass of the object. If you don't believe me:

formula for gravitational force: F= G(m1*m2/r^2)

formula for acceleration: a=F/m

a=G(m1*m2/r^2)/m2

the m2s cancel

a=G(m1/r^2)

where m1 would be the mass of earth, G is a constant and r is the distance between the centre of the two masses.

So the only thing that would affect the acceleration due to gravity for an object on earth would be to change your distance from its centre.
though things like drag will also affect the overall acceleration.

edit: ninja'd by 5 minutes... the shame...

Exiled.

[McBumLuv]

Hehehe

Yup, all that's dependent is on how long it takes for the bullet to get there along the horizontal plane, which would then effect how far the bullet drops vertically.

Kinda crazy thinking that they drop around ten meters in a kilometer if they cover it within a second.

[[R-DEV]Ninja2dan]

Originally Posted by McLuv

Unless that affects the bullet's friction against the air or slows it down somehow, I don't see how it affects drop, seeing as gravity's constant and independent of the lateral movement.

Below is the data on the two rounds fired from the LAV-25, or at least US ammunition. I would assume Canadian rounds are identical or pretty close.

M792 HEI-T/SD: Steel thin-walled projectile loaded 30.2 g HE and with red tracer; fuzed PD M758 with self-destruct between 6.2 and 19 seconds; projectile, 184 g; MV 1,100 m/s

M919 APFSDS-T: A 98 g depleted uranium long rod penetrator, fin-stabilised, carried in an alloy/plastic sabot. Rear of fin unit bored for tracer. 132 g; MV 1,390 m/s. Time of flight to 2,000 m 1.56 seconds; remaining velocity 1,149 m/s

As you can see, the projectile of the HE round is only slightly heavier than the AP round. The main reason though that the AP round doesn't drop as much is because of its design. The HE projectile remains intact until impact, while the AP round has a sabot design that sheds its firing shell shortly after leaving the muzzle. This leaves a very thin and fast penetrator. Not only does the AP projectile have less drag in flight and retain its velocity better, but a good chunk of the projectile weight is made up of the discarded sabot shells.

So both weight and drag/resistance of the projectile are what affects the drop differences between the two rounds. This is the case with most heavy munitions, including those fired from MBT's.

[Th3Exiled]

Originally Posted by McLuv

Hehehe

Yup, all that's dependent is on how long it takes for the bullet to get there along the horizontal plane, which would then effect how far the bullet drops vertically.

Kinda crazy thinking that they drop around ten meters in a kilometer if they cover it within a second.

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.

[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.

[Alex6714]

Cool stuff, thanks.