Please repeat the test using a Russian Bombsite.
Bomb ground strike position relative to the bombsite aiming angle is the point to note also, testing shows this is wrong/different OAT air density or something is having an effect ?
If so new bombsite calculation tables are needed for v4.11 >
I have been a bit too busy to test again with new data this week but this is from a guide pre v4.11, hopefully the same theory applies to current game versions and the problem is else where.
Equations of motion
v = velocity, u = initial velocity, a =
acceleration, s = distance, t = time.
s = ut + ½at2
v = u = s/t (unaccelerated)
Fig 1.2.
The aircraft is heading from right to left at speed, v, when it releases a bomb at A.
Initially the bomb continues to move with the aircraft, but starts to drop as gravity accelerates it downwards.
The bomb follows a parabolic path, represented by the blue line.
AB is the height, h, of the aircraft and BC is the forward throw, R, of the bomb.
If the bomb takes time, t, to reach the ground, g is the
acceleration due to gravity and we ignore air resistance for the moment, then the equations of motion give us the following:
AB = h = ½gt² -- eq 1
BC = R = vt -- eq 2
We know the height, h, speed, v, and g is a constant 9.81 m/s², so we can find t from eq 1 and substitute for t in eq 2 to find the forward throw, R:
R = v 2h/g) -- eq 3
The angle between the horizontal – the dotted line in fig – and the point of impact at C is, of course,
the angle, a, from fig 1.1 above.
This is what we want to know when we come to aim the bomb and is the same angle as ACB in fig 2.
As we now know 2 sides of the triangle, AB and BC, we can find the angle:
Tan ACB = AB/BC = h/[v 2h/g)] -- eq 4
DIVE-BOMBING
In the previous example we used the height to find the time the bomb is in the air (equation 1) and
then used equation 2 to find the forward throw of the bomb.
From these two pieces of information we could deduce the angle corresponding to the amount the bomb drops from the horizontal.
We can do exactly the same when the bomb is released from a dive at an angle, degrees.
The situation is slightly more complex, however, because the bomb now has an initial downwards velocity (v.sin ) and the horizontal velocity (v.cos ) is not quite the same as the airspeed.
The equations for AB and BC now become:
AB = h = vt.sin + ½gt² -- eq 5
BC = R = vt.cos -- eq
Equation 5 results in a quadratic equation, which can be solved to find t.
Equation 6 can then give the forward throw, R, which allows us to find AB/BC and then the sight angle.
This would be quite tedious to solve for each combination of dive angle, airspeed and release height, but it is not too
difficult to produce a spreadsheet, which will do the sums for us once we enter the desired
parameters.
The dive angle, , is found by noting height, h, and range, D, to the point on the ground under the
cross hairs at the moment of bomb release.
Tan = h/D -- eq 7
Air resistance. We have not yet considered the effect of air resistance on the falling bomb.
Once the bomb leaves the aircraft it will start to slow down.
However, so long as the bomb does not produce any lift (either up or down), the drag will only act back along the bomb’s direction of travel.
The bomb will slow down and take longer to reach the end point, C in the above fig, but it will follow the same path and still reach that point.
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