TN9 Hadleigh Mystery Cache

This cache is NOT at the given coordinates. Read the description for more details.

TN9 Hadleigh, 1, 2, 3 Fire!

The TN9 Hadleigh World War II Heavy Anti- Aircraft gun site (6th AA Division) was on Sandpit Hill, South West of Hadleigh. It is thought that the four 4.5 inch guns were operational from 1940, whilst the four 5.25 inch guns came into operation during the course of 1944.
The 4.5" gun positions are centred on Sandpit Hill (N51º 32.904’ E00º 35.697). There are four octagonal emplacements in a semi-circle facing east. Each has five ammunition recesses and an external integral shelter. In the centre of the semi-circle is a number of buildings and structures including the command post.
These 5.25" gun positions are centred on Adders Hill (N51º 32.963’ E00º 35.977), in a square formation. The circular gun platforms can be seen with their integral rectangular structures containing the computer, power house, war shelter and gun store. These 5.25" gun emplacements have been filled in with soil and grassed over although the area is easily distinguished by its high, rounded profile. One part of an emplacement, the north-easternmost of the four shows the outer rim of the ammunition gallery above ground level although not enough can be seen to determine overall details.
To the west of the 5.25" site is a combined Operations Room/Generator Block. This survives intact, measuring 50' long with the narrower, Operations, end 14' wide. It is built of heavy concrete with steel-framed, shuttered windows
Around 30 structures can still be identified on the site including the rare Operations Room /Generator Block.

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To find this cache you need to calculate where a projectile will land after being fired from the 5.25" heavy gun position. Use the following information / data:-

The motion of an object moving near the surface of the earth (ignoring air friction) can be described using the equations:

where x_o = initial horizontal position, v_xo = initial speed in horizontal direction, t = time

(2): y = y_o + v_yo • t - 0.5 • g • t²

where y_o= initial vertical position, v_yo = initial vertical speed, g = acceleration due to gravity.

Solving these two simultaneous equations gives a description of the ballistic trajectory. The horizontal and vertical components of initial velocity are determined from:

v_xo = v_o • cos θ

v_yo = v_o • sin θ      where θ is the launch angle, in degrees

Find the time to reach the maximum height by finding the time at which vertical velocity becomes zero:

v_y = v_yo - g • t

t_rise = v_yo / g

Maximum height is obtained by substitution of this time into equation (2).

h = y_o + v_yo • t - 0.5 • g • t²

Next, the time to fall from the maximum height is computed by solving equation (2) for an object dropped from the maximum height with zero initial velocity.

0 = h - 0.5 • g • t²

t_fall = √(2 • h / g)

The total flight time of the projectile is then:

t_flight = t_rise + t_fall

From this, equation (1) gives the maximum range:

range = v_xo • t_flight

The projectile speed at impact v_f is determined by applying the Pythagorean Theorem:

v_f = √(v_xf² + v_yf²)

In which:

v_xf = v_xo

v_yf = -g • t_fall

FH

• Bearing = 252° True
• Launch Angle (gun elevation) = 53.139°
• Launch Speed = 135m/s
• Acceleration due to gravity = 9.8m/s⊃2;
• Elevation is 15 metre (from gun to cache position)

But to make life a lot easier, here is a handy tool for working it out 😁