An RAF Boeing-Vertol CH-47 Chinook helicopter declared a general emergency near London on Wednesday night, transmitting the universal 7700 squawk code. The aircraft, identified as SHF482, made the declaration at approximately 10.15pm on October 7 at around 2,000 feet.
What the 7700 squawk code means
The code 7700 is the universal transponder code, known as a "squawk code", that signals a general emergency and is recognised worldwide. Squawking 7700 also alerts all nearby air traffic controllers in the surrounding areas to the situation with the flight. The codes are used globally as a standard way to manage flights.
Specific details of the emergency are currently unknown.
Aircraft background
The aircraft involved, a Boeing CH-47 Chinook, is a twin-engine, tandem-rotor heavy-lift helicopter developed by Vertol and manufactured by Boeing. Military forces across the world, including the UK, use the aircraft. It can carry up to 24,000 pounds of internal cargo or sling-load nearly 26,000 pounds externally.
Recent focus on RAF
There has been a focus on the RAF in recent weeks following a "terror" incident at the Fairford base in Gloucestershire on September 27. Counter-terror police opened an investigation after vans were found near the base. Five British nationals, all in their mid-20s, arrested near the base have since been released on bail. A sixth man, a dual UK-Iranian national aged 25, was arrested last week and has also been released on bail.
US military planes have all been removed from the base "urgently" following the incident, believed to have been "linked to Iran". Since the incident, No10 has insisted RAF Fairford is safe despite the US withdrawing fighter jets from the base due to security concerns.
The Express has contacted the RAF for further details on why the emergency was declared.
