Originally posted by F4UDOA
If an aircraft has a 3G stall speed at it's gross weight of 160MPH and another has a 3G stall of 200MPH which one should be able to complete a 360 degree turn faster?
I did 3 360 degree turns then divided by 3 to get an average.
Tested the F4U-1D vrs the FW190A5 and found that they are virtually identical both with 75% fuelin both and a heavy load for weapons in the 190.
The F4U-1D = 20.8
The FW190A5= 21.06
How is this possible?
Does anyone the FW190 3G stall numbers? Annecdotally from Navy test the F4U could get on the tailof the 190 in 1.5 turns.
Hi F4UDOA
Here is an Aces High Energy Maneuverability diagram for the F4U-1D overlaid with the Fw190-A5.

You can see that the sustained turn rates are very close to each other with a slight edge to the F4U. You will also notice that the F4U-1D can also achieve a slightly smaller turn radius than the Fw190-A5 when they are both fighting on the edge of the envelope. The diagram also shows that the F4U-1D has an advantage in instantaneous and sustained turns at all speeds below corner velocity, but the advantage is only small.
You can use the that diagram to read off the 3g stall speed for the Aces High aircraft by following that line of constant g until it cuts the stall limit curve on the envelope for the each aircraft.
To answer your original question, the aircraft with the lower stall speed should be able to complete its turn more quickly, because it will have a higher instantaneous turn rate, speed for speed, than the other one. However, you have to be careful in general terms, because if the aircraft in question has a very poor sustained turn, it may lose its energy very quickly then be out turned by an aircraft with a higher stall speed.
That's why energy managment is considered to be such a valuable skill in aircombat, and also why you need to consider instantaneous and sustained turning ability together... that's why EM diagrams are so useful
Hope that helps
Badboy